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See discussions, stats, and author profiles for this publication at: https://www.researchgate.net/publication/260284994 Basics of Photonics and Optics Book · December 2012 CITATIONS READS 3 1,963 1 author: Srini Vasan University of New Mexico 52 PUBLICATIONS 35 CITATIONS SEE PROFILE Some of the authors of this publication are also working on these related projects: Biophotonics Training View project Optics, Lasers & Semiconductor Processing View project All content following this page was uploaded by Srini Vasan on 24 February 2017. The user has requested enhancement of the downloaded file. Basics of Photonics and Optics Srini V. Vasan Current affiliation: Albuquerque Technical Vocational Institute Albuquerque, NM, USA © 2004 by Srini Vasan Published by Srini Vasan All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the written prior permission of the publisher. (Front cover: Photograph of an argon ion laser passing through an amber colored acrylic filter and casting interference patterns on a wall. Photo courtesy of Robert Hall) Preface The Book This book entitled “Basics of Photonics and Optics” is an algebra-based book suitable for post-secondary technical/vocational education, as well as for high school photonics academy programs. It requires that the student be familiar with concepts of high school algebra as well as basic elements of geometry. This book deals with just the appropriate material needed to enable a student succeed in foundation courses leading to a two year Associates Degree program or a one year Certificate program in Photonics Technology. The book is organized as follows. Chapter 1 reviews algebra fundamentals and graphing. Chapter 2 describes characterization of a photon. Chapters 3 and 4 discuss aspects of photonics/laser safety, laser operation and types of lasers. Chapter 5 describes rectilinear propagation of light to explain reflection and refraction phenomena. Chapter 6 provides a description of light propagation in an optical fiber. Chapter 7 discusses polarization of light. Absorption and scattering of light with an emphasis on optical filters are addressed in Chapter 8. Chapter 9 describes laser gain in depth, characterizes pulsed lasers and looks at some applications. Chapters 10 and 11 pertain to geometric optics, lenses, ray tracing, aberrations and optical instruments. Chapter 12 looks at interference and diffraction phenomena in wave optics and discusses applications. The outstanding features of this book lie in the logical manner in which the chapters are laid out, a brief description of concepts of algebra and graphing early on in the book, conciseness, explanation of physical concepts in a simple, yet elegant, fashion, numerous examples and exercises, and suggested laboratory experiments at the end of each chapter. A set of thought provoking questions are provided under “Group Discussion Questions.” Some of the questions require the student to search the web, find the answers and discuss in small groups. Wherever applicable, a review of calculator use is provided. Answers to oddnumbered exercise problems and “End of Chapter Exercises” are provided at the end of the book. Appendices at the end of the book provide quick reference to useful conversion factors, physical constants, general calculator review, metric prefixes and formulas useful in geometry, trigonometry and fiber optics. The Author’s Background The author, Srini V. Vasan, has a Ph.D. in Chemical Engineering from Clarkson University, awarded in 1986. His dissertation was entitled, “Thermal-, Photon- and Plasma-Induced Processes in Polymers.” Additionally, he has a Master’s degree in Social Work from the University of Texas in Austin in 1996. He has over twelve years of industrial experience in the petrochemical and electronics industries. He has over seven years of experience in teaching, of which over three years have been in various technologies at a community college. He has three patents, two in the United States and one in Europe, as well as over a dozen publications in refereed technical journals, over forty industry-internal publications and disclosures, and many presentations. Some of these publications are in excimer laser applications and photolithography. Acknowledgements The author would like to thank Dr. Marc Nantel from Toronto, Canada, and Dr. Sydney Sukuta from San Jose, CA, USA, for critiquing several chapters of this book. Thanks are also due to Ms. Sue Sujka for proofing and critiquing portions of the book, and to Mr. Joel Gellman, Ms. Sue Sujka, and Dr. Gordon Bennett for their support, review, ideas and suggestions. The author acknowledges Mr. Don Goodwin (Dean), Mr. Steve Benavidez (Associate Dean) and Mr. Robert Hall (Director) for their support and encouragement which made this work possible. Thanks are also due to Mr. Kevin Ryan for technical assistance with some of the laboratory experiments. Finally, I extend my gratitude to my wife, Joy, for her patience throughout this endeavor. Topic Page Table of Contents Topic Page 1. Elements of Basic Algebra 1 1.1 Decimal, Scientific and Metric Notation of Numbers 1 1.2 Graphing 12 1.3 Various equation forms for straight lines 26 1.4 Formula Transformation 33 Chapter Summary 37 End of Chapter Exercises 39 Group Discussion Questions 40 2. Elements of Light Propagation Theories 2.1 Wave Nature of Light 2.2 Particle Nature of Light 2.3 Energy of a Photon 2.4 Photon Characterization 2.5 Photoelectric Effect 2.6 Laboratory Experiments Chapter Summary End of Chapter Exercises Group Discussion Questions 41 41 47 50 56 57 58 59 59 60 3. Photonics Safety 3.1 Laser Classifications 3.2 Basics of the Human Eye 3.3 Power and Irradiance Calculations 3.4 Safety Precautions to be Observed During Laser Operation 3.5 Electrical Safety, Electrostatic Discharge 3.6 Fiber Handling Safety 3.7 Optical Power Meters 3.8 Laser Safety Laboratory Chapter Summary End of Chapter Exercises Group Discussion Questions 61 61 63 66 71 73 75 75 77 78 78 79 4. Basics of Laser Operation Theory 4.1 Energy Levels in an Atom 4.2 Spontaneous Emission 4.3 Stimulated Emission 4.4 Light Amplification 4.5 Gas and Ion Lasers 4.6 Solid State Lasers 4.7 Laboratory Experiments Chapter Summary End of Chapter Exercises Group Discussion Questions 81 81 84 86 87 90 93 95 95 96 97 5. Rectilinear Propagation of Light 5.1 Fundamentals of geometry and trig. 5.2 Laws of Reflection 5.3 Refraction and Snell’s Law 98 98 101 115 5.4 Total Internal Reflection 5.5 Dispersion 5.6 Lenses 5.7 Laboratory Experiments Chapter Summary End of Chapter Exercises Group Discussion Questions 130 139 141 143 145 146 148 6. Basic Principles of Fiber Optics 6.1 History of Fiber Optics 6.2 Elements of a Fiber Optics Cable 6.3 Modes of Propagation in a Fiber 6.4 Acceptance Angle of a Fiber 6.5 Power Losses in a Fiber 6.6 Properties of Light Sources 6.7 Fiber Optics Connectors 6.8 Laboratory Experiments Chapter Summary End of Chapter Exercises Group Discussion Questions 150 150 153 157 164 169 176 179 182 183 184 185 7. Polarization 7.1 Electric and Magnetic Fields 7.2 Polarization, Types of Polarization 7.3 Methods of Polarizing Light 7.4 Law of Malus 7.5 Birefringence 7.6 Laboratory Experiments Chapter Summary End of Chapter Exercises Group Discussion Questions 187 187 189 191 201 206 208 209 210 211 8. Absorption and Scattering of Light 8.1 Absorption of Light 8.2 Filters 8.3 Scattering of Light 8.4 Optical Windows 8.5 Laboratory Experiments Chapter Summary End of Chapter Exercises Group Discussion Questions 213 213 221 226 231 232 234 235 236 9. Pulsed Laser Characterization 9.1 Laser Gain 9.2 Pulsed Lasers 9.3 Pulsing Techniques 9.4 Laboratory Experiments Chapter Summary End of Chapter Exercises Group Discussion Questions 238 238 245 251 255 256 257 258 10. Geometric Optics: Part I 10.1 Lens Maker’s Equation 259 259 Topic Page 10.2 Thin Lens Equation 10.3 Imaging with Single Lens 10.4 Imaging with Multiple Lenses 10.5Laboratory Experiments Chapter Summary End of Chapter Exercises Group Discussion Questions 264 272 277 282 283 284 285 11. Geometric Optics: Part II 11.1 Imaging using Apertures 11.2 Optical Instruments 11.3 Aberrations 11.4 Laboratory Experiments Chapter Summary End of Chapter Exercises Group Discussion Questions 286 286 310 323 334 335 337 338 12. Introduction to Wave Optics 12.1 Huygen’s Principle 12.2 Interference 12.3 Diffraction 12.4 Laboratory Experiments Chapter Summary End of Chapter Exercises Group Discussion Questions 339 339 341 353 369 371 372 373 Topic Page Appendix I Useful Conversions 374 Appendix II Scientific System and Metric Prefixes 375 Appendix III Calculator Review (General) 376 Appendix IV Basic Geometric and Trigonometric Definitions and Identities 380 Appendix V Additional Useful Formulas in Fiber Optics 383 Answers to Odd-Numbered Problems 384 Index 413 Laboratory Experiments near the end of each chapter are merely suggestions. Safety precautions must always be observed especially when dealing with lasers and electrical circuits. 1 Chapter 1 Elements of Basic Algebra 1.1 Decimal, Scientific and Metric Notation of Numbers 1.2 Graphing Data 1.3 Various equation forms for straight lines 1.4 Formula Transformations Chapter Summary End of Chapter Exercises Group Discussion Questions Chapter Objectives After completing this chapter, the student will be able to (i) Convert numbers among regular decimal, scientific and metric engineering forms (ii) Plot points on a graph paper and connect those using straight lines or smooth curves (iii) Apply slope-intercept and point-slope equations of straight lines to experimental data. (iv) Transform algebraic formulas Key words: metric prefix, scientific notation, slope, intercept, “best-fit,” logarithmic scale, function 1.1 Decimal, Scientific and Metric Notation of Numbers1 In order for the student to converse in the language of the professionals working in lasers, fiber-optics, optics and related fields in photonics, it is necessary to understand the basics of number notation and measurement units. The system of counting numbers uses the base ten. Whole numbers can have any number of digits and can be positive, negative or zero, for example: -332, 0, 223,000, etc. The system of fractions came about to denote numbers that are not whole. Examples of fractions include 23 , −54 or 2 23 . Fractions have their usefulness in certain applications, for example: calculating one-third off sale at a store, three-quarters of the job completed, etc. However, in the scientific world, the decimal system is extensively used, and in certain cases, percentages are preferred over fractions. The decimal system of numbers may or may not have the decimal point that we are familiar with. Examples of decimal numbers include 3, 17, -5, -243, 23, 400, 0.001, -0.00234, 3.1416, -2.71828, etc. With whole numbers, the decimal point resides to the right of the last digit and is, hence, “invisible.” However, when one performs multiplication and division of whole numbers, the location of the decimal point becomes important since it could shift to the right or left. Let us analyze a few numbers based on the decimal system. 2 Example 1.1 Consider the number 23, 465 (twenty three thousand four hundred and sixty-five). It comprises of 5 6 4 3 2 in the unit’s place (1) 5x1 ten’s place (10) 6 x 10 hundred’s place (100) 4 x 100 thousand’s place (1000) 3 x 1000 ten thousand’s place (10,000) 2 x 10,000 = = = = = Value 5 60 400 3000 20000 When we add the values in the last column, 5+60+400+3000+20000, we get 23,465. Remember the “invisible” decimal point is after the number 5, the units place digit. Example 1.2 Consider the number 0.0234. Unlike the previous example involving a whole number, this number is clearly less than one since the starting digit in the unit’s place is zero (0). Thus, there is no need to consider values in the unit’s, ten’s, and hundred’s place, etc. as before. However, look at tenth’s place, hundredth’s place, thousandth’s place, etc. In this example, It comprises of 0 2 3 4 in the tenth’s place (0.1) hundredth’s place (0.01) thousandth’s place (0.001) ten thousandth’s place (0.0001) 0 x 0.1 2 x 0.01 3 x 0.001 4 x 0.0001 = = = = Value 0.0 0.02 0.003 0.0004 When we add up the values in the last column, 0+0.02+0.003+0.0004, we get 0.0234. With numbers that are too large or too small, it is often clumsy to represent them using the decimal system. For example, if I won $ 14,232,782 in a lottery, while I can boast my accomplishment to the nearest dollar to my friends, it is sufficient to say that I won about $14 million or about $14.2 million in a lottery without losing much meaning. This process is called rounding off. The original lottery amount is significant to 8 places or digits, while $14 million is significant to 2 places or digits and $14.1 million is significant to 3 places or digits. The rule of rounding off involves looking at the number to the right of the significant digit. If that number is 0,1,2,3 or 4 leave the significant digit as it is. If that number is 5,6,7,8 or 9 add one (1) to the significant digit. The following examples will illustrate these concepts. Example 1.3 Round off the following to three significant digits (a) 234,952 (b) 234,499 (c) 0.54287 (d) 0.0054287 (e) –1.085 (f) 23.02858 (g) 1,005,285 (h) 0.00005246 3 (a) 234,952. The third significant digit is 4. Look at the 4th significant digit. It is 9. Therefore add one to 4, making it 5. The answer is 235,000. (b) 234,499. The third significant digit is 4 and the fourth significant digit is also 4. Therefore the answer is 234,000 (c) 0.54287. The location of the decimal is irrelevant as far as counting significant digits. The third significant digit is 2 and the fourth is 8. Therefore we need to add one to two. The answer is 0.543. (d) 0.0054287. The location of the decimal or the number of zeros immediately following the decimal point is irrelevant as far as counting significant digits. The third significant digit is 2 and the fourth is 8. Therefore we need to add one to two. The answer is 0.00543. Note that the two zero’s immediately following the decimal point are kept intact. (e) –1.085. The minus sign will remain intact in the answer. The fourth significant digit is five. Therefore, the answer is –1.09. Note that the zero immediately following the decimal point is counted as a significant place since there is a non-zero number to the left of the decimal. (f) 23.02858. The fourth significant place is 2. Therefore the answer is 23.0 (g) 1,005,285. The fourth significant digit is 5. Therefore add one to the zero preceding it. The answer is 1,010,000. (h) 0.00005246 Again, don’t be baffled by the four zeros immediately following the decimal point. The fourth significant digit is six. Thus, add one to the four before. The answer is 0.0000525. Note again that the four zeros remained intact. Example 1.4 Round off the following to four significant digits (a) 234,952 (b) 234,499 (c) 0.54287 (d) 0.0054287 (e) –1.085 (f) 23.02858 (g) 1,005,285 (h) 0.00005246 (i) 1.0204 Verify that you get the following answers: _ (a) 234,952 Answer 235,000 (b) 234,499 Answer 234,500 (c) 0.54287 Answer 0.5429 (d) 0.0054287 Answer 0.005429 (e) –1.085 Answer –1.085 (f) 23.02858 Answer 23.03 (g) 1,005,285 Answer 1,005,000 (h) 0.00005246 Answer 0.00005246 and _ (i) 1.0204 Answer 1.020 (A bar above a zero indicates that the particular zero is the last significant digit.) 4 Example 1.5 Round off the following to two significant digits (a) 234,952 (b) 234,499 (c) 0.54287 (d) 0.0054287 (e) –1.085 (f) 23.02858 (g) 1,005,285 (h) 0.00005246 (i) 1 Verify that you get the following answers: (a) 234,952 Answer 230,000 (b) 234,499 Answer 230,000 (c) 0.54287 Answer 0.54 (d) 0.0054287 Answer 0.0054 (e) –1.085 Answer –1.1 (f) 23.02858 Answer 23_ (g) 1,005,285 Answer 1,000,000 (h) 0.00005246 Answer 0.000052 (i) 1 Answer 1 Throughout this text, unless otherwise specified, all answers will be rounded off to three significant digits. See the example below to understand the answer for (g) above. Example 1.6 How would you distinguish between say (a) 1,000,000 which is accurate to, say, four significant places and (b) 1,025,999 also accurate to four significant digits? _ (a) 1,000,000 = 1,000,000; the bar over the zero indicates that it is significant up to that digit; the zero’s following it arise from round offs. (b) 1,025.999 = 1,026,000. Here there is no bar. Six is the fourth and last significant digit. Significant digits and round offs are important in science and engineering because it is only proper to denote a measurement to the accuracy with which it can be measured. For example, if you weighed six potatoes of equal weight on a scale capable of reading up to an oz, and the total weight was 32 oz, then each potato weighs 32 divided by 6 or 5.33333….oz. Writing the answer to as many figures as the calculator or computer provides does not mean that a potato can be weighed to that accuracy. Since the scale can weigh only up to an oz., it is proper to round off the answer to 5 oz. On the other hand, if the scale were graduated to read in units of half-oz and the total weight is still 32 oz., the weight of each potato will be rounded to 5.5 oz. since 0.333… is closer to 0.5 than zero. Exercise 1.1A Express answers to three significant digits: 1. 988.82 2. 0.05687 3. 3.1446 4. 0.0000058754 6. 2.74828 7. –0.0245 8. 22.000 9. 872,499.998987 5. –32.17 10. 101,101 5 Express answers to two significant digits: 11. 988.82 12. 0.05687 13. 3.1446 16. 2.74828 17. –0.0245 18. 22.000 14. 0.0000058754 19. 872,499.998987 15. –32.17 20. 101,101 Express answers to four significant digits: 21. 988.82 22. 0.05687 23. 3.1446 26. 2.74828 27. –0.0245 28. 22.000 24. 0.0000058754 29. 872,499.998987 25. –32.17 30. 101,101 While the decimal system of numbers serves our purpose adequately to represent numbers and perform computations, expressing quantities outside the range of 1 to 10 (what humans can count with ten fingers) may tend to be cumbersome. For example, if a number has many zeros at the end or after the decimal point it is easy to miscount the number of zeros and arrive at a wrong answer. To avoid such errors and to write numbers in a more compact form, the scientific notation is often used. Scientific Notation of Numbers The system is still based on base 10 of the decimal system and uses the decimal point as necessary. However, in addition to a decimal number, it is followed by a multiplication sign which itself is followed by 10 to the power or exponent of a positive or negative integer. In order to understand the powers of ten refer to the table below: Positive exponents (or powers) of ten: 10 1 = 10 x 1 10 2 = 10 x 10 10 3 = 10 x 10 x 10 10 4 = 10 x 10 x 10 x 10 10 5 = 10 x 10 x 10 x 10 x 10 10 6 = 10 x 10 x 10 x 10 x 10 x 10 = = = = = = 10 100 1000 10,000 100,000 1000,000, etc. Negative exponents (or powers) of ten: 10 -1 = 0.1 x 1 10 -2 = 0.1 x 0.1 10 -3 = 0.1 x 0.1 x 0.1 10 -4 = 0.1 x 0.1 x 0.1 x 0.1 10 -5 = 0.1 x 0.1 x 0.1 x 0.1 x 0.1 10 -6 = 0.1 x 0.1 x 0.1 x 0.1 x 0.1 x 0.1 = = = = = = 0.1 0.01 0.001 0.0001 0.00001 0.000001, etc. Note that 100 is always one (1). Thus, for positive powers of ten, the exponent indicates the number of zeros following the significant digits. For example 2 x 10 5 is 200,000 (i.e. five zeros following 2.) For negative exponents, the value of the exponent is the number of places that the significant digit starts after the decimal point. For example, 2 x 10 -5 is 0.00002, where the significant digit is at the fifth place from the decimal point. 6 Example 1.7 Convert the following into scientific notation: (a) 0.0128 (b) –235,087 (c) 0.000589 (d) 5387 (e) 5,000,020 Scientific notations are always expressed as a single digit followed by a decimal point, followed by rest of the digits, ultimately multiplied by a positive or negative power of 10. (a) 0.0128 has to start with 1.28. Note that the decimal point is after the first significant digit, which is one. The decimal point of 1.28 is two places to the right (i.e. 100 times larger) compared to the original number of 0.0128. Thus 1.28 has to be multiplied by a number 100 times smaller to maintain its value. In other words, it is to be multiplied by 10 -2. The answer is 1.28 x 10 -2 (b) –235,087. Ignore the negative sign for the moment. 235,087 will have to be changed to a number starting with 2.35… The decimal point of this number has moved five places to the left compared to the original number of 235,087. Thus 2.35087 should be multiplied by 10 5 to maintain its value. The answer is –2.35087 x10 5, remembering to tag the negative sign back. (c) 0.0005893 = 5.89 x something, which should be 10 –4 yielding us the answer of 5.893 x 10 –4 (d) 5387 = 5.387 x 10 3 (e) 5,000,020 = 5.000020 x 10 6 Exercise 1.1B Convert the following into scientific notation: 1. 0.008758 2. 235.87 3. –45.236 4. 3.1416 5. 978.23 6. 62.4 7. 236,658.23 8. –0.000258 9. –689.25 10. 101,101 11. 22,400 12. 1024 13. 0.00000000000000000016 14. 235,000,000,000,000,000 15. –0.00005875 16. 858,564,458 17. 3,256.875 18. –875.5 19. 1,928 20. –32,400 21. 0.00256 22. 0.27858 Continuing the discussion on significant digits, the same rules apply in scientific notation. Example 1.8 Express the answers to each problem of Example 1.7 to three significant digits (a) 1.28 x 10 -2 = 1.28 x 10 -2 –4 (b) –2.35087 x 10 = –2.35 x 10 –4 –4 (c) 5.893 x 10 = 5.89 x 10 –4 (d) 5.387 x 10 3 = 5.39 x 103 and _ (e) 5.000020 x 10 6 = 5.00 x 106 Exercise 1.1C Convert the following into scientific notation rounding off to three significant digits: 1. 0.008758 2. 235.87 3. –45.236 4. 3.1416 5. 978.23 6. 62.4 7. 236,658.23 8. –0.000258 9. –689.25 10. 101,101 11. 22,400 12. 1024 7 13. 0.00000000000000000016 14. 235,000,000,000,000,000 15. –0.00005875 16. 858,564,458 17. 3,256.875 18. –875.5 19. 1,928 20. –32,400 21. 0.00256 22. 0.27858 Now that we have a more compact way of writing cumbersome numbers using the scientific notation in conjunction with rounding off to a few significant digits, let us refine the process further so that we have simplified notations for denoting certain powers of ten. This is accomplished by using prefixes in place of powers of ten. Many of us have already heard of words like kilometer, kilogram, milliliter, megabyte, milli-inches (mils), etc. The following prefixes cover powers of ten from a thousandth to thousand. Prefix name (symbol) milli (m) centi (c) deci (d) base unit (none) deca (da) hecto (h) kilo (k) Value (power of 10) Value (decimal) 10-3 0.001 -2 10 0.01 10-1 0.1 100 1 1 10 10 102 100 3 10 1000 The prefixes take on the multiples of the base unit. For example 1 mm (millimeter) is 10-3 or 0.001 of the base unit of m (meter). Similarly 1 kg (kilogram) = 103 or 1000 of the base unit g (gram). While the prefix notation is part of the metric system, it is further desirable to use only prefixes that use positive or negative integral powers of 1000. We will call these metric engineering prefixes due to its common use in electronics, engineering and technology. In this system all numbers are expressed as a decimal number whose value lies between 1 and 999 followed by a metric prefix. The following metric engineering prefixes are commonly used, arranged from small to large values: Prefix name Prefix symbol Value (powers of ten) Value (decimal) atto a 10-18 0.000 000 000 000 000 001 -15 femto f 10 0.000 000 000 000 001 pico p 10-12 0.000 000 000 001 -9 nano n 10 0.000 000 001 -6 micro µ 10 0.000 001 milli m 10-3 0.001 base unit none 100 1 3 kilo k 10 1000 mega M 106 1,000,000 9 giga G 10 1,000,000,000 tera T 1012 1,000,000,000,000 15 peta P 10 1,000,000,000,000,000 exa E 1018 1,000,000,000,000,000,000 8 An easy way of remembering the prefix symbol is to notice that all prefixes less than 1 in value have lower case symbols (f, p, n, µ,m) whereas all prefixes greater than 1 other than kilo (k) have upper case symbols (M,G,T,P) Powers of ten are easy to multiply and divide. Let us look at some examples. Example 1.9 Multiply 106 by 105 In decimal notation, 106 is 1,000,000 and 105 is 100,000. Thus the product is 1,000,000x100, 000 which is 100,000,000,000. This number is 1 followed by eleven zeros or 1011 which we realize can be simply obtained by adding the exponents, i.e. 106+5 = 1011 Example 1.10 Multiply 106 by 10-4 Using the principle above the product is 106-4 = 102 Example 1.11 Divide 106 by 105 In decimal notation, 106 is 1,000,000 and 105 is 100,000. The quotient is 1,000,000 divided by 100,000, which yields 10 or 101, which is simply obtained by subtracting the exponent of denominator from the numerator, i.e. 106-5 = 101 Example 1.12 Divide 103 by 10-3 Using the principle above the quotient is 103-(-3) which is 103+3 = 106 These examples reveal that when prefixes multiply or divide it is possible to simplify or collapse them into different prefixes instead of multiplying or dividing individual prefixes using powers of ten. Prefix multiplications: (a) milli x milli = 10-3 x 10-3 = 10-6 which is micro (b) milli x micro = 10-3 x 10-6 = 10-9 which is nano (c) micro x micro = 10-6 x 10-6 = 10-12 which is pico (d) milli x kilo = 10-3 x 103 = 1 which is the base unit (e) milli x mega = 10-3 x 106 = 103 which is kilo (f) micro x mega = 10-6 x 106 = 1 which is base unit (g) micro x giga = 10-6 x 109 = 103 which is kilo (h) micro x kilo = 10-6 x 103 = 10-3 which is milli (i) kilo x kilo = 103 x 103 = 106 which is mega =micro x milli =kilo x milli =mega x milli =mega xmicro =giga x micro =kilo x micro These are just a few sample short-cuts. You are encouraged to derive others. Also notice the reciprocal relationships among prefixes. 9 The other useful property of powers of ten is the ease with which they can be squared, cubed, etc. For example (103)2 is the same as 103 x 103 which is 103+3 = 106. This can be simply obtained by multiplying the two exponents i.e., 2 x 3 = 6. In the same way, (103)-3 = 10-9. Note that this is very different from multiplying 103 by 10-3, which would give 103-3 = 100 = 1 One (1) divided by milli is kilo (1 ÷ 10-3 = 103 which is kilo); similarly one (1) divided by kilo is milli; micro and mega are reciprocals of each other; nano and giga are reciprocals of each other and the same for pico and tera. We can now line up the prefixes on a straight line as shown in Figure 1.1, remembering that the neighbor on the right is 1000 times higher. //-|------|------|------|------|------|------|------|------|------|------|------|------|--//-----10-18 10-15 10-12 10-9 10-6 10-3 100 103 106 109 1012 1015 1018 a f p n µ m base k M G T P E Figure 1.1 The real number line There are two main categories of problems that we will deal with here. The first type is converting a decimal number into its most appropriate engineering metric prefix unit. The second type involves converting from one metric unit into another. Always remember to add the symbol for the base unit after the prefix (example m for meter, W for Watt, etc.) Just like common factors and common units in the denominator and numerator cancel each other out, common prefix symbols also do the same. Example 1.13 Power is measured in watts (W) as the base unit. Convert 0.00045 W to the most appropriate metric unit. Rewrite it as 0.000 450 W. Clearly, this value is much smaller than one. Therefore the first clue is that the prefix may be m,µ n, p, or f. The number 0.000 450 can be rewritten as 450 x 10-6 W or 450 µW (Answer). Note that the answer is the most appropriate since 450 lies between 1 and 999. Example 1.14 Frequency is measured in Hertz (Hz) as the base unit. Express 560,000,000,000,000 Hz in the most appropriate metric unit and round it off to two significant figures. "560" is followed by 12 zeros (or x 1012). 1012 corresponds to the metric prefix Tera (T). 560 THz (Answer). We know that this is the most appropriate answer since 560 lies between 1 and 999. 10 Example 1.15 Convert 3.3 µm to nm The base unit is m. 1 µm is 10-6 m and 1 nm is 10-9 m. The conversion factor is obtained by dividing the starting prefix by the ending prefix, i.e. 10-6 / 10-9 = 103 = 1,000 Thus, 3.3 µm = 3.3 x 103 nm = 3300 nm Answer Note that the starting prefix was the most appropriate metric unit. However, quite frequently, one is required to express the answer in a different metric or non-metric unit. In terms of length units, the following are the most commonly encountered units. Some of them belong to the metric engineering prefixes, others do not. “A” with a small circle above it denotes the Angstrom unit. Length measurement (Base unit = meter, symbol m) o 1A 1 nm 1 µm 1 mm 1 cm 1 km = = = = = = 10-10 m 10-9 m 10-6 m 10-3 m 10-2 m 103 m Time measurement (Base unit = second, symbol s.) 1 fs = 10-15 s 1 ps = 10-12 s 1 ns = 10-9 s 1 µs = 10-6 s 1 ms = 10-3 s Frequency measurement (Base unit = Hertz, symbol Hz.) 1 GHz = 109 Hz. 1 THz = 1012 Hz. 1 PHz = 1015 Hz. Power measurement (Base unit = Watt, symbol W.) 1 mW = 10-3 W 1 µW = 10-6 W Energy (work) measurement (Base unit = Joule, symbol J.) 1 mJ = 10-3 J 1 µJ = 10-6 J Example 1.16 Convert 25 µm (micrometers) to Angstrom units and also to mm (millimeters). 11 1 µm = o 10-6 m and 1 A= 10-10 m The conversion factor is obtained by dividing the starting prefix by the ending prefix, i.e. 10-6 / 10-10 = 104 = 10,000 25 µm = 25 x 104 o = 250, 000 A Answer For the second part, realize that 1 µm = 10-6 m and 1 mm = 10-3 m The conversion factor is obtained by dividing the starting prefix by the ending prefix, i.e. 10-6 / 10-3 = 10-3 This gives us 25 x 10-3 mm or 0.025 mm as the answer. Example 1.17 Frequency (ν) measured in Hz (Hertz, meaning per second) and time period T measured in s (seconds) follow a reciprocal relationship. (a) Find ν for T of 10 µs and 10 fs (b) Find T for ν of 300 THz, and 20 GHz (a) ν = 1 ---T Base unit of T is s, base unit of ν is Hz; thus one over second is Hertz. ν= 1 -----10 µs Step 1: Evaluate the reciprocal of the number first i.e. one over 10 is 0.1 Step 2: The reciprocal of prefix µ is M and the reciprocal of s is Hz Step 3 Thus ν =0.1MHz. MHz is not the most appropriate unit since 0.1 is below 1. We know that there are 103 kHz per 1 MHz. Thus, ν = 0.1 MHz x 103 kHz ------------1 MHz giving us an answer of 100 kHz Find ν for T=10 fs Step 1: Reciprocal of 100 is 0.1 Step 2: Reciprocal of f is P and reciprocal of s is Hz Step 3: Combine steps 1 and 2 to write the answer as 0.1 PHz or more appropriately 100 THz (b) Evaluate T for ν =300 THz Step 1: Reciprocal of 300 is 0.003333. Step 2: Reciprocal of T is p and reciprocal of Hz is s. 12 Step 3: T=0.00333 ps which converts to 3.33 fs as the most appropriate metric unit. T for ν =20 GHz Step 1: Reciprocal of 20 is 0.05 Step 2: Reciprocal of G is n and Hz is s Step 3: T=0.05 ns which convert to 50 ps Exercise 1.1D Convert the following to the most appropriate metric engineering unit: (In the following W stands for Watts, J for Joules, g for grams, s for seconds, Hz for Hertz, and m for meter) 1. 0.0010 W 4. 0.000022 s 7. 0.032 W 10. 105,600 Hz 13. 1000GHz 16. 0.050 s 19. 1 x 10-13 s 22. 1.2 x 104 g 25. 0.025 ms 28. 0.1PHz 2. 0.033 J 5. 0.000001 m 8. 0.33 mJ 11. 2200 GHz 14. 1100 J 17. 0.0026 mW 20. 22 x 10-16 s 23. 3.3 x 107 Hz 26. 25 x 105 Hz 29. 10,000 pg 3. 0.010 m 6. 0.002200µs 9. 0.100 g 12. 165,000,000,000 Hz 15. 3300 kW 18. 1.6 x 1011 W 21. 3.32 x 1015 Hz 24. 2000 mg 27. 3750 ns 30.1.37 x 104 kHz Exercise 1.1E Perform the following conversions to the units indicated for problems 1-20: 1. 0.01 ms = _______µs 3. 3300 kg = _______g 5. 1.2 kg = _______ mg 7. 0.00003 s= _______ µs 9. 2320 g = _______mg 11. 2320 k$ = ________M$ 13. 3.3 x 102 W = _______ kW 15. 0.0085 W = _______mW 17. 236,000 Hz = _______MHz 3 19. 6.023 x 10 ms = ________s 21. T=100 ps to ν = ______ 23. ν =125 THz to T= ______ 25. T=22000 ns to ν =_______ 27. ν = 0.1 Hz to ν = _______ 29. T=3.3 ms to ν = _______ 2. 0.027 µs = _______ ns 4. 2GHz = ________MHz 6. 1.1 x 103 Hz = _______MHz 8. 220 km = _______ m 10. 2 x 10-3 s = _______ms 12. 0.025 ms = _______ s 14. 1000 nJ = ________kJ 16. 1800 x 10-12 s = _______ µs 18. 5 x 10-3 J = _______mJ 20. 0.99 m = _______cm 22. T=20 ms to ν = _______ 24. ν =66.7 kHz to T=_______ 26. ν =4.3 x 1014 Hz to T= ____ 28. T=0.033 fs to ν = ________ 30. ν =355 THz to T= _________ 1.2 Graphing Data1 As a technician working in photonics and laser/electro-optics fields, it is necessary to know how to graph data properly, as well as read information from graphs, charts and 13 tables. While we will not delve deeply into algebraic functions, we will discuss linearity and non-linearity of functions, increasing or decreasing trend (i.e., direct and inverse proportions), setting up appropriate linear and logarithmic scales, drawing smooth curves and interpreting slopes of straight lines. Rectangular coordinate system and notations The topic deals with plotting points on a graph paper, drawing lines and curves and understanding trends. The most useful tool is the graph paper, which is basically divided into small squares. The squares result from drawing horizontal and vertical lines. We call the principal horizontal line as the x-axis or abscissa and the principal vertical line as the y-axis or ordinate. We also choose these axes to be located somewhat centrally with respect to the graph paper. Thus, in essence the x- and y-axes divide the graph paper into four sections or quadrants. The intersection of these axes is known as the origin, typically denoted as (0, 0) meaning x=0, y=0, The coordinate system is not unique to courses in mathematics. In fact, the entire earth is divided into latitudes and longitudes, which are angular representation of a place location with respect to its origin, which is, zero degrees latitude and zero degrees longitude. U.S. cities use a consistent numbering system for house numbers, always starting from an origin and going north, south, east and west in blocks of 100. The Geographical Information System (GIS) also depends on its coordinate system to locate places. The rectangular coordinate system is so called because the two axes that are fundamental to this system are perpendicular to each other. Other systems in two dimensions may use a polar coordinate system (distance from an origin and angle with respect to a fixed line), while more complicated three dimensional systems can be represented by rectangular (x, y, z coordinates), cylindrical (distance, angle, height) and spherical (distance and angles from two axes) coordinates. In this chapter we will deal only with the twodimensional rectangular coordinate system. For drawing graphs, it is irrelevant that we use a graph paper marked in inches, centimeters or any arbitrary measurement unit. What is important is that the distance from one marking to its adjacent marking on the graph up, down, right or left be the same. As shown in Figure 1.2, the first quadrant comprises of positive values of both x and y, the second quadrant has positive y values and negative x values, the third quadrant contains negative values for both x and y, and the fourth quadrant is composed of positive x values and negative y values. From a map reading standpoint quadrant I is in the northeast (NE), quadrant II is in the northwest (NW), quadrant III is in the southwest (SW) and quadrant IV is in the southeast (SE), all with respect to the origin. Example 1.18 Locate the following points on the graph and mark them with the respective letters: (A) (1, 3), (B) (5,3), (C) (-4,5), (D) (-3,-2), (E) (1,-2), (F) (3,0), and (G) (0,-4) 14 Example 1.18 illustrates how to plot points on a graph paper (Figure 1.3). If the numbers (or coordinates) are not whole as in the previous example, one has to read between the values to locate the point. For example if one coordinate is 1.5, we know that it lies halfway between 1 and 2. If another coordinate is –2.7, it lies between –2 and –3, however approximately two-thirds of the way from –2 and a third of the way from –3. Second Quadrant Third Quadrant First Quadrant 5 4 3 2 Origin (0,0) 1 y -5 -4 -3 -2 0 -1 -1 0 1 2 3 4 5 -2 -3 Fourth Quadrant -4 -5 x Figure 1.2 Illustration of rectangular the coordinate system. C 5 4 B 3 2 1 y -5 -4 -3 -2 0 -1 -1 0 A 1 2 3 4 5 -2 -3 -4 -5 x G D F E Figure 1.3 Illustration of Example 1.18 15 Example 1.19 Plot the following points and determine visually if they lie on a single straight line: A (1,2), B (0,1) and C (-3,-2). 3 2 1 y 0 -2 -4 -1 0 2 -2 -3 x Figure 1.4 Illustration of Example 1.19 The three points A, B and C are shown in the Figure 1.4. They all appear to lie on one straight line. Exercises 1.2A Plot the following sets of points on a graph paper. In each case visually determine of they fall on a straight line. 1. (2,1), (1,1), (0,1) 3. (1,2), (2,4), (3,9), (4,16) 5. (-2,-3), (3,7), (5,11) 7. (3,0), (4,0), (5,0), (-1,0) 9. (-5,-1), (-2,5), (0,9), (2,13) 11. (2,-7), (3,-8), (-4,-1), (-2,-3) 13. x y 14. x y 15. x y 16. x y 17. 2. (-2,3), (-1,4), (0,5), (3,8) 4. (-5,4), (-2,7), (1,10) 6. (4,5), (5,7), (6,5) 8. (-2,2), (3,3), (4.5,4.5) 10. (6,-2), (-2,2), (4,-1), (-4,3) 12. (2,-1), (7,4), (-4,4), (3,5) -3 15 -2 12 -1 9 0 6 1 3 2 0 3 -3 4 -6 -3 -4.5 -2 -2.5 -1 -0.5 0 1.5 1 3.5 -3 145 -2.5 137 -2 127 -1.5 116 -1 -0.5 0 102 81.1 1.5 0 0 50 60 100 170 150 180 200 250 300 350 200 300 320 420 2 3 4 5.5 7.5 9.5 16 x y 0 0 25 60 100 120 175 180 200 225 300 325 240 300 360 420 Graphing straight lines and curves1 We already had a preview of plotting points and connecting them. In the previous example, the points A, B and C happened to lie on a straight line. However, in other cases, points plotted may lie on a curve. In this section, we will learn how to draw straight lines and curves. Sometimes, when we collect data in the laboratory and plot them on a graph, all the points may not line up to give a nice straight line. Rather, some points may show “scatter” depending upon the accuracy of data collection process. In this case, the theory becomes useful. For example, if a resistor obeys Ohm’s law and all resistors we use in the laboratory do, then a graph of current vs. voltage should be a straight line. In this case, the current and voltage obey a linear relationship; the word “linear” derived from the word “line.” Similarly, if an object travels at constant speed, the relationship between distance it travels and time will also be linear. On the other hand, if you throw an object such as a ball at an angle at a certain speed, the path taken by the ball, known as its trajectory, is not a straight line. Rather, it is curved as it reaches a peak height and then starts loosing height as gravity takes over and pulls the ball towards the ground. Therefore, while a baseball fielder may or may not be aware of the equation of the trajectory of a “fly ball” in a baseball game, he/she should know to anticipate where the ball would descend and take the catch. Let us practice drawing straight lines and graphs. Example 1.20 Data Series 1: (3,5), (2.2,6.6), (-4,19), (1,9), (-1,13),(-2,15) and Data Series 2: (3,4.5), (2.2,2.4),(-4,8),(1,0.5),(-1,0.5),(-2,2) are given. Plot each data series on the same graph. 20 15 Series1 Series2 10 5 0 -5 0 Figure 1.5 Illustration of Example 1.20 5 17 Examine Figure 1.5. Clearly, data set A falls on a straight line with a negative slope, while data set B is curved with a minimum occurring at x=0. Notice that for data set B, while there are six individual points, one does not join adjacent points by straight lines. Rather, a smooth curve is drawn accommodating all six data points. Example 1.21 Plot the following points: x -4 -3 -2 -1 0 1 2 3 4 y 0.7 1.1 1.8 3.0 5.0 8.2 13.6 22.4 36.9 40 30 20 10 0 -6 -4 -2 0 2 4 6 Figure 1.6 Illustration of Example 1.21 Again, a smooth curve is drawn through the nine data points, as shown in Figure 1.6. This is an example of an exponentially increasing function. A real life example may be the population of an animal/bird species in a certain area where food is plenty and there are few predators or diseases. Here the x-axis represents time, perhaps in centuries, and the y-axis the population in units of thousands. We know that x comes before y in our alphabet. Hence, we always locate the xcoordinate first, then the y-coordinate while plotting points. x is called the independent variable and y the dependent variable. It implies that x causes y to happen, not the other way around. “x causes y to happen.” In electronics, voltage causes current flow. That is why we plot voltage on the x-axis and current on the y-axis. In mechanics, the x-axis is chosen to represent horizontal motion and y-axis for vertical motion. Time is normally plotted in the x-axis and a variable that depends on it such as distance, voltage, and 18 money or body weight is plotted on the y-axis. In some cases, distance is plotted on the x-axis, while power, speed or magnification may be plotted on the y-axis. Example 1.22 Plot the following points: x 0 1 2 3 4 y 1000 819 670 549 449 1200 1000 800 600 400 200 0 0 5 6 7 8 9 2 4 368 301 247 202 165 6 8 10 Figure 1.7 Illustration of Example 1.22 Figure 1.7 is an example of an exponential decay function characteristic of nuclear materials. It may also represent discharging of a capacitor where the charge on the capacitor is decreasing with time, or irradiance of a light beam measured at various distances from the source. A final example deals with a situation where the y values go up and down, i.e. cyclical. Example 1.23 Plot the following points: x y 0 15 30 45 60 75 90 x 0 1.3 2.5 3.5 4.3 4.8 5 y 135 150 165 180 195 210 225 x 3.5 2.5 1.3 0 -1.3 -2.5 -3.5 y 270 285 300 315 330 345 360 x -5 -4.8 -4.3 -3.5 -2.5 -1.3 0 y 405 420 435 450 465 480 495 3.5 4.3 4.8 5 4.8 4.3 3.5 19 105 120 4.8 4.3 240 255 -4.3 -4.8 375 390 1.3 2.5 510 525 2.5 1.3 Solution: 6 4 2 0 -2 0 -4 200 400 600 -6 Figure 1.8 Illustration of Example 1.23 Figure 1.8 is an example of a wave, light wave, sound wave or an electrical voltage from an alternating current (AC) source. The x-axis here represents an angle in degrees. It may also be translated to represent time. Let us now create graphs using functions. These functions will be of one independent variable only and could be linear (straight line) or non-linear (curved.) The function is denoted by the letter f. For example, f(x) means function of x. It does not represent f multiplying x. Some examples of functions are 1. f(x) = 2. f(x) = 3. f(x) = 4. f(x) = 5. f(x) = 6. f(x) = 7. f(x) = functions. 2 (a constant; i.e. the function does not change and it is always 2) 3x+2 (a linear function in x; i.e. a straight line when graphed.) 2x2 –4x + 1 (a quadratic function- it will be a curve) k/x, or k/x2, inverse functions, k=constant 4e-4x, an exponential function log x, logarithmic function A(sinx), sine function, (A=constant) one of several trigonometric For this chapter we will confine ourselves only with functions of the types (1)-(4) above. A discussion of other types of functions and their graphs will be presented in later chapters. To draw graphs, choose a value of x, substitute it in the function to obtain the value of the function. The value thus obtained is the y-coordinate, i.e. y=f(x). Choose additional xvalues and compute the corresponding y-values. Graph the ordered pair of (x, y) values on the graph and join adjacent points by a smooth curve or line as the case may be. The interval of x-values plotted is known as the domain of the function whereas the corresponding interval of y-values is called the range of the function. The true domain 20 and range of many functions are really large, i.e. they may occupy all real values from negative infinity to positive infinity. However, for technical applications, we work with limited values of the domain and range because they are of practical use to us in solving real life problems. Example 1.24 Draw the graph of the function f(x) = -3x + 4 It is always a good idea to choose x=0 and compute the corresponding y=f (0). In this case f (0) = -3(0) + 4 = 4. Thus, (x ,y) = (0,4) Choose x=1. f (1) = -3(1) + 4 = - 3 + 4 = 1. Thus (x, y) = (1,1) While two points are adequate to draw a straight line, it is always a good idea to use a third point to ensure that the calculations are correct. If an error was made in any of the two previous steps, then the three points will not lie on a straight lie. Choose x=3. f(3) = -3(3) + 4 = - 9 + 4 = -5. Thus, (x, y) = (3,-5). Create a table as follows. x 0 1 3 y 4 1 -5 Plot the three data points on a graph paper. The result should look like this (Figure 1.9). 6 4 2 0 -2 0 -4 -6 1 2 3 4 Figure 1.9 Illustration of Example 1.24 Redraw the graph from Figure 1.9. Call the original data set A. Plot a second set of data points as per the following table (Data set B): x y 0 1 1 2 At what points do the two lines intersect? Plot both sets of data. The graphs will look like this (Figure 1.10): 3 4 21 6 Data set B 4 2 0 -2 0 1 2 3 4 Data set A -4 -6 Figure 1.10 Illustration of Example 1.24 The two lines intersect at x=0.75, y=1.75. We used a domain from x=0 to 3 and were able to find the intersection point. Had we used a domain of x=2 to 3 for the second function, we would not have noticed the intersection point! Example 1.25 Draw the graph of the function f(x)= 2x2 – 3x + 4, for the domain x=-5 to x=5. This is a quadratic function and thus will be curved. Thus, it is essential to use sufficient number of points that will enable a smooth curve to be drawn. Let us use at least seven data points. The table of (x, y) values is given below obtained via substitution of x-values in the above function. As an example, for x=5, y= f(5) = 2(5)2 –3(5) + 4 = 50 –15 + 4 = 39. x -5 -1.5 -0.75 0 0.75 1.5 5 y 69 13 7.4 4 2.9 4 39 We are considering here a limited domain of x=-5 to x=5. The corresponding limited range is y=f(x)=4 to y=f(x)=69. (See Figure 1.11). It is seen from Figure 1.11 that the function goes through a minimum at x=0.75 Example 1.26 Frequency (ν) and time period (T) are reciprocals of each other. Choose T from 0.1 s to 1 s, in steps of 0.1 s and then T from 1 to 10 s in steps of 2 s. Calculate the corresponding ν values (in cycles per second or Hertz) and make a plot of ν vs. T. T(s) ν Hz) 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 3 5 7 9 10 10 5 3.33 2.50 2.00 1.67 1.43 1.25 1.11 1.00 0.33 0.20 0.14 0.11 0.10 22 The table above was generated by substituting for various values of T (horizontal axis) and calculating ν (vertical axis) using the formula ν = 1/T. The resulting graph is shown in Figure 1.12: -5 80 70 60 50 40 30 20 10 0 -1 -3 1 3 5 Figure 1.11 Illustration of Example 1.25 12 10 8 6 4 2 0 0 2 4 6 8 10 Figure 1.12 Illustration of Example 1.26 It is clear that as the time period is decreased the frequency increases drastically and as the frequency is decreased, the time period increases dramatically. For this particular behavior, it is more instructive to use a logarithmic graph paper the details of which will be discussed later in this section. Exercises 1.2B Plot the following points on a graph paper. Join adjacent points by a smooth curve (or straight line, if applicable). 1. x y 0 25 50 50 100 75 150 100 200 250 300 350 125 150 175 200 23 2. x y 0 250 50 100 150 200 250 300 50 -150 -350 -550 -750 -950 3. Data set A: x -4 -3 -2 -1 0 1 2 3 4 5 y 1.5 2.5 3.5 4.5 5.5 6.5 7.5 8.5 9.5 10.5 Data set B: (-1, 0.5), (0,0), (2, 2), (4,8), (-3, 4.5), (1, 0.5), (-4,8), (-2,2), (3, 4.5), (-4,8), (5,12.5) 4. Data set A: x -4 -3 -2 -1 0 1 2 3 4 y -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 Data set B: (-4, -12.8), (-3, -5.4), (-2, -1.6), (-1,-0.2), (0,0), (1,0.2), (2,1.6), (3,5.4), (4,12.8) 5. Data set: (-3,6), (-2,2), (-1,0), (-0.5,-0.25),(0,0), (0.5,0.75), (1,2), (2,6), (3,12), (4,20) 6. Data set x -3 -2 -1 -0.5 0 0.5 1 2 3 4 y 12 6 2 0.75 0 -0.25 0 2 6 12 7. Data set x 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 y -19.0 -18.4 -17.3 -15.5 -12.6 -7.8 0.1 13.1 34.6 70.0 8. Data set T 25 30 35 40 45 50 55 R 100 102 106 112 120 130 142 (Hint: Plot T along the x-axis and R along the y-axis.) 9. Data set x -180 -150 -120 -90 -60 -30 0 30 60 90 120 150 180 y -1.00 -0.87 -0.50 0.00 0.50 0.87 1.00 0.87 0.50 0.00 -0.50 -0.87 -1.00 10. Data set x -180 -150 -120 -90 -60 -30 0 30 60 90 120 150 180 y 0.00 -0.50 -0.87 -1.00 -0.87 -0.50 0.00 0.50 0.87 1.00 0.87 0.50 0.00 11. Data set I (mA) 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 P 0.100 0.225 0.400 0.625 0.900 1.23 1.60 2.03 2.50 3.03 (mW) (Hint: Plot I on x-axis and P on y-axis) 24 12. Data set (Hint: Plot λ(nm) on the x-axis and E (eV) on the y-axis). λ (nm) 400 425 450 475 500 525 550 575 600 625 650 675 700 E (eV) 3.11 2.92 2.76 2.62 2.48 2.37 2.26 2.16 2.07 1.99 1.91 1.84 1.77 13. Data set d 0.25 0.5 0.75 1 1.25 1.5 1.8 2 Stress 510 127 56.6 31.8 20.4 14.2 10.4 7.96 (Hint: Plot d on the x-axis and Stress on the y-axis) 14. Plot f(x) = x and f(x) = -x + 2 on the same graph. Label them with the functions. 15. Plot f(x) = 0.1 x + 0.3 for x=10 to x=20. What is the corresponding range? 16. Plot f(x) = 2x – 4 in the range of f(x)=-2 to f(x) = 16. What is the corresponding domain? 17. Plot f(x) = -x + 1 and f(x) = x + 1. At what point do the lines intersect? 18. Plot f(x) = -0.5x + 1.0 and f(x) = 3(0.2-x). Use x-values in the domain of –1 to 3. At what points do they intersect? What is the y-value at that point? What is the range? 19. Plot f(x) = 2x2 in the domain of –2 to 2 for x. What is the corresponding range? 20. Plot f(x) = -x2 + 3x + 10 for x=-1 to x=+3. What is the corresponding range? 21. Plot f(x) = -x2 and f(x) = 2x on the same graph. At what point do they intersect? Use x-values from –5 to 0. 22. Plot f(x) = x2 + 2x + 1 and f(x)=x2-2x +1. (Where) do they intersect? 23. Plot the function I = 100/R, with R on the x-axis and I on the y-axis for the following values of R: 100, 220, 330, 470, 560, 680, 810 and 1000. 24. Plot the function P = 100I2 for the range of values from I=0.1 to I=1.0 in steps of 0.1. Plot I on the x-axis and P on the y-axis 25. Plot the function s’ = fs/(s-f), given f=5 for the following values of s: 20, 15, 10, 8, 6, 5, 4, 3,2 and 1. What happens at s=5? 26. Plot E (eV) = 1241/λ for values of λ from 300 to 1000 in steps of 100. 27. Plot V=P/I, given P=0.025 and for values of I from 1x10-3 to 0.01, in steps of 1 x 10-3. 25 28. At what points does the function y = (x-4)(3-2x) intersect the x-axis? Does the curve have a maximum or a minimum anywhere? If so, at what point(s)? Solve the problem graphically. (Choose your limited domain carefully so that the required answers will be seen within the bounds of the graph.) Logarithmic scale For situations where the x or y values extends over several orders of magnitude, a logarithmic scale is preferred. While the details of exponentials and logarithms are presented later in connection with fiber losses and absorption, the method of using logarithmic scales is illustrated by redoing Example 1.26 using a logarithmic scale. Example 1.27 Frequency (ν) and time period (T) are reciprocals of each other. Choose T from 0.1 s to 1 s, in steps of 0.1 s and then T from 1 to 10 s in steps of 2 s. Calculate the corresponding ν values (in cycles per second or Hertz) and create a graph of ν vs. T. T(s) ν Hz) 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 2 4 6 8 10 10 5 3.33 2.50 2.00 1.67 1.43 1.25 1.11 1.00 0.50 0.25 0.17 0.13 0.10 This is the same data as in Example 1.26, however logarithmic scales will be used both for time period and frequency. 10 1 0.1 1 10 0.1 Figure 1.13 Illustration of Example 1.27 Notice several differences between Figure 1.13 (logarithmic scale) and Figure 1.12 (linear scale). The data now lie on a straight line. All data points are almost evenly distributed here and much easier to read. The scales themselves have been adjusted to enable this. Each major division is ten times higher than its previous division on both axes. The above is an example of 2 decades by 2 decades graph paper. Adjacent minor divisions are not equally spaced unlike in a linear scale. For example, the distance between 0.1 and 0.2 is much higher than the distance between 0.2 and 0.3, and so on. However, the distance between 0.1 and 0.2 is the same as the distance between 1 and 2. 26 Example 1.28 Data collected from power measurements are shown below. When power P was measured close to the light source, one obtains a higher power reading while the power reading drops off drastically at larger distances d from the source. Using logarithmic scales, plot the data. d(cm) 10 20 30 40 60 80 100 150 200 500 P(mW) 5 1.25 0.5556 0.3125 0.1389 0.0781 0.05 0.0222 0.0125 0.002 The graph is shown in Figure 1.14. It requires 4 decades on the vertical axes and two decades on the horizontal axis to accommodate all data points. Both the above examples use a log-log paper, meaning logarithmic scales on both axes. In some cases, one uses a semi-log paper, i.e. only one axis is logarithmic, the other linear. The choice depends on the extent of the data on either axis, i.e., if the data extends over several orders of magnitude (tenths, units, tens, hundreds, etc.) it is advisable to use a logarithmic scale for that axis. Realize that one cannot plot zero (0) on a logarithmic axis. Therefore, only non-zero values can be plotted. 10 1 10 0.1 100 1000 0.01 0.001 Figure 1.14 Illustration of Example 1.28 1.3 Various equation forms for straight lines1 At this point, we should be experts drawing straight lines and curves given either the individual data points or the associated function between x and y denoted by f(x). In this section, we will discuss various ways of characterizing a straight line. There are four ways of doing this: (i) the slope-intercept form, (ii) the point-slope form, (iii) the two points form and (iv) the dual intercept form. In any of the four cases, one needs two pieces of information and then the functional form of the straight line can be easily generated. We will discuss the first two forms here. 27 (i) The slope-intercept form: As you might have guessed, slope denotes how slanted a line is. A line with a “small” slope is considered shallow whereas one with a “large” slope is considered steep. The steepest slope is the case of a vertical line. Examples of vertical lines on a graph are the y-axis itself, given by the equation x=0, or any other vertical line given by the equation x=c, where c is any constant, positive or negative, whole number, decimal or fraction, so long as it is real. Lines with no slope (zero slope) can also be drawn on the graph paper. These are horizontal lines, which lay perfectly flat. Examples of horizontal lines include the x-axis itself, given by the equation y = 0 or any other horizontal line given by y = c, where c is any constant, positive or negative, whole number, decimal or fraction, so long as it is real. Thus, x = c or y = c are the simplest forms of straight lines discussed thus far. They are, respectively, vertical or horizontal. The y-intercept, hereinafter referred to as the intercept, is the vertical height from the origin where the line crosses the y-axis. For a vertical line, the intercept is infinity since the vertical line and the y-axis are parallel to each other and therefore will never intersect, except at infinity. For a horizontal line y = c, the intercept is c. For all other lines, the intercept can be calculated by substituting x = 0 in the equation and solving for y. The value of y thus obtained is the intercept. Example 1.29 Find the intercepts of the following straight lines: (a) 3x-2y=12, (b) y=2.2x-5, (c) –3y = 5x (a) Substitute x=0. 3(0) – 2y = 12 -2y = 12 Dividing both sides by –2, we get y = 12/(-2) = -6 Answer (b) Substituting x=0, we have y = 2.2(0) –5 = 0 – 5 = -5 Answer (c)Substitute x=0, we get –3y = 5(0) =0 gives us y=0 Answer (i.e. no intercept; line passes through the origin) The graphs are shown in Figure 1.15. Thus far, we know the slopes of two extreme classes of straight lines. A vertical line has a slope of infinity and a horizontal line has a slope of zero. All other lines will have a slope in between and can be positive or negative. A line with a positive slope travels “uphill” as one travels from low values of x to high values of x. A line with a negative slope travels “downhill” as one travels from low values of x to high values of x. The slope itself can be calculated by the formula m = rise / run. Here m is the slope; rise is the difference in y-values (vertical rise) for a given movement in the horizontal direction (run.) If we denote the starting point by the coordinates (x1, y1) and the ending point by the coordinates (x2,y2), then the slope is given by 28 -2 -1 4 2 0 -2 0 -4 -6 -8 1 a b c 2 Figure 1.15 Illustration of Example 1.29 m = y2 – y1 ---------x2 – x1 This formula holds for all straight lines, horizontal, vertical, slanted upwards or downwards. Example 1.30 Calculate the slope of each straight line passing through the gives sets of points: (a) (1, 2) and (3,5), (b) (2,1) and (3,-5), (c) (-3,-3) and (4,2), (d) (4,-2) and (4,-4) and (e) (-4,2) and (4,2) In each case, assign (x1, y1) to the first point and (x2, y2) to the second point. (a) Using m = y2 – y1 ---------x2 – x1 m = 5-2 3-1 (b) m = -5 –1 = 3–2 = -6 Answer (c) m = 2 – (-3)= 4 – (-3) = 5/7 Answer (d) m = -4 – (-2) = -2 = Infinity Answer 4–4 0 = 3 2 = 1.5 Answer 29 (e) m = 2–2 = 0 4 –(-4) 8 = 0 Answer Once we know how to calculate the slope and intercept, the equation of the straight line in the slope-intercept form is y = mx + b with m the slope and b the intercept. In the following examples, the slope and intercept will be given. One needs to evaluate the equation for the straight line. Example 1.31 For each case, find the equation of the corresponding straight line. Then evaluate y for x=2. (a) Slope=2, intercept=3 (b) Slope=-4, intercept=1.5 (c) Slope = 0, Intercept=-3 (d) Slope=-1, intercept=0. (a) y = mx Substitute m=2, b=3. y = 2x Find y when x=2. y = 2(2) + b + 3 Answer + 3 = 4+3 = 7 Answer (b) m=-4, b=1.5, thus y = -4x + 1.5 Answer For x=2, y= -4(2) + 1.5 = -6.5 Answer (c ) m=0, b=-3. thus y = -3 Answer Thus no matter what x is including x=2, y is always –3 Answer. (d) m=-1, b=0, thus y = -x Answer For x=2, y = -2 Answer Example 1.32 For each of the following straight lines, find the slope and intercept. (a) y = -2x + 3 (b) 3x - 4y = 6 (c) y = 2.72 In all the cases, if the equation looks like the mx + b form, one can, by comparison determine m and b easily. Otherwise, the equation should be manipulated to the slopeintercept form. (a) y = -2x + 3 looks like y = mx + b. Thus, m=-2 and b=3 Answer by comparison. (b) 3x - 4y = 6 needs transformation + 4y +4y ----------------------Adding 3x = 6 + 4y Subtract -6 -6 ----------------- 30 3x – 6 = 4y Dividing both sides by 4 gives y = (3/4) x – (3/2), thus m= ¾ and b= - 3/2 Answer (c) y = 2.72 has no x term, thus m=0 and b=2.72 Answer (ii) The point-slope form In this case the slope and the coordinates of a point on the straight (rather than the intercept) will be given. If the slope is m and the straight line passes through (x1, y1), then the equation of the straight line is y – y1 = m(x – x1) Once the values of m, x1 and y1 are substituted, one must simplify it. Example 1.33 If a straight line has a slope of –2 and passes through (3,-3), what is its equation? Here m=-2, x1=3, y1=-3. Substituting in y – y1 = m(x – x1) y – (-3) = -2(x-3) y + 3 = -2x + 6 - 3 = - 3 Subtracting, y = -2x + 3 Answer (Thus the straight line has an intercept of 3) Determining slope and intercept from experimental data Experimental data collected on variables inherently would exhibit variability or scatter due to measurement and calibration errors and accuracy of the measuring devices or tools used. For example, the following is a set of data collected from measurements of applied voltage (independent variable) and device current (dependent variable). Example 1.34 Graph the data. Find the slope and intercept. V 10 20 I (mA) 1.25 2.4 31 30 40 50 60 70 80 3.8 5 6.4 7.2 9 10.8 The graph in Figure 1.16 is an incorrect way of joining the experimental data points. I 12 10 8 6 4 2 0 0 20 40 60 80 100 V Figure 1.16 Illustration of Example 1.34 The reason why this is incorrect is because adjacent data points are connected by straight lines resulting in a graph with “kinks.” One needs to plot the points and create a straight line that best fits all the data points to the extent possible. The manual technique is called “eye ball” method. Programs such as Microsoft Excel® can find the best fit straight line given a set of data points. Figure 1.17 illustrates the best-fit line. One could choose two sets of points that lie on the straight line (not the original experimental data points). For example, the points (16, 2) and (56, 7.2) lay perfectly on the straight line. Note that while these points did not come from the original measurements, they are useful because it is easy to read them off the graph since they are coincident with a grid line on the graph. Notice that the straight line passes through the origin and thus the intercept, b is zero. We calculate the slope, m as m = (7.2 – 2) / (56-16) = 0.13 Remember that the equation of the straight line in slope-intercept form is y = mx + b. Here I represents y, V represents x. Thus the equation of the best-fit straight line is I = 0.13V Answer 32 (This means that the current in mA can be obtained by multiplying voltage by 0.13). Original data points corresponding to V=60 and V=80 were in considerable error as shown by their deviation from the straight line. The straight line was drawn to give maximum preference (weight) to the data points that automatically fall close to the best fit line and to give less preference (weight) to those that fall farther from it (i.e. the ones exhibiting scatter.) Correct graphing method I 12 10 8 6 4 2 0 0 20 40 60 80 100 V Figure 1.17 Illustration of Example 1.34 Exercises 1.3A Find the intercept for each of the following straight lines 1. y = 2x – 5 2. 2x + 4y = 10 3. –2x + 3y = 15 5. 2x + 6y = 3 6. 1.5x + 2.2y = 4.3 7. 2x + 3y = 13 4. 2y = 7x + 14 8. 2y – 3x – 18 = 0 Find the slopes of the straight line connecting each of the following pairs of points. 9. (1,1), (0,0) 10. (0,2), (-4, 4) 11. (-1,1), (-1,4) 12. (4,3), (3,4) 13. (-5, 4), (-1,12) 14. (2.2, 3.3), (10,20) 15. (2,0), (4, -10) Find the equation of the line in each case given the slope and intercept 16. m=-1, b=4 17. m=0, b=1 18. m=-2, b=4 19. m=1.5, b=2.2 20. m=20, b=400 21. m=1000, b=-5000 Find the slope and intercept for the following lines: 22. y= x + 1 23. y= 2x+3 24. 2y=3x + 4 26. 2x-4y-13=0 27. –x-y-1=0 28. 4x – 3y=45 25. –5y+3x=32 29. –2x + 7y = -28 Find the equation of the straight line passing through the given point with the given slope. 30. (1,1) with m=1 31. (-3,5) with m=-1.5 32. (0,0) with m=4 33. (-2,-4) with m=-2 34. (2,-5) with m=2.5 35. (0,-2) with m=0 36. (-2,9) with infinite slope 37. (1,-5) with m=-3 38. (-6,4) with m=-6 33 39. (1.3, 2.6) with m= –3.9 40. (¾ , ¾) with m=¾ Graph the following data sets on separate graph sheets, draw a best-fit line in each case, and then determine the slope and intercept for each data set. 41. Data set: Time (s) Distance (mm) 42. Data set: Energy (eV) Depth (nm) 43. Data set: Voltage(V) Current (mA) 5 10 15 20 25 30 35 40 45 50 12 18 30 35 45 58 77 85 90 95 1 2 3 4 5 10 15 20 25 50 1.4 3.1 4.8 5.8 7.2 17 25 29 40 72 20 40 60 80 100 120 1.7 2.8 4.8 5.8 8 140 160 180 200 8.8 11.2 11.8 14.3 14.8 44. Data set: (All values in degrees) Angle of incidence 5 10 15 20 25 30 35 40 Angle of reflection 5 10 14.5 20.5 24 31 33.5 41.5 45 50 43 52 1.4 Formula Transformations: Quite frequently, one is often required to manipulate formulas and express a variable in terms of other variables. In such cases, the final answer will be an algebraic expression of other variables. If the values of those variables are given, then one can substitute them into the algebraic expression and give a final numeric value to the problem. Example 1.35 Given 1 + 1 = 1 s s’ f Solve for (a) s (b) s’ and (c) f in terms of the other variables. This is the thin lens equation that one often encounters in optics. More details are available in Chapter 10. On the left hand side (LHS), the common denominator of s and s’ is ss’. Rewrite the original expression as 34 or s’ + ---ss’ s’ + s ---------ss’ s = ---ss’ = 1 ---f 1 --f Cross-multiplying, we get f(s’ + s) = ss’ We will use this expression to solve for (a) s, as follows Using the distributive law, fs’ + fs = ss’ Subtracting ss’ from both sides yields fs’ + fs –ss’ = 0 Factoring the s term on the LHS, s (f-s’) + fs’ = 0 Subtracting fs’ from both sides, s (f-s’) = -fs’ Dividing both sides by (s’-f), s = fs’ --------- Answer s’-f f(s’ + s) = ss’ We will use this expression to solve for (b) s’, as follows (a) Using the distributive law, fs’ + fs = ss’ Subtracting ss’ from both sides yields 35 fs’ + fs –ss’ = 0 Factoring the s’ term on the LHS, s’ (f-s) + fs = 0 Subtracting fs from both sides, s (f-s’) = -fs Dividing both sides by (s-f), s’ = fs --------- Answer s-f To solve for (c), start from f(s’ + s) = ss’ Divide both sides by (s’+s), gives f = ss’ -------- Answer (s+s’) The above equation is used extensively in optics where s denotes distance of object from a lens, s’ the distance of the image formed from the lens and f the focal length of the lens. Replacing s by R1, s’ by R2 and f by RT, one gets a similar equation valid for finding total (or equivalent) resistance RT of two resistors R1 and R2 in a parallel circuit. Example 1.36 Find m the slope if the equation of a straight line is y = mx + b Notice that m is associated with a term to which b is added. The first step is to undo the addition operation by subtracting b from both sides of the equation y = mx + b -b -b ---------------------------y – b = mx. Now, x multiplies m, therefore to undo the multiplication operation, we need to divide both sides by x. 36 y – b = mx. ----------- -------x x Thus, m = (y-b) / x Answer Example 1.37 Power P is given by V2 / R, where V is the voltage, R the resistance. Rearrange this equation and solve for V. P = V2 R Notice that R is dividing the V term. Undo this by multiplying both sides with R, giving PR = V2 We need to solve for V, not V2. Undoing the square operation on V is the same as taking the square root (or one-half power). Taking the square root on both sides, we get ____ V = √PR Answer. Example 1.38 The volume of a sphere is given by V = 4 π R3 3 Find R, the radius. V = 4 π R3 3 The term (4/3) π multiplies R3. Undo this operation by first multiplying by (3/4), then dividing by π, giving us 3V = R3 ----4π To solve for R take the cube-root (i.e. one third power) on both sides, yielding __ 3 R = √3V or (3V/4π)1/3. Answer 3 √4π Formula substitution involves plugging in the values of the unknowns on the RHS of the equation and evaluating the variable on the LHS. For example, for the last example, if V was given as 100 cm3, we get, 37 R = [(3 x 100)/(4 x π)]1/3 R = 2.88 cm to three significant figures. Answer Formula transformations involving logarithms and exponentials are explained in further chapters. Exercises 1.4A Solve for the variables indicated 1. v2-u2 = 2as for (a) v and (b) s 3. s = ut + (1/2) at2 for (a) u (b) a 5. RT = R1 + R2 + R3 for R3. 7. P = F / A for A 9. g = (4π2l/T2) for T 11. k=1.6m for m 13. R = R0(1+ αT + βT2) for (a) β, (b) α (c) R0 15. V = πr2 h for (a) r (b) h 2. v = u + at for (a) t (b) u 4. Ffr = µN for µ 6. P = I2R for I 8. τ = V/v for V 10. F = 1.8C + 32 for C 12. R = ρl/A for ρ 14. A = πr2 for r 16. A2 = s(s-a)(s-b)(s-c) for (a)a (b)b and (c) c Chapter Summary • Positive exponents (or powers) of ten: • 10 1 = 10 x 1 • 10 2 = 10 x 10 • 10 3 = 10 x 10 x 10 • 10 4 = 10 x 10 x 10 x 10 • 10 5 = 10 x 10 x 10 x 10 x 10 • 10 6 = 10 x 10 x 10 x 10 x 10 x 10 = 10 = 100 = 1000 = 10,000 = 100,000 = 1000,000, etc. Negative exponents (or powers) of ten: • 10 -1 = 0.1 x 1 • 10 -2 = 0.1 x 0.1 • 10 -3 = 0.1 x 0.1 x 0.1 • 10 -4 = 0.1 x 0.1 x 0.1 x 0.1 • 10 -5 = 0.1 x 0.1 x 0.1 x 0.1 x 0.1 • 10 -6 = 0.1 x 0.1 x 0.1 x 0.1 x 0.1 x 0.1 = 0.1 = 0.01 = 0.001 = 0.0001 = 0.00001 = 0.000001, etc. Note that 10 0 is always one (1). 38 Prefix name(symbol) • milli (m) • centi (c) • deci (d) • base unit (none) • deca (da) • hecto (h) • kilo (k) value (power of 10) 10-3 10-2 10-1 100 101 102 103 METRIC ENGINEERING NOTATION Prefix name Prefix symbol Value (powers of ten) atto a 10-18 femto f 10-15 pico p 10-12 nano n 10-9 micro µ 10-6 milli m 10-3 base unit none 100 kilo k 103 mega M 106 giga G 109 tera T 1012 peta P 1015 exa E 1018 • • • • • • Value (decimal) 0.001 0.01 0.1 1 10 100 1000 Value (decimal) 17 0’s between “.” & 1 0.000 000 000 000 001 0.000 000 000 001 0.000 000 001 0.000 001 0.001 1 1000 1,000,000 1,000,000,000 1,000,000,000,000 1,000,000,000,000,000 1 followed by 18 0’s The rectangular coordinate system comprises of four quadrants: I (x>0, y>0), quadrant II (x<0, y>0), quadrant III (x<0, y<0) and quadrant IV (x>0, y<0). The origin of this system is (0, 0). One plots points on a piece of graph paper using the (x, y) coordinates. If the points appear to fall on a straight line, they must be joined by the best-fit straight line. Otherwise, a smooth curve should be drawn accommodating all points. A function defines the relationship between the independent variable (x) and the dependent variable (y) or f(x). If this relationship is linear, i.e. a function with x in its first or 0th power, one obtains a straight line. If not, one obtains a curve. The span of the x variable is denoted as the domain of the function and the corresponding y=f(x) values are denoted as the range. While theoretical domain and ranges are often large, one is usually interested in a limited set of domain and range values which are of practical interest to the technician. The slope of a straight line is given by the ratio of the rise to the run. Slopes can be positive, negative, zero, or infinite. Straight lines can be described using the (i) slope-intercept form, (ii) slopepoint form, (iii) two-points form and (iv) the dual intercept form. 39 • • A straight line with zero slope is horizontal, one with infinite slope is vertical, one with positive slope slants upward and one with negative slope slants downward. Formula transformation involves algebraic manipulation to result in a desired variable expressed as a function of the other variables. End of Chapter Exercises Convert the following into scientific notation and round off to three significant digits: 1. 0.007738 5. 928.53 2. 735.17 6. 62.9 3. –15.936 7. 236,499.99 4. 3.7416 8. –0.0002585 Convert the following to the most appropriate metric engineering unit: (In the following W stands for watts, J for Joules, g for grams, s for seconds, Hz for Hertz, m for meter) 9. 0.0077 W 12. 0.000052 s 15. 0.038 W 18. 705,600 Hz 10. 0.003 J 13. 0.0000001 m 16. 0.47 mJ 11. 0.0010 m 14. 0.03200µs 17. 0.100 s Perform the following conversions to the units indicated for problems 19-22: 19. 0.002 ms = _______µs 21. 5300 kg = _______g 23. T=500 ps to ν = ______ 25. ν =625 THz to T= ______ 20. 0.007 µs = _______ ns 22. 5.5GHz = ________MHz 24. T=60 ms to ν = _______ 26. ν =33.7 kHz to T=_______ Plot the following sets of points on a graph paper. In each case visually determine if they fall on a straight line. 27. (7,1), (1,-5), (0,-6) 29. (1,2), (2,8), (3,27), (4, 64) 28. (-2,3), (-1,4), (0,5), (3,8) 30. (-5,4), (-2,7), (1,10) Plot the following points on a graph paper. Join adjacent points by a smooth curve (or straight line, if applicable). 31. x y 0 25 32. x y 0 250 40 50 80 75 120 100 160 200 240 280 125 150 175 200 50 100 150 200 250 300 50 -150 -350 -550 -750 -950 40 33. Plot the function s’ = fs/(s-f), given f=10 for the following values of s: 30, 20, 15, 10, 5, and 1. What happens at s=10? 34. Plot E (eV) = 1241/λ for values of λ from 400 to 750 in steps of 50. Find the slope and intercept for the following lines: 35. y= 2x + 1 36. y= -2x+3 37. 2y=7x + 6 38. –7y+3x=35 Find the equation of the straight line passing through the given point with the given slope. 39. (-2,-3) with m=-1 40. (2,-5) with m=3.5 41. (2,-2) with m=0 Graph the following data sets on separate graph sheets, draw a smooth line in each case, and then determine the slope and intercept for each data set. 42. Data set: Time (s) Distance (mm) 12 18 30 35 45 58 77 85 90 95 43. Data set: Energy (eV) Depth (nm) 1 2 3 4 5 10 15 20 25 50 1.3 3.2 4.7 5.9 7.0 19 24 29 39 75 4 11 14 19 25 29 36 41 45 51 Group Discussion Questions These require the students to form small groups of three or four and discuss each question. Some questions may require you to do a search on the Internet and find the answers. 1. Why do we have three systems of numbers, decimal, scientific and metric engineering? 2. We have looked at prefixes from 10-18 to 1018. This does not imply that other prefixes do not exist. Using the web, find out the names of prefixes outside this range. Are some of these prefixes used in physics and technology? 3. Why are we not able to plot zero on a logarithmic scale? 4. How do you relate the slope of a straight line (or the slope of the tangent to a curve at a point) to concepts from calculus? 5. Other than using logarithmic axes, how can you plot the graph of say y=x3 so that it appears as a straight line? 6. We have dealt with rectangular coordinate system for plotting graphs. Do other systems exist? What are they? Where are they used? Reference Vasan, S.V. “Technical Mathematics (with Applications in Electronics and Photonics)” Victoria: Trafford Publishing, 2003. 1 41 Chapter 2 Elements of Light Propagation Theories 2.1 Wave Nature of Light 2.2 Particle Nature of Light 2.3 Energy of a Photon 2.4 Photon Characterization 2.5 Photoelectric Effect 2.6 Suggestions for Laboratory Experiments-Use of a Spectrometer Chapter Summary End of Chapter Exercises Group Discussion Questions Chapter Objectives After completing the chapter, the student should be able to (i) Distinguish between wave and particle nature of light (ii) Characterize a photon through its wavelength, frequency, energy and wave number (iii) Perform conversions among wavelength, frequency, wave number and energy (iv) Explain the phenomenon of photoelectric effect. Key words: photon, Planck’s constant, wavelength, frequency, wave number, energy Light is electromagnetic radiation. The word light is used to denote radiation that may be visible to the human eye or invisible. In fact, as we will see later, only a very small portion of the electromagnetic spectrum lies in the visible range. While radiation is not the model used to describe light, we have different models to describe it. 2.1 Wave Nature of Light This theory posits that light travels in waves. Waves can be highly localized, modeled as resulting from the superposition of many sine waves. Then they are termed as wave packets. Waves can also be non-sinusoidal in nature. However, describing the wave as a simple sine wave is easy mathematically and facilitates rough comparison to harmonic waves such as ripples on a pond. If a rock is thrown into the middle of a quiet pond, waves propagate from the source of disturbance and travel outwards to the periphery of the pond. Each wave itself is characterized by a crest (peak) and a trough (ebb or valley). If we throw a cork on top of the waves, it bobs up and down as the wave goes up and down. Thus, at any location on the wave, there is an up and down motion as time passes by. The time it takes for the wave between two successive crest (or trough) appearances is termed the time period2, T. At any instant in time, if one looks at the series of waves from the center of the pond outwards, one would notice a repeating pattern of each wave with many crests and troughs; however, each single wave is identical to its neighbor. The distance at which the wave repeats itself is called the wavelength, denoted by the Greek letter λ (lambda.) Wavelength can be measured from crest to crest, trough to trough or between 42 corresponding points on neighboring (adjacent) waves. On an average, the sine wave is zero since it goes through a positive half-cycle above the reference line and a negative half-cycle below the reference line as shown in Figure 2.1 1,2,3. 0 20 40 60 80 100 1 Distance from starting point Figure 2.1 Illustration of wave motion in space. In Figure 2.1, the wavelength is about 31 units Similarly, one can look at a particular location and plot the up and down movements with respect to time. As an example one might find a graph that looks like Figure 2.21,2,3: 1 0.5 0 0 5 10 15 20 -0.5 -1 Time starting from arbitrary reference Figure 2.2 Illustration of wave motion in time. In Figure 2.2, the time period T is about 5.6 units. 43 What is the relationship between the two graphs, Figure 2.1 and Figure 2.2? One measures displacement of wave with respect to a reference plane as it varies with distance, and the second, as it varies with time. How fast do light waves move? In vacuum, the speed of light is 3 x 108 m/s. This value is the same for all types of light waves, visible or invisible. In one complete cycle, the wave advances a distance λ during a time period T. From early on, we know that Speed = distance divided by time. Using this relationship, we obtain c = λ T where c is the speed of light (in vacuum). When an event occurs often, we say that it happens frequently. If the time between successive occurrences is small, the frequency at which the event happens is high. If we denote the frequency using the Greek letter γ (pronounced “new”), then ν = 1 T T = 1 ν or When T is measured with the base unit of second (s), the reciprocal ν is in reciprocal seconds (s-1), or per second, or cycles per second (cps), or Hertz (Hz). All of these units mean the same thing. Typically, the frequency of light in the visible range is of the order of hundreds of Terahertz (THz.). Also, typically, the time period of visible light waves is in the femtoseconds (fs) regime. Substituting for T in the equation for c, we obtain c = λν or ν = c or λ = λ c ν Only one of these three equations is independent. However, the three forms of equations, particularly the last two, are useful in characterizing light waves1,3. Remember that while we have exaggerated the waves in Figure 2.1 and Figure 2.2, in reality, the wavelength is quite small. For example, visible light occupies a wavelength 44 range of 400-700 nm (1 nm = 1 billionth of a meter.) Wavelengths in the neighborhood of 100-400 nm belong to ultraviolet light. Example 2.1 Calculate the frequency and time period of light emanating from a 1.06 µm wavelength Nd:YAG laser. (Note that this light is not visible to the human eye). Use ν = c λ with c=3 x 10 m/s and λ =1.06 x 10-6 m 8 ν = 3 x 108m/s 1.06 x 10-6 m = 2.83 x 1014 Hz or 283 THz Answer To calculate the time period, use T = 1 ν T = 1 . 2.83 x 1014 Hz = 3.53 x 10-15 s or 3.53 fs Answer Example 2.2 One of the prominent frequencies of an argon-ion laser is about 584 THz. What is the wavelength of light? Is it visible? If so, what may be its approximate color? Use λ = λ = 3 x 108m/s 584 x 1012 Hz = 514 nm Answer c ν 8 with c=3 x 10 m/s and ν =584 x 1012 Hz Since the wavelength lies between 400 and 700 nm, the light must be visible. Answer The 400 nm corresponds to violet and 700 nm to red. In between are blue, green, yellow and orange. We can guess the 514 nm wavelength to be near the green portion of the visible spectrum. Answer 45 Radar TV/radar TV AM radio FM/AM FM radio microwave microwave microwave microwave visible-IR UV- X-rays X-rays Gamma- 1.00E+21 1.00E+18 1.00E+15 1.00E+12 1.00E+09 1.00E+06 1.00E+03 1.00E+00 Gamma- Frequency, Hz The entire electromagnetic spectrum is shown in Figure 2.3 with approximate values of wavelength and frequency1. Notice that the shorter wavelengths are towards the left, and the longer wavelengths are to the right, whereas the shorter frequencies are to the right and larger frequencies to the left. Radar TV/radar TV AM radio FM/AM radio FM radio microwave microwave microwave microwave visible-IR UV- X-rays X-rays Gamma-rays Gamma-rays Light Wave Type 1.00E+05 Wavelength, m 1.00E+03 1.00E+01 1.00E-01 1.00E-03 1.00E-05 1.00E-07 1.00E-09 1.00E-11 Light Wave Type Figure 2.3 The electromagnetic spectrum. In Figure 2.3 2, “UV” represents ultra-violet radiation and “IR” denotes infra-red radiation. What the human eye can see is a very small portion of the electromagnetic spectrum. Light waves that fall anywhere in the spectrum have the same speed of 3 x 108 m/s, a vast majority of them are invisible to us. Note that we use the word “light” loosely to denote both visible and invisible waves. For the sake of clarity, the visible spectrum is shown in Figure 2.4: 46 Wavelength, nm Frequency, THz 750 700 650 600 550 500 450 400 800 750 700 650 600 550 500 450 400 Violet Blue Green Yellow Orange Approximate Color Wavelength Red Frequency Figure 2.4 The electromagnetic spectrum detailing the 400-700nm region. Frequency designations of FM radio stations are in MHz. For example, "107.1" denotes that the station’s frequency is 107.1 MHz. AM radio stations express their frequency in kHz. For example "880" on the AM dial corresponds to 880 kHz frequency. Example 2.3 Characterize an IR light wave with a wavelength of 2 x 10-6 m by its (a) λ in nm, (b) λ in µm, (c) λ in Angstroms (d) ν in its most appropriate metric units. (b) λ = 2 x 10-6 m = 2 µm Answer (a) λ = 2 x 10-6 m = 2 x 10-6 m 1 x 10-9 m per nm = 2000 nm Answer (c) λ = 2 x 10-6 m = 2 x 10-6 m o 1 x 10-10 m per A = 20000 A Answer (d) Use ν c o = λ with c=3 x 10 m/s and λ =2 x 10-6 m ν = 3 x 108 m/s 2 x 10-6 m = 150 THz Answer 8 Exercise 2.1A Complete the following Table. Round off all answers to three significant figures. Express the frequency and time period in the most appropriate metric units. 47 Where required, indicate approximately in what portion of the electromagnetic spectrum the light wave can be found under the column “Wave Type.” Problem 1 2 93.3FM 3 4 5 6 7 8 9 10 11 12 740 AM 13 14 15 16 17 18 19 20 λ (µm) λ ν T 477 ? ? ? ? ? ? ? ? ? 157 ? ? ? 2.2 ? ? ? ? ? ? 0.633 ? ? (Angstrom) ? ? ? 5550 ? ? ? ? 125 ? ? ? ? ? ? ? 333 THz ? ? 1.1 PHz ? ? ? ? ? ? ? ? ? 1 fs 82.7 fs ? ? ? ? ? ? ? ? ? ? ? 1064 ? ? ? ? ? ? 0.193 ? ? ? ? ? ? 4050 ? ? ? ? ? ? 509 THz ? ? ? ? ? ? ? ? ? 5140 750 THz ? ? ? ? ? 0.353 fs 7.22 x 10-14 s ? ? Wave Type ? ? ? ? ? ? ? ? ? ? ? ? λ (nm) 2.2 Particle Nature of Light In this theory, light is considered as “particles.” We know that if we shoot a pool ball, it travels relatively straight along a line unless it encounters an obstacle. A pool ball can be considered to be a large particle. This is a rough comparison between a particle of light and a mechanical particle such as a ball. We do not want to extend the analogy beyond this because a light particle is quantum mechanical in nature. A light particle has a distinct level of energy, and each such individual particle is called a photon. Despite the differences in energy among different types of photons, amazingly they travel at the same speed, designated by the symbol c. As mentioned previously, the value of c is 3 x 108 meters per second1. For all practical purposes, we consider the speed of photons in atmospheric air the same as that in vacuum. In reality, it is slightly smaller. Obviously, when one deals with air at high pressure in a chamber, one can expect the speed of the photons to be slower than in atmospheric air. This is very roughly comparable to a high pressure chamber where particles in the form of oxygen and nitrogen molecules that make up pure air are slowed down due to collisions, etc., compared to regular atmospheric pressure air. 48 Are all photons visible? The answer is no. As shown in Figure 2.3 and Figure 2.4, visible photons occupy only a small portion of the entire electromagnetic spectrum. While all photons which belong to this spectrum travel at the same speed of 3 x 108 m/s, not all are visible. Gamma rays, X-rays, ultra-violet rays, infra-red rays, microwaves, radio and TV radiation are invisible to the human eye. The only photons that the human eye can perceive belong to the colors of the rainbow, which from top to bottom are red, orange, yellow, green, blue, and violet. The formation of rainbow colors through a process known as dispersion will be discussed in a later chapter. Different photons in different part of the electromagnetic spectrum have different energies. For example, a gamma photon, which emanates from stellar radiation, is more energetic than an x-ray photon, which is more energetic than a visible photon, which, in turn, is more energetic than a microwave photon, and so on. A third way of looking at light: Let us take a detour from the particle model. In many cases, one can approximate the behavior of light by considering it to travel in a straight line. This is a good approximation in many instances especially if we deal with relatively large size objects such as a coin, a pencil or a building. This is known as rectilinear propagation of light and will be discussed further in Chapters 5, 10 and 11. A simple illustration is shown in Figure 2.5. Table Light Bulb Object Screen Figure 2.5 Illustration of rectilinear propagation of light. A reasonably well-defined shadow is cast by an object illuminated by a light bulb onto a screen behind the object. Similarly, buildings and trees cast well defined shadows when illuminated by the sun. Rectilinear propagation of light, the phenomenon that light travels in a straight line, is the basis for geometric optics. Using a light source such as a candle or a light bulb, depending upon how close the object is to the source, one could obtain a shadow of a size much larger than the original object itself. This is because light travels in all directions from an isotropic source. While in the previous section, we considered light to be traveling as waves, here we look at light as traveling in straight lines. How can we reconcile the two, almost diametrically opposing, theories? The answer is simple. It depends. If the dimensions that light encounters are much larger than the wavelength of light, the theory of rectilinear propagation of light can be used. One can explain reflection and refraction of light 49 (discussed in detail in later chapters) using the theory and geometric optics. On the other hand, when the dimensions that light has to pass through approach the wavelength of light, one considers its wave nature to explain phenomena such as diffraction (bending of light), interference, etc. (See Chapter 12 for details). Example 2.4 A circular pendant is suspended from the ceiling by a thin wire. The diameter of the pendant is 50 mm. The pendant is located 30 cm from a small light bulb directly in line with it. What is the diameter of the shadow cast by the pendant on a wall 2.5 meters behind the pendant? For the sake of calculations, assume that the wire diameter and bulb dimensions are negligible. X L P W o Y Figure 2.6 Illustration of Example 2.4. L denotes light source, P the side view of the pendant and W the wall. The dashed line represents the optical axis, i.e. a line from the center of the lamp and passing through the center of the pendant and is extended to the wall. Consider two rays of light, both originating from the center of the lamp, one just missing the top of the pendant, and the second just missing the bottom of the pendant. The extent of the shadow on the wall is shown by the line XY, which is the diameter of the shadow. Equating the ratio of distances (to the wall) to the ratio of diameters, we have the following direct proportionality XY = 50 mm LW LP XY = 50 mm 280 cm 30 cm XY = XY = 50 x 280 mm 30 467 mm Answer or or Thus, the shadow of the pendant is over nine times larger in diameter than the pendant! 50 Similar arguments are used to explain the phenomenon of how one could see practically the entire landscape when viewed through a pinhole held close to the pupil of the eye, or, using “trick” photography, the sun (or the Niagara Falls) can be made to appear as though it is being held on someone’s palm! Further discussion of the particle nature of light will be provided in later sections and chapters. The words photon and wave will be interchangeably used in further discussions. 2.3 Energy of a Photon We know that light, visible or invisible, also has energy, as exhibited by phenomena such as photosynthesis in plants, x-ray imaging of human bone and tissue, heating food in a microwave oven, listening to music carried by radio waves, to name a few. It is to be noted that while a photon has no mass, it has energy. The photon energy is given by2,3 or Ep = hν Ep = hc λ where Ep represents the energy of a photon of wavelength λ, or frequency ν. In this equation h is known as Planck’s constant, in honor of the scientist Max Planck. Its value is 6.625 x 10-34 J.s. When the wavelength is expressed in meters or the frequency in Hertz, using this value of h would give us the photon energy in Joules (J.) Joule is a unit of measuring energy. A Joule corresponds very roughly to the energy expended in lifting about 100 grams (3.5 oz) of mass against gravity for a distance of 1 meter (3 ft 3 in ) From these relationships, we see that shorter the wavelength (or higher the frequency), larger is the photon energy. This explains why gamma rays and X-rays possessing relatively short wavelengths, in comparison to visible light or radio waves, are highly energetic. Example 2.5 Calculate the photon energy of the output of a He-Ne laser (λ=632.8 nm) and a CO2 laser (λ=10.6 µm.) Using Ep = hc we have Ep = 6.625 x 10-34 x 3 x 108 632.8 x 10-9 3.14 x 10-19 J for the He-Ne laser photon Answer = Ep = = λ 6.625 x 10-34 x 3 x 108 10.6 x 10-6 1.88 x 10-20 J for the CO2 laser photon Answer 51 We see that the He-Ne laser photon has higher energy than the CO2 laser photon because the wavelength for the former is shorter. Photon energies, expressed in Joules as above, for the UV-visible-IR range is of the order of 10-18 to 10-20 J. A more convenient way of expressing photon energies is in the form of electron volts (eV) units of measurement. An eV is the energy expended in moving an electron through a one-volt potential difference. The relationship between Joules and eV is as follows: 1 eV = 1.602 x 10-19 J This simply results from the fact that an electron has a charge of 1.602 x 10-19 Coulombs (C) and is multiplied by 1 V. In electronics, a Volt multiplied by a Coulomb results in a Joule. Example 2.6 Convert the photon energies in Example 2.5 into eV. For the He-Ne laser photon, Ep = 3.14 x 10-19 J 1.602 x 10-19 J per eV Ep = 1.96 eV Answer For the CO2 laser photon, Ep = 1.88 x 10-20 J 1.602 x 10-19 J per eV Ep = 0.117 eV Answer As you can see photon energies expressed in eV usually range from a few tenths of an eV to a few eV in the UV-visible-IR range of the spectrum, which is of interest to students in photonics. A third way of expressing photon energy exists. Just as the time period and frequency are reciprocals of each other, one can devise a system of units which expresses the reciprocal of the wavelength. For this, the unit of centimeter (cm) is used to express the wavelength. It then poses the question of how many waves exist in a 1 cm distance. This is denoted as the wave number2,4, k. k = 1 λ where λ must be expressed only in cm. 52 As illustrated in Figure 2.7, a greater number of waves from a He-Ne laser can be accommodated in 1 cm distance than waves from a CO2 laser. Note that the figures are not to scale and are shown for purposes of comparison only. 1 cm 1 cm -------------------------- -------------------------- 1 0.5 0 0 2 4 6 8 10 12 14 16 18 20 -0.5 -1 1 0.5 0 0 2 4 6 8 10 12 14 16 18 20 -0.5 λ -1 λ = 632.8 nm = 10.6 µm Figure 2.7 Illustration of the concept of the wave number. The wave number k expressed in cm-1(inverse cm, or per cm, or waves per cm) is directly proportional to the photon energy. Example 2.7 Convert the wavelengths in Example 2.5 into their respective wave numbers. (a) λ = 632.8 nm to be converted first to cm. nm cm -9 10 divided by 10-2 = 10-7 is the conversion factor λ = 632.8 x 10-7 cm k = 1 -----------------632.8 x 10-7 cm = 15,800 cm-1 Answer (b) λ = 10.6 µm to be converted first to cm. µm cm -6 10 divided by 10-2 = 10-4 is the conversion factor λ = 10.6 x 10-4 cm k = 1 ----------------10.6 x 10-4 cm = 943 cm-1 Answer 53 Unlike frequency, the wave number is expressed only in the regular decimal form. The same is true of photon energy in eV. Photon energy in J is always expressed in scientific form. These are some standard conventions that came about when the branch of photonics was carved out from the realm of physicists, chemists and electronics engineers. Clearly, there is direct proportionality between photon energy and wave number. Therefore, Table 2.1 was created summarizing the conversions. To convert FROM E(J) E(eV) E(J) k(cm-1) E(eV) k(cm-1) TO E(eV) E(J) k(cm-1) E(J) k(cm-1) E(eV) Table 2.1 Multiply Divide E(J) by 1.602 x 10-19 1.602 x 10-19 5.03 x 1022 5.03 x 1022 8060 8060 E(eV) E(J) k(cm-1) E(eV) k(cm-1) Exercise 2.3A Complete the following Table: Problem 1 2 93.3FM 3 4 5 6 7 8 9 10 11 12 740 AM 13 14 15 16 17 18 19 20 477 ? ? ? ? ? ? ? ? ? 193 ? ν ? ? ? ? 333 THz ? ? 1.1 PHz ? ? ? ? E(J) ? ? ? ? ? 5 x 10-19 7 x 10-20 ? ? ? ? ? E(eV) ? ? 2.2 ? ? ? ? ? ? 0.633 ? ? k(cm-1) ? ? ? 15000 ? ? ? ? 12500 ? ? ? 1064 ? ? ? ? ? ? ? ? ? ? 509 THz ? ? 750 THz ? ? ? ? ? 1.5 x 10-19 2 x 10-18 ? ? ? 5.5 ? ? ? ? ? ? ? ? 4050 ? ? ? ? 5140 λ (nm) 54 Shown in Figure 2.8 is a graph that spans over a wide range of wavelengths, from gamma rays to radio waves and the corresponding photon energy in eV. Figure 2.8 is drawn on a log-log scale to cover this large range. Photon Energy (eV) 100 1 1.00E-10 1.00E-07 1.00E-04 1.00E-01 1.00E+021.00E+05 0.01 0.0001 0.000001 0.00000001 Wavelength (m) Figure 2.8 Wavelength versus photon energy. The enlarged view of the UV-visible-IR region is shown in Figure 2.9 for clarity. Wavelength (m) 1.0E-07 1.0E-06 1.0E-05 1.0E-04 1.0E+01 Photon Energy (eV) 1.0E+00 1.0E-01 Figure 2.9 Wavelength versus photon energy. Similarly, a plot of wave number against wavelength for the same region is shown in Figure 2.10. Thus far we have characterized a photon in terms of its energy. Is energy the most important parameter, or is it the rate at which energy is transferred more important? Both are equally important. The rate at which energy is transferred is referred to as power1. 55 Wavelength (m) 1.0E-07 1.0E-06 1.0E-05 1.0E-04 100000 - 1 10000 Wave Number (k) 1000 100 Figure 2.10 Wavelength versus wave number (cm-1). Power = or P Energy Time = E t where P represents power, E the energy, and t the time of energy transfer. We know that the fundamental measurement unit in the metric system for energy is Joule. The fundamental unit of measuring time is second (s.) Based on these units, the measuring unit for power is Watt (W.) One watt of power is expended if 1 Joule of energy is transferred in 1 second. Thus, a Watt is a Joule per second. We can relate to Watts in terms of wattage of light bulbs, for example 60 W, 75 W, 150 W, etc. Higher the power (or wattage), brighter is the light bulb. We know, from photon energy calculations, that a photon by itself has very little energy. Thus, a collection of photons is required for a beam to have sufficient power. Let us perform some calculations. Example 2.8 Calculate the number of photons that need to be emitted per second in the beam of a 200 mW power argon ion laser giving out 514 nm wavelength radiation. We know that P = E/t or E = P.t Given P=200 mW and t=1 s, we obtain E = 200 mJ Obviously we need many photons for the individual photon energies to add up to 200 mJ. Let us say we need n number of photons. 56 We calculate the individual photon energy as Ep -19 3.87 x 10 = n = hc λ = 3.87 x 10-19 J after substituting for the known quantities. 200 x 10-3 Solving for n yields, n=5.17 x 1017 photons Thus, the laser should emit 5.17 x 1017 photons/ second to maintain 200 mW of power Exercise 2.3B 1. A He-Ne laser outputs 6.2 x 1015 photons per second of wavelength 632.8 nm? What is the beam power? 2. An excimer laser gives out pulses each of which lasts 14 ns. If each pulse has 200 mJ of energy, what is the pulse power? 3. How long does one have to expose a tissue to result in 1.5 mJ of energy transferred when exposed to a 750 µW power laser beam? 4. How many photons are emitted in a minute from the transmitting tower of a 1 kW power radio station with code letters KPHO93.3? 5. A source emits 200 eV X-ray photons. What is the wavelength? 6. A material requires 50 mJ of energy for each square centimeter of surface area. If it receives 7.5 mW of power for each square centimeter of surface area, how long should the exposure be? 2.4 Photon Characterization This section summarizes some of the previous sections in terms of characterizing a photon using its wavelength, frequency, time period, energy and wave number. Given that the speed of the photon is the same in vacuum no matter what location of the electromagnetic spectrum the photon is in, if we know any one of the parameters λ, ν, Ep or k then all others can be calculated easily. As will be discussed in the chapter on lasers, a photon may be absorbed by an atom, resulting in the loss of the photon, and the atom moves to a higher energy level. Or, if an atom at a higher energy level comes down to a lower energy level without an external stimulus, then it can emit a photon whose energy corresponds to the loss of energy for the atom. This phenomenon is called spontaneous emission and accounts for much of the light we receive from conventional light sources such as an incandescent lamp or a mercury vapor lamp. A number of different wavelengths make up the output spectra of such lamps. As a result, the light appears as white, which is a combination of the different colored photons plus invisible photons in the UV and IR regions. 57 Exercise 2.4A In each of the following cases, a parameter characterizing the photon is given. Calculate the remaining parameters. Your answer should include wavelength, time period and frequency expressed in the most appropriate metric prefix unit, photon energy (J) in scientific form and wave number and photon energy (eV) in regular decimal form. 1. λ = 589 nm 2. ν = 355 THz 3. T= 4 fs 4. k=15600 cm-1 5. Ep=2.2 x 10-19 J 6. Ep=3.33 eV 7. λ = 1.06 µm 8. ν = 680 THz 9. T= 5.2 fs 10. k=1100 cm-1 11. Ep=1.5 x 10-20 J 12. Ep=15 eV 2.5 Photoelectric Effect When UV light shines on certain metals, it results in electric current flow in that material. While this phenomenon has been known for many years, it was not until 1905 that Einstein provided an explanation. The existing wave theory of light was unable to explain the photoelectric effect since increasing the intensity of light did not increase the energy of the electron emitted; rather, it increased the number of electrons emitted. However, Einstein explained that light travels in packets of energy called photons. As the frequency of the photon was increased, the energy of the electron increased. As we know, a UV photon has relatively larger energy than a visible photon. For a single photon absorption process, a certain minimum amount of energy is required before current flow occurs. This minimum depends on the material and is called the work function or threshold energy4. If the photon energy exceeds the threshold energy value, then current flow results. Photoelectrons are created by this process and the phenomenon is called photoelectric effect. Clearly, there is a cut off frequency, νc, above which current flow will occur, with is a corresponding cut-off wavelength, λc. Solar cells use the photoelectric effect principle. However, if the beam is powerful enough, there will be electron flow due to multi-photon absorption processes in which case a wavelength longer than λc may be used for illumination. Example 2.9 A certain material has threshold energy of 3.48 eV. What is the minimum wavelength of radiation required to illuminate the material before current flow starts? If the wavelength of light corresponds exactly to 3.48 eV of photon energy, this is where the photoelectric effect starts happening. 58 3.48 eV = 5.57 x 10-19 J hc Ep = 6.625 x 10-34 x 3 x 108 = 357 nm Answer 5.57 x 10-19 Unless radiation of 357 nm wavelength or lower is used, there will be no photo electricity using this material. If we use violet light at 400 nm, there will be no electron flow in the material unless the beam is powerful enough. An illustration is shown in Figure 2.11. Ep λ = = UV light Ep=hν Load Ammeter Current flow Figure 2.11 Illustration of photoelectric effect. What if one uses photons of wavelength much shorter than that required to initiate photoelectron flow? The energy over and above the threshold will be carried by the photoelectrons exhibiting larger voltage across the load. How would one increase the current flow? The answer is to increase the number of photons hitting the target every second. Thus, theoretically one could obtain a moderate flow of current at a reasonable voltage, provided the appropriate wavelength of light at an appropriate power level is used to illuminate the target. 2.6 Suggestions for Laboratory Experiments In this chapter, we have learned that the electromagnetic spectrum encompasses a large range of wavelengths (and corresponding frequencies.) A spectrometer enables determination of the wavelengths present3. A visible spectrometer has graduations from around 400 nm to 750 nm in steps of 10 nm. A diffraction grating resolves the incident light into its different wavelength components as shown in Table 2.2. Table 2.2 |-i-i-i-i-|-i-i-i-i-|-i-i-i-i-|-i-i-i-i-|-i-i-i-i-|-i-i-i-i-|-i-i-i-i-| 400 450 500 550 600 650 700 750 By viewing through the eyepiece of the spectrometer and adjusting the slit that allows a portion of the light to enter the spectrometer, one can view spectral lines of the incoming 59 radiation. As an example, a few sample lines are shown above. In the above sample spectrum, four lines, one each at 425 nm, 510 nm, 575 nm and 690 nm are visible. In many cases, the lines will not be as sharp and well-defined as shown here; however, they appear as a wide band encompassing one or two colors. The visible spectrometer should face the light source. By adjusting the height of the spectrometer, the slit opening and the height of the light source, a clear spectrum can be obtained. Different light sources may be viewed through the spectrometer including the fluorescent lamp, an incandescent lamp, a candle and several plasma tubes (example argon, neon, helium nitrogen, xenon, carbon dioxide, etc.) Each photon corresponding to an intense line of the spectrum may be characterized using the methods described in section 2.4. • • • • • • • • Chapter Summary A photon is a light particle with a distinct energy given by hν or hc/λ Light propagation may also be visualized in wave form, each photon possessing a distinct wavelength and frequency. The relationship between frequency and wavelength is given by ν= c/λ The photon energy may be characterized in Joules or eV. The photon energy may also be characterized by the wave number, k, always expressed in cm-1 units only. All electromagnetic radiation travels at the same speed of 3 x 108 m/s in vacuum. When UV light strikes certain metal surfaces, the excess energy above and beyond a minimum threshold is converted into electrical energy and appears as electric current. This is photoelectric effect. Conventional light sources emit a variety of wavelengths of light, which can be analyzed by a spectrometer. End of Chapter Exercises Complete the following table. Express E (J) using scientific notation and E (eV) and k in regular decimal form. Express λ in the unit indicated. Express in ν metric form. (“A” represents Angstrom units under the columnλ) ν k(cm-1) E(J) E(eV) λ 1. 356 nm 2. nm 350 THz 9,600 3. µm 4. nm 3.33 x 10-19 5. nm 2.5 6. A 1.65 -19 1.75 x 10 7. µm 8. A 14,400 9. 589 nm 10 nm 475 THz 11. 6500 A 12. 2.2 µm 60 13. A 95.5 FM station broadcasts sending out 1 x 1029 photons every 6 seconds. What is its power? 14. How may photons per second emanate from a 2 mW He-Ne laser of wavelength 632.8 nm wavelength? 15. If it takes 1 hour for light from Star A to reach earth, how far is Star A from the earth in km? 16. Gold has a threshold for photoelectron emission at 4.78 eV. What is the minimum wavelength of radiation required to illuminate a gold target so that photoelectric effect can be seen? 17. If 375 nm wavelength light is used to irradiate a gold target, would it result in photoelectric current flow? Show by calculations. Group Discussion Questions These require the students to form small groups of three or four and discuss each question. Some questions may require you to do a search on the Internet and find the answers. 1. To simplify the math, assume that the light wave is described by a simple sine function. What is its equation then? Explain what each term stands for. 2. How is the intensity of a wave related to its amplitude? 3. Convert the speed of light into miles per hour. 4. What wavelengths emanating from the sun reach the earth’s surface? 5. Explain how elements of the earth’s outer atmosphere protect us from harmful radiation from the sun? 6. Describe how a typical solar cell is constructed and how it works. References Sears, F.W, M. W. Zemanksy and H.D. Young. “College Physics.” Reading: AddisonWesley Publishing Company, 1991. 2 Jenkins,F. and H. White. “Fundamentals of Optics.” Singapore: McGraw Hill, 1981. 3 Welford, T. “Optics.” Oxford: Oxford University Press, 1988. 4 Glasstone, S. and D. Lewis. “Elements of Physical Chemistry.” London: MacMillan & Co, 1970. 1 61 Chapter 3 Photonics Safety 3.1 Laser Classifications 3.2 Basics of the Human Eye 3.3 Power and Irradiance Calculations 3.4 Safety Precautions to be Observed During Laser Operation 3.5 Electrical Safety, Electrostatic Discharge 3.6 Fiber Handling Safety 3.7 Optical Power Meters 3.8 Laser Safety Laboratory Chapter Summary End of Chapter Exercises Group Discussion Questions Chapter Objectives After completing the chapter the student should be able to (i) Define laser classifications based on levels of hazards posed (ii) Explain which human tissues are susceptible to laser radiation damage (iii) Calculate beam irradiance (iv) Operate a laser safely (v) Explain the additional hazards associated with Class IV lasers (vi) Explain electrical hazards present around lasers (vii)Handle glass fibers safely Key words: retina, cornea, irradiance, thermal damage, laser power, laser classification 3.1 Laser Classifications In the next five sections, we will discuss safety aspects of dealing with a laser. The details of the laser operation will be presented in Chapter 4. It is sufficient to know the laser basics in this paragraph, for the moment. LASER stands for Light Amplification by Stimulated Emission of Radiation. In essence, laser output is an intense beam of light, highly directional in nature and can cause more damage to human tissue than ordinary light. In the next two sections, we will define clearly which human tissue is susceptible to damage because of laser exposure. The wavelength of the radiation is an important consideration. For continuous wave (CW) lasers, the average power and the time of exposure are critical. These are also true for pulsed lasers with a high pulse repetition frequency (PRF). For pulsed lasers, the pulse energy, the pulse width (duration of pulse) and PRF need to be considered. The American National Standards Institute1 (ANSI) presents guidelines for use of lasers in standard ANSI Z136.1-1993 (also 2000, and later versions). Accordingly, lasers are classified into four main groups (classes) based on the hazards posed. • Class 1 lasers are low-power lasers. They do not normally pose a hazard. Therefore, they are also known as exempt lasers. A laser printer is an example. 62 • • • Class 2 lasers are also low-power visible lasers or laser systems. They cannot cause eye damage normally unless they are viewed directly for an extended duration (1000 seconds.) A bar code scanner is an example. Class 3 lasers are medium-power lasers and laser systems. They can cause eye damage with short duration exposures (<0.25 s) to the direct or reflected beam off shiny surfaces. Class 3 lasers are further subdivided into Class 3a and Class 3b. Class 3a lasers normally do not present a hazard if viewed for a very short time with an unprotected eye, but may present a hazard if the light is collected and viewed using collecting optics, such as telescope or fibers. Class 3b lasers can pose a hazard if the beam or its reflections from shiny surfaces are viewed directly. (The ANSI 2000 standard redefined Class 3a to include visible lasers only and Class 3b to include non-visible lasers.) Class 4 denotes high-power lasers and laser systems. These may cause severe eye injury with short duration exposure (<0.25 s) to the beam or reflections from either shiny or non-shiny surfaces. They may also cause severe skin damage and pose a fire hazard. Mirrors and polished metal surfaces (i.e. shiny surfaces) reflect much of the light incident upon them, following the laws of reflection presented in a later chapter. This type of reflection is called specular (meaning mirror-like) reflection, shown in Figure 3.1. Mirror (or shiny surface) Original laser beam Reflected beam Eye Figure 3.1 Illustration of specular reflection. The reflecting surface only changes the direction of the beam; it does very little to decrease the power. If the human eye is in the path of the reflected beam, depending upon the power of the laser, the damage can be as severe as viewing the original beam directly. The latter is referred to as intra-beam viewing. Intra-beam exposure results if the eye or skin is directly in the path of the laser beam. Thus, lasers in classes 2, 3 and 4 have potential for damage, to various degrees, because of intra-beam viewing, whereas lasers in classes 3 and 4 pose hazards with specular reflections as well. Additionally, class 4 lasers pose the hazard with reflections off non-shiny surfaces such as a paper or paint on a wall. This type of reflection is called diffuse reflection, as illustrated below. In Figure 3.2, the original (or incident) laser beam is striking the paper surface as shown by the “up” arrows. Depending upon how curved the surface is at the point of contact, the reflected beam comes off in all different directions. In general, since diffuse reflectors spread the reflected light in many different directions, the portion of light that reaches the eye from diffuse reflectors is much smaller than from specular reflectors. 63 Exaggerated view of the surface of paper Figure 3.2 Illustration of diffuse reflection. Also, the hazards from diffuse reflectors are not significant with lasers of Class 3 or lower, however, they can pose danger with Class 4 lasers. Additionally, Class 4 lasers can pose a fire safety hazard, due to ignition potential of surfaces that the beam irradiates. Additionally Class 4 lasers can cause skin damage. The output power of various class lasers is shown in Table 3.1 below1: Table 3.1 Laser Class Power Class 1 below 0.4 µW Class 2 below 1 mW Class 3a* 1 mW – 5 mW Class 3b* 5 mW – 500 mW Class 4 above 500 mW *The ANSI standard published in year 2000 classified lasers in the 1mW-500 mW range to be Class III, with III(a) for visible lasers and III(b) for non-visible lasers. The 1993 classification labels are widely used still and therefore mentioned in the Table. The 1993 classification using laser power is based on visible lasers such helium-neon or argon ion. The latest ANSI standard strives to harmonize with the International Electrotechnical Commission Standard IEC 60825-1, the latter itself being very similar to European Standard EN 60825-11b. 3.2 Basics of the Human Eye While this is not a text book on human physiology, it is instructive to learn about the basic construction of the human eye so that elements of laser safety can be better appreciated. Only the bare elements of the eye that need to be considered in terms of potential hazards to each are presented here skipping the others. A cross section of the human eye illustrating a few elements is shown in Figure 3.3 2,3. The iris is an adjustable “shutter” for the human eye. Its function is to allow varying amounts of light into the eye. During day time, when the illumination level outside is usually high, the iris is practically closed allowing only a minimum amount of light through (just enough for us to see.) On the other hand, during night time, or when the illumination level is low, the iris is nearly fully open to its maximum extent of 8 mm in diameter. 64 Retina Cornea Iris Eye Ball Fovea Optic Nerve Lens Figure 3.3 Cross section of the human eye (only certain elements are shown). When one is driving at night on an undivided two-lane highway, the iris is nearly fully open. When the headlight of an oncoming car approaches, the eye receives elevated levels of illumination. The iris responds by restricting the size of the pupil to about 2 mm in diameter. This process is slower than the closing of a shutter in a digital electronic device. When the oncoming car has gone by, the level of illumination is back to the low level. The iris responds by opening the pupil size back to the nearly 8 mm in diameter. This process takes up to seven seconds. We refer to this phenomenon as the “blinding effect.” Factors such as fatigue and impairment/intoxication, due to alcohol and/or other drugs, may slow down the time it takes for the iris to open fully. Depending upon the iris opening, differing amounts of light then pass through the lens. In conjunction with the cornea, the lens focuses the light received into a sharp image which is cast on the “screen” located on the back part of the eye, called the retina. The retina is light- sensitive. The most light-sensitive part of the retina, directly behind the lens, is called the fovea. One can imagine that if very high levels of illumination are focused through the lens onto the retina for extended periods of time, temporary or permanent damage to the retina may occur. A case in point is looking at the sun directly for a few seconds without eye protection. It may take a few seconds to a few minutes before normal vision is restored. The damage in this case is often temporary. The outer transparent layer that surrounds the eye ball is called the cornea. Inside the cornea is a darker brown layer called choroid (not shown in the figure above) whose function is to minimize internal reflection within the eye ball by selective absorption of different colors of light2,3. The optic nerve conveys the message to the brain so that the brain can make sense of the image captured by the lens and cast on the retina. The optic nerve itself lacks light sensitive elements. 65 What types of wavelengths affect which portions of the eye? Table 3.2 illustrates the effects3,4: Table 3.2 Region of spectrum Wavelength Tissue that may be damaged UV (UV-C 200-280 nm Cornea causing cataract and UV-B 280-315 nm keratitits; damage caused by UV-A) 315-400 nm photochemical processes Visible 400-700 nm Retina damaged by photochemical and thermal means Near IR called IR-A 700-1400 nm Retina burn, cataract Mid-IR (IR-B 1400-3000 nm Corneal burn, cataract Far IR IR-C) 3000nm – 1mm Corneal burn If the power density is below the threshold for retinal damage due to thermal burn, provided that the retina has sufficient time between exposures to cool off, the damage is not cumulative. On the other hand, if the damage is caused by photochemical processes such as with UV lasers, repeated exposures will be cumulative. In terms of effects on the human skin2, the UV lasers, in general, can cause skin cancer (UV-C) and skin burn (UV-A). Visible lasers can cause skin burn and pigments to darken, and IR lasers can cause skin burn. The cornea is transparent to visible and near infra-red wavelengths. Consequently, much of the light passes through with minimum damage to the cornea. The retina receives the focused light and can be damaged. Figure 3.4 illustrates formation of image on the retina2,5. Note that in Figure 3.3 and 3.4, light is shown as traveling from right to left. This is an exception to the rule that we will predominantly use elsewhere, which states that light travels from left to right for purposes of modeling and analysis. Portion of the retina Lens Object Image Figure 3.4 Illustration of image formation on the retina due to a positive lens. This is an example of how an object, such as a building, is viewed by the lens of the eye. The lens converges rays of light from the object and forms a small image on the retina. The object, such as a building illuminated by the sun, is an extended source of reflected light, i.e., each point in the building acts as a small, single source of light. One can imagine that if the source is concentrated to a small parallel beam such as that from a laser, the image formed on the retina will be focused to a tiny spot by the lens. As will be 66 illustrated in the next section, even if the laser power is a few milliwatts, the concentration of energy on the retina will be so great that it can lead to thermal damage of the retinal tissues. This is from intra-beam viewing2. Also, when one is using a lens to focus a laser beam, the lens surfaces will reflect a portion of the beam back. These are known as Fresnel reflections. Fresnel reflections coming off a concave surface converge to a point. Care must be exercised not to place the eye in the path of these Fresnel reflections. Remember that even a flat piece of glass can reflect about 4% of the light back through Fresnel reflection for each surface for near-normal incidence. On the other hand, the cornea absorbs the ultra-violet and infrared regions of the spectrum readily, so that less light reaches the retina to cause any damage. In this case, the damage is restricted to the cornea2. We are limiting our discussion to lasers in the UV to IR wavelength. Lasers exist at other wavelengths as well. For example, X-ray lasers have wavelengths of the order of a few nm to few tens of nanometers in wavelength. They can cause damage to tissue and sometimes DNA, leading to cancer. X-rays are so powerful that they can create a plasma (mixture of atoms, radicals and ions). Additionally, high powered lasers, when focused on a target, can create a plasma, which in turn can emit radiation. High voltage electrical equipment associated with the laser, especially those operating at 15 kV or higher can themselves be a source of X-ray radiation2. In terms of laser classifications3,5, Class 1 lasers are considered exempt since normally they do not produce radiation levels that are damaging to the eye. A laser printer is an example. Class 2 lasers are such that the normal human defense response, such as turning away of the head in response to accidental exposure to the beam, is adequate. These lasers are limited to 1 mW power. If the beam is viewed directly for a long period of time, Class 2 lasers can pose a hazard. With Class 3 lasers, even the defensive mechanism ceases to protect the eye. 3.3 Power and Irradiance Calculations Power, as defined earlier, is the rate at which work is done, or rate at which energy is transferred. The actual energy coming from a laser source or a light bulb is less important compared to the rate at which energy can be transferred. Laser classifications group lasers according to power. Irradiance, also known as power density or energy flux, is defined as the power divided by area6. The area itself may correspond to the area of cross section of the beam. For many of the lasers that we use, such as the helium-neon or argon-ion, we approximate the beam to be a circle. In many of these beams, the power itself is not uniform across the entire beam; rather, it is highest in the center and drops off drastically when one moves outwards. When one looks at the beam pattern of a typical lower power visible laser on a wall or piece of white paper, it is difficult to estimate the diameter of the beam. About 86% of the energy is located within a certain circle. The human eye cannot perceive the energy outside of this circle since the balance of only about 14% is located outside it. 67 Assuming that one is able to estimate the diameter of this circle, one can calculate the irradiance of the laser beam. I = P A where I is the irradiance (power density or energy flux) measured in W/cm2 or sometimes mW/cm2, P is the laser beam power (mW or W) and A the area of cross section. For a beam of circular cross section, the area A is given by πd2 4 where d is the diameter of the beam. This is derived from the familiar equation for area that we know from middle school, namely, A= πr2 where r is the radius of the circle. In practice, it is always easier to measure a diameter than a radius because it is harder to know where exactly the center of a circle is and then to measure the radius starting from the center. A = Using these two relationships, one can calculate the beam irradiance easily as shown in the following examples. A typical laser beam entering the eye can be focused down to a spot size of 10 to 20 µm in diameter5, sometimes as small as 7 µm. Example 3.1 Using the average of 15 µm for the spot diameter, compare the irradiance on the retina for a fully dilated pupil of 8 mm opening for a 1 mW power laser. With the pupil fully dilated, the entire laser beam (up to a diameter of 8 mm) can enter the lens and be focused on the retina. Using d=15 µm = 15 x 10-4 cm (since 1 µm = 10-4 cm), A = A = A = Using I = I = = πd2 4 π(15 x 10-4 cm )2 4 1.77 x 10-6 cm2 P A 1 x 10-3W 1.77 x 10-6 cm2 566 W/cm2 Answer 68 Compared to this, the retinal irradiance, when one looks directly at the sun at noon5 is about 10 W/cm2. Thus, the retina receives a light intensity from a 1 mW laser about fifty-seven times larger than from the sun. Example 3.2 For the same laser beam as in Example 3.1, if the raw beam diameter is estimated to be 2 mm, what is the beam irradiance as it strikes the lens? A = A = A = Using I = I = = πd2 4 π(0.2 cm )2 4 0.0314 cm2 P A 1 x 10-3W 0.0314 cm2 0.0318 W/cm2 (31.8 mW/cm2)Answer Comparison of the irradiances from these two examples, the irradiance or power density of the focused laser beam on the retina is about 18,000 times higher! Clearly with laser powers higher than 1 mW, the retinal irradiances can be correspondingly higher. You may be wondering why nothing drastic results, other than minor discomfort, when you stare at a 100 W light bulb for a few minutes. The major differences, which will be explained in detail in the next chapter, are due to the differences between laser light and ordinary light. The laser beam is a concentrated beam of energy, with photons of the same wavelength, traveling in the same direction and traveling in unison (i.e., in phase), which makes laser light very different from regular light. The bulb, on the other hand, lacks any of these properties. Besides, the light from the bulb is distributed in all directions (4π steradians of solid angle), resulting in a much smaller irradiance reaching the cornea to begin with. If the laser is continuous wave (CW), it emits radiation continuously and therefore its power is constant. However, if it is a pulsed laser, the laser energy is delivered in pulses, each pulse lasting from a few hundreds of femtoseconds to milliseconds duration. How often the pulses occur is determined by the pulse repetition frequency7 (PRF.) For example, a PRF of 10 Hz implies that ten pulses are emitted every second. If let us say that the pulse duration (often referred to as pulse width) is 15 ns, it means that the pulse lasts only 15 ns. One can calculate the instantaneous power of a pulse or the average power during a one second period for pulsed lasers. 69 (Peak) Power of a pulse = Energy of a pulse Pulse width A pulsed laser with a high PRF may result in poor heat dissipation from thermal burns of eye tissue and can cause thermal damage when compared to a pulsed laser with a lower PRF2. Average power of a pulsed laser = Energy of all the pulses emitted in one second combined = (Energy of an individual pulse) x PRF In any case, no matter if the laser is CW or pulsed and what class it belongs to, common sense tells us never to point a laser directly into anyone’s eyes. Example 3.1 illustrated the situation for a converging beam. Example 3.2 considered a raw laser beam (i.e., neither converging or diverging). Sometimes one expands the laser beam using a beam expander or a collimator. This results in the final beam diameter to be much larger than the original beam. The irradiance in this case drops off drastically, as illustrated in Example 3.3. Example 3.3 If the laser beam of Example 3.1 is expanded to 1 cm in diameter, what is the irradiance? πd2 4 π(1 cm )2 4 A = A = A = 0.785 cm2 Using I = I = P A 1 x 10-3W 0.785 cm2 = 1.27 x 10-3 W/cm2 or 1.27 mW/cm2 Answer Diffuse reflections off surfaces such as paper or paint on the wall resulting from a low power laser essentially are similar to an expanded beam, with photons now radiating in all directions. In this case, the portion of diffusely reflected light passing through a fully dilated pupil is small enough that the focused irradiance on the retina may not be hazardous. This is not true with Class IV lasers where diffuse reflections can be hazardous. 70 Example 3.4 A pulsed laser possesses 150 mJ for each pulse. If the pulse duration is 10 ns, what is the peak pulse power? If the PRF is 20 Hz, what is the average power? Power of a pulse = Energy of a pulse Pulse width Power of a pulse = 150 x 10-3 J = 15 MW Answer 10 x 10-9 s Yes, it is 15 Megawatts for each pulse! This instantaneous power is very high and hence pulsed lasers are used in applications such as ablation, etching, etc. Average power of a pulsed laser = Energy of the pulses in one second combined = 20 x 150 x 10-3 J/s = 3 W Answer Quite often, the technician is required to convert from one set of irradiance units to another. Example 3.5 provides such an exercise. Example 3.5 Convert 1.33 x 10-4 W/cm2 to µW/mm2. Use the same conversion principle discussed in Chapter 1. The conversion factor is obtained by writing out the prefix values and squaring them, as needed. W/cm2 ----------------- µW/mm2 1/(10-2)2 divided by 10-6/(10-3)2 1/10-4 divided by 10-6/10-6 or 104 divided by 1 which is 104 (This conversion factor must multiply 1.33 x 10-4 to get the answer in the final units required). 1.33 x 10-4 W/cm2 = 1.33 x 10-4 x 104 = 1.33 µW/mm2. Answer Exercise 3.3A 1. The diameter of a 5 mW laser beam is approximately 3 mm. If it illuminates a target, what is the irradiance? 2. A 2mm beam has 7 W of power. What is the irradiance? 3. An excimer laser puts out 14 ns pulses of 200 mJ energy each. If the PRF is 10 Hz, and the rectangular cross-sectional beam itself is 25 mm x 16 mm in dimension, calculate (a) peak power, (b) peak irradiance and (c) average power. 4. If the laser beam in problem 1 is expanded to a 2 cm diameter, what is the irradiance? 5. The laser in problem 1 is focused using converging optics to a spot of diameter 20 µm. What is the irradiance of the focused spot? 71 6. A 2.5 mW diverging laser beam of diameter 22 mm is strikes a detector connected to a power meter. If the active area of the detector is 1 cm2, what would the power meter read in mW? 7. A 2 W power laser beam is focused to a spot size of 100 µm in diameter. What is the irradiance? 8. If the laser beam of problem 7 hits a target for 5 seconds, what is the energy density in J/cm2? (Hint: Energy = Power x time.) 9. Convert 1.23 W/mm2 to mW/cm2. 10. Convert 0.00256 W/cm2 to µW/m2. 11. Convert 850 W/m2 to µW/cm2. 12. Convert 5.5 x 10-5 mW/cm2 to W/µm2. 3.4 Safety Precautions to be Observed During Laser Operation2,3 The ANSI standard1 defines Maximum Permissible Exposure (MPE) for different wavelengths of laser radiation. When one is building an optical system with collecting optics, one must be cognizant of the MPE and provide shielding to keep the radiation level below the MPE. The same is true in general of the radiation level in the laser area. Quite frequently, a student in photonics or laser/electro-optics is required to use a Class 3 helium-neon (He:Ne) laser for conducting experiments or for purposes of aligning another laser. The He:Ne used typically emits red light at 632.8 nm wavelength. Much of the precautions are common sense in nature. However, they must be observed for the safety of the operator and those around. All of the following precautions must be followed for a Class 4 (high power) laser; those subsections more pertinent only to high power lasers are indicated by an asterisk (*) following the precautionary statement. 1. One should not look directly into the beam, at the beam, or at specular or Fresnel reflections, no matter what class laser it is. 2. The laser beam path should be below the eye level. All chairs or stools that place the operator’s eye level at or below the beam level should be removed from the laser area. The room illumination should be kept purposefully high to minimize the human pupil size. 3. The laser beam must be terminated at a location immediately after it has served its useful purpose. The beam should never be directed at entrances or windows or into aisle ways. 72 4. The laser should be secured firmly to minimize gross movements of the beam. Relative motion between the beam, the target or intervening optics must be minimized, and, if warranted, must be slow and deliberate. 5. Warning signs must be posted at each entrance to the laser area, which contains a cautionary message, along with laser power, laser class and hazard – all conforming to ANSI specifications. 6. Only personnel trained in the use of lasers and knowledgeable about the hazards should operate them. Traffic must be minimized and access to laser operating areas controlled and restricted (*). 7. The laser area should have a warning (red) light outside the entrance to make personnel outside of the laboratory aware when a laser is in operation. With higher power lasers, an interlock system should be built in that shuts down the power to the laser in less than fifteen seconds when the laboratory entrance is opened. 8. Accidents must be immediately reported to medical personnel, seeking an ophthalmologist’s help in case of eye exposure to radiation. 9. Electrical safety is as important, if not more so than radiation safety. It has been reported that more fatalities happen yearly due to electrical accidents with laser systems than from radiation exposure. Depending upon the type of the laser, high voltages and/or currents are present in the power supplies that drive the laser. One must be extremely careful when working on a laser with its case open. Precautions noted in section 3.5 must be followed. 10. Operators should wear appropriate Personal Protective Equipment (PPE.) Depending on the type of laser, its power and class, this includes, but is not limited to safety glasses, hearing protection and breathing masks. With high power lasers, wearing of safety glasses is mandatory since even diffuse reflection may be hazardous. Use safety goggles suitable for the wavelength and power of the laser. Remember if you can still see the beam through the safety glasses, you are probably not safe! With high power lasers and some pulsed lasers, hazardous chemicals may emanate from target surfaces upon irradiation2,4, and, therefore, the need for breathing protection. With some pulsed lasers, the popping sound associated with the pulse impingement on the target is sometimes loud(*). 11. Avoid wearing jewelry or neckties. Ensure that a body part does not accidentally traverse the laser beam(*). 12. If at all possible, enclose the beam(*). 13. With high power lasers, do not view diffuse reflections (*). 73 These precautions also hold for pulsed lasers. When the laser is used for cutting operations, one must be cognizant of the by products of the reactions and wear breathing protection, if warranted. Some of the contaminants may include cyanides and aldehydes from some materials, benzene from PVC cutting, and some heavy metals from etching reactions2,4. In addition, one must watch for potential contaminants coming off of lenses and other optics when high energies are used. In addition to breathing masks, one must use a good exhaust process as well as keeping personnel far away from the source of contamination. Personnel should consult Materials Safety Data Sheet (MSDS) of the chemicals involved to understand the exposure limits allowed by the Occupational Safety and Health Administration (OSHA.) With gas lasers, one must be aware of the hazards associated with high pressure gases stored in cylinders. Some of the gases used for an excimer laser operation such as fluorine, chlorine and bromine are toxic, and it is important that they are bought from the manufacturer buffered with an inert gas. Also, cylinders must be labeled and secured properly. Gas lines must be properly labeled and color coded. Different kinds of gases such as flammables, oxidizers, inerts and toxics must be stored separately. It must be ensured that the regulators do not release gas into the atmosphere2. Isolation valves which will help during purge, connect, or disconnect of gases should be provided. Various dyes are used in the operation of dye lasers. One must exercise caution when handling dyes during preparation. Use fume hoods for such purposes. Additionally, when a company or a university has several different lasers with different hazards and classes, it is recommended and sometimes it is mandatory that they hire a Laser Safety Officer (LSO). The LSO is a qualified individual who performs laser safety training, administer a safety program for the company and conducts hazard analysis7. 3.5 Electrical Safety, Electrostatic Discharge As mentioned previously, more accidents and deaths happen from electrical shocks than due to exposure to radiation. When working with the laser enclosure open, be aware of hazards especially when working with systems that operate at 50 Volts or above2. Use a barrier system which would protect personnel from accidental contact with high voltage devices. The operator should not be complacent because of low operating voltages. As shown in Figure 3.5, the current plays an important role in determining the severity of the accident8. The mA (milli-Amperes) values will vary depending upon the weight and gender of the person, if the current is AC or DC, and if AC, the frequency4,8. In some cases, Class 4 lasers are enclosed with interlocks that shut the laser power off if the enclosure is opened. Realize that the operator is putting himself/herself at risk when they defeat the interlocks. In water cooled lasers, condensation of water may result on the outside of cooling tubes. Either this case or the case of a water hose rupture may pose hazards especially if it is located close to high voltage devices. The following recommendations apply for all classes of lasers2: 74 Current (mA) Heart Paralysis Tissue Burning 10000 1000 100 10 Mild sensation Shock/ no pain Heart fibrillation Shock, pain P 1 0.1 Perception Person lets go Muscle contraction Possible death Result Figure 3.5 Threshold current for various phenomena when current passes through the body. 1. Ensure that the equipment is installed as per OSHA and NEC (National Electrical Code.) including use of a three-wire plug for the laser and/power supply that has a ground or a hard-wired system. 2. Whenever feasible, unplug the equipment before performing maintenance or repairs, or when the equipment is not in use. 3. It is preferred that two persons be present when working with live electrical systems. The personnel should be knowledgeable of CPR and first aid. 4. Remove all conductive material from personnel such as jewelry, pens, etc. 5. Use only one hand when working with live circuits. Place the other hand behind your back. This would minimize risk of an electric current flowing through the heart. 6. Capacitors which are part of the power supply for pulsed lasers contain high voltage. For example, an excimer laser may use 33 kV or higher. Discharge these capacitors completely to prevent electrical shocks during maintenance or repairs. Make sure that large capacitors in the spare parts bin are shorted that they don not accidentally start charging from air! 7. Use tools with insulated handles. 8. Equipment that produces even a slight shock should be isolated, tagged, and sent for repairs. 9. Use insulated matting, suitable for electrical work, on floors, when working with high voltage systems. 10. Do not work with live circuits if the floor is wet or when the hand or body is moist, sometimes due to sweating. 11. Avoid working with electrical equipment if fatigued, stressed, or on medication that slows down the body and/or the mind. 12. If the equipment is hard-wired, follow the appropriate lockout/tag out procedures. 75 13. Know where emergency power shut off and emergency exits are located in your work area, as well as the location of an emergency phone. 14. Seek medical help even when you feel fine after receiving an electrical shock. Incidents of a delayed shock syndrome, where the victim appeared unaffected immediately following receiving the electric shock, however died later, have been reported. 15. Do not defeat electrical interlocks unless absolutely necessary for repairs or troubleshooting. 3.6 Fiber Handling Safety Thus far, we have considered hazards associated with laser radiation and electric shocks. Remember that if the laser light is routed through a fiber cable, one must be cognizant of the same radiation hazards present at the exit end of the fiber as from the original laser itself. The same precautions valid for the original laser hold for the laser beam exiting from the fiber optic cable. Additionally, fibers pose other hazards, mostly mechanical in nature. 1. Wear safety glasses during cleaving fibers. 2. Wear suitable gloves when handling fibers and avoid puncture wounds. 3. Clean up the work place of glass shards or fiber pieces using a sticky tape. Wrapping masking tape outside a gloved hand, with sticky side of the tape out, is an easy method of accomplishing this task. 4. Dispose of glass debris and shards in suitable containers and label them. The reason behind items (2) and (3) above is so that small pieces of fiber, usually of the order of a few micrometers to few tens of micrometers in diameter, inadvertently get into the human blood stream through a puncture wound and then travel through the capillary, vein and artery system throughout the body, potentially causing bleeding. 3.7 Optical Power Meters Always observe all safety precautions prior to and during the operation of a laser. Detectors can be broadly classified into photo detectors and thermal detectors. The former class converts the incident energy into a current or voltage which is read out on a display. It is usually calibrated at a definite wavelength to result in an automatic display of the incident beam power. Photodiodes and photoconductors are two examples. The second class of detectors converts the incident energy into heat and temperature change, which would change an electrical property such as resistance. Thermistors and pyroelectric detectors are two examples9. A typical optical power meter has a silicon detector hooked to a power meter is used to measure the power of visible laser beams. Germanium or indium gallium arsenide detectors are typically used to measure the power of near infra red lasers. The detector converts light energy into a voltage and the digital multimeter (DMM) (or analog 76 multimeter) displays the voltage, which in turn can be interpreted as a power reading. A block diagram of a typical set up is shown in Figure 3.6. Laser Beam Power Supply Detector Multimeter 888 To 120 VAC Figure 3.6 Block diagram illustrating typical power meter hook up. Depending on the type of power meter, one has to follow the specific instructions for that model. For the detector, one must use a correction graph (known as the spectral response curve) to obtain the correct reading10. If one uses a PIN-10 diode photo detector to measure power from a He:Ne laser beam, and if the detector is calibrated at 632.8 nm wavelength, the power meter reading obtained is correct without any modifications. However, for all other wavelengths, one must use a correction factor10. For example, for a certain laser wavelength, if the correction factor is about 1.9, and if the observed power on the meter is 2.0 mW, then the true power = 2.0 mW x 1.9 = 3.8 mW. Several commercially available sources can calibrate your detector using National Institute for Standards and Technology (NIST) procedures. Sometimes, if, instead of a correction factor, the responsivity of the detector is shown on the vertical axis, then, one must divide the observed power reading by the responsivity factor. In any case, the conversion between, say 1 W of actual laser beam power and the corresponding mV reading of the voltmeter must be supplied. Secondly, one must not exceed the threshold irradiance for damage of the detector9. Therefore, before using the meter, one must perform irradiance calculations to ensure this. A focused laser beam can easily exceed the threshold. An expanded laser beam is sometimes used if the irradiance from the original beam exceeds the threshold for damage. When the laser beam is expanded and its area of cross section exceeds the active area of the detector, the beam is said to overfill the detector. In this case, power from only the portion of the beam that is incident on the detector is measured by the power meter. To calculate the irradiance in this case, one must divide the power reading by the active area of the detector. The active area of the detector must be specified by the manufacturer. In many cases, this is one square centimeter. The correction graph for a typical photodiode detector is shown in Figure 3.7 11. This detector is calibrated for around 900 nm. There is no correction required if power measurements are made at λ=900 nm. All others require a multiplication by the correction factor at the corresponding wavelength. 77 Correction Factor 4 3.5 3 2.5 2 1.5 1 300 400 500 600 700 800 900 1000 1100 Wavelength, nm Figure 3.7 Correction factor (calibration) graph for a typical silicon photodiode. To emphasize the safe use of detectors, one must not subject the power meter to high irradiances, such as from a focused laser beam. When the power of the raw beam is high, such as in pulsed lasers, one may need to expand the beam to known dimensions and use the detector to sense the power of a portion of the expanded beam and then calculate the beam power. Pulsed lasers such as an excimer laser often require a Joule meter to measure the pulse energy. The joule meter may use a thermopile type (disc calorimeter) sensor or pyroelectric head to measure pulse energies from nanojoules to several joules9. Pyroelectric detectors have fast response times, and therefore, are most suitable for measuring pulse energies. Calorimeters, on the other hand, have response times in the one to three second range and are thus suitable for measuring CW laser powers9. Many of these are provided with a black coating that absorbs the radiation and converts the energy into heat. While these coatings can handle average irradiances of 500 W/cm2, the coating is ablated to some extent by excimer laser pulses and will have to be recoated after extended use. The process involves converting the pulse energy into thermal, measuring the temperature rise and converting back to the pulse energy in joules. Further details of pulsed laser characterization are provided in a later chapter. 3.8 Laser Safety Laboratory 1. Place the laser on the lab jack. 2. Observing all the safety precautions, turn on the laser. 3. Turn all lights in the room to maximum output (i.e. make the room as bright as you can.) 4. Describe what you see 5. Turn all room lights off. 78 6. Describe what you see. Compare to step 4. 7. Adjust room lighting to a comfortable level and look at the beam on the wall with He-Ne goggles. 8. Describe what you see. Compare to steps 4 and 6. 9. Hook up the elements of the optical power meter. Measure the power of your laser. Make sure that the beam reaches the center of the detector for an accurate reading. Measure the beam diameter using a graph paper12. Calculate the area of cross section of the beam. Calculate the beam irradiance. 10. What is the laser safety classification for your lab laser? • • • • • • Chapter Summary Lasers are classified based on the level of hazard posed. With visible lasers, retinal tissue damage is the most common form of hazard. Suitable eye protection must be worn to avoid accidental exposure to the laser beam. With high powered lasers, intrabeam viewing, specular reflections, Fresnel reflections and diffuse reflections can all be hazardous. In addition to skin and eye damage, several electrical hazards exist when operating lasers and electrical safety procedures must be adhered to. One should take chemical safety precautions when dealing with laser gases and chemicals as well as products generated as a result of laser beam irradiating a target material. The student must be able to measure the power of the beam correctly and perform irradiance calculations taking into account correction factors if the laser wavelength is different from what the meter is calibrated for. End of Chapter Exercises 1. Match an item in Column A correctly to an item in Column B Column A Column B 1. IR-C a. 400-700 nm 2. IR-B b. 200-280 nm 3. IR-A c. 3000 nm – 1 mm 4. Visible d. 315-400 nm 5. UV-C e. 280-315 nm 6. UV-B f. 1400-3000 nm 7. UV-A g. 700-1400 nm 2. Match an item in Column A correctly to an item in Column B Column A(wavelength) Column B(tissue affected) 1. 200-400 nm a. Retina 2. 400-700 nm b. Cornea 3. 700-1400 nm c. Cornea 4. 1400 nm – 1mm d. Retina 3. For each class laser, write down the range of power. 79 4 The diameter of a 3.5 mW laser beam is approximately 2.8 mm. If it illuminates a target, what is the irradiance? 5. A 2mm beam has 7 W of power. What is the irradiance? 6. An excimer laser puts out 20 ns pulses of 250 mJ energy each. If the PRF is 5 Hz, and the rectangular cross-sectional beam itself is 27 mm x 18 mm in dimension, calculate (a) peak power, (b) peak irradiance and (c) average power. 7. If the laser beam in problem 4 is expanded to a 3.14 cm2 area, what is the irradiance? 8. The laser in problem 4 is focused using converging optics to a spot of diameter 18 µm. What is the irradiance of the focused spot? 9. If the laser beam of problem 4 hits a target for 5 seconds, what is the energy density in J/cm2? (Hint: Energy = Power x time.) 10. The power meter monitoring the beam power of a Nd: YAG laser (λ=1.06µm wavelength) reads 2.3 W. What is the actual power of the beam? Use the calibration graph provided. 11. A power meter measures the laser radiation from a laser as 3.34 W. What is the true power reading, if the correction factor is 1.9? Group Discussion Questions These require the students to form small groups of three or four and discuss each question. Some questions may require you to do a search on the Internet and find the answers. 1. Using Internet search for three of the most bizarre laser radiation exposure accidents that ever occurred. 2. The ANSI Standard Z136.1 is constantly revised. Search the web and determine the significant changes that have occurred since the publication of this book. 3. What are the duties of a Laser Safety Officer (LSO)? 4. Which is more important – voltage or current in determining the harm it can do to a person. 5. A fiber optics cable carriers light from a 10 mW visible laser. List all the hazards associated with that fiber cable. References ANSI Standard Z136.1-1993 (also 2000 and later), American National Standard for the Safe Use of Lasers, 1993, 2000. 1b see for example, Biooptica Laser Safety, http://www.biooptica.co.uk/product_safety. 2 Princeton University Laser Safety Training Guide, http://www.princeton.edu/~ehs/laserguide/laserguideTOC.htm, 2001 1 80 3 University of Pennsylvania Laser Safety Guide, http://www.ehrs.upenn.edu/programs/laser/manual.html 4 Weber, M.J. “CRC Handbook of Laser Science and Technology Volume I – Lasers and Masers.” Boca Raton: CRC Press, 1982. 5 Myers, G. Elements of Laser Safety. http://gary.myers.net/elements.htm, 1998 6 Luxon, J. “Industrial lasers and their applications.” Englewood Cliffs: Prentice Hall, 1985. 7 Laser Institute of America, http://www.laserinstitute.org/ 8 Floyd, T.L. “Electronics Circuits Fundamentals.” Englewood Cliffs: Prentice Hall, 2002. 9 Froman, D. and G. Shelmire, Scientech, Inc. Power Meter Literature http://scientechinc.com 10 UDT Sensors, Inc. Product Catalog http://www.udt.com/pdf/Detector-FilterComboSeries.pdf. 11 MACAM Photometrics Ltd. PM203 Product Catalog http://www.macam.com/ 12 Gottlieb, H. “Experiments Using a Helium Neon Laser.” Bellmawr: Metrology Instruments, Inc. (1981) 81 Chapter 4 Basics of Laser Operation Theory 4.1 Energy Levels in an Atom 4.2 Spontaneous Emission 4.3 Stimulated Emission 4.4 Light Amplification 4.5 Gas Lasers 4.6 Solid State Lasers 4.7 Suggested Laboratory Experiments: Power Meter Use Chapter Summary End of Chapter Exercises Group Discussion Questions Chapter Objectives: After completing this chapter the student should be able to (i) Describe the Bohr model of an atom (ii) Explain spontaneous and stimulated emissions (iv) Describe population inversion, light amplification and lasing (v) Describe the principles behind gas, liquid and solid state laser operation Key Words: spontaneous emission, stimulated emission, pumping, excitation, population inversion, lasing, feedback, high reflectance mirror, output coupler, active medium 4.1 Energy Levels in an Atom The electron, as we may know, is the fundamental current carrying unit in an electrical circuit. It has negligible mass and carries a charge of 1.602 x 10-19 Coulomb (C.) The electron itself is originally part of an atom. The simplest method of looking at an atom is using the model developed by Niels Bohr1. While the real atom does not exhibit stationary orbits as implied by the Bohr model, the latter is simpler to illustrate at this level. Accordingly, electrons rotate around a central nucleus, occupying different orbits, similar to the planets revolving around the sun at different distances from it. An example of a sodium atom is shown in Figure 4.1. The path of each orbit is roughly circular. Electrons, shown by grey spheres, rotate in their respective orbital. There are a total of 11 electrons, made up of two, eight, and one respectively in the orbits. The nucleus, consisting of 11 protons and several neutrons, is shown by a dark sphere. The charge on a proton is the same as that on an electron, however, a proton is positively charged, as opposed to an electron which is negatively charged. The neutrons provide stability to the nucleus. The electrons are held in orbit by the attractive forces from the protons. The outermost electron in the case of sodium is called the valence electron, since it can participate in bonding with other kinds of atoms such as chlorine or oxygen. The valence electron can also be excited energetically using an electric voltage, so that it can start conducting electricity, providing that the energy gap between the conduction state and valence state is relatively small. 82 Nucleus containing protons and neutrons Figure 4.1 The Bohr model. One needs to know quantum mechanics to describe accurately energy levels of electrons. Electrons occupy different electronic levels based on the orbital. Additionally, there are sub-classes of energy levels within each electronic energy level that are known as vibrational levels. Within each vibrational level are rotational levels. Using photons of different energies one can excite electrons of different electronic, vibrational and rotational levels to higher energy states. An increase in the energy of electrons results in an increase in the atom’s energy. Since an atom is composed of electrons that occupy different energy levels, if we add the electron energies at a certain configuration, we get the atom’s (or molecule’s) total energy for that configuration. Only certain discrete atomic energy levels are allowed. Consequently, quantum mechanics allows certain energy transitions to occur within electronic states. Thus, one can look at an atomic energy level diagram for any atom. The reason for doing this is to understand the different possibilities for energy transitions for the atom. If the atom occupies the lowest energy allowed, then it is said to be in the ground state. A ground state atom is designated to have zero electron volts (0 eV) of energy. This is similar to choosing a ground for an electronic circuit. Quite frequently, one is unable to connect a point in the circuit to the earth ground. For example, in an automobile, the chassis is chosen as the ground and all voltages that run devices in a car are measured with respect to this ground. For atoms, ground state just implies that it has the minimum allowable energy1 based on the dictates of quantum mechanics. Thus, an atom can occupy any of the allowed energy states at any time. However, more atoms occupy the lower energy states and fewer atoms occupy the higher energy states. 83 Thus, one normally expects a high population of atoms in the ground state (i.e. the lowest allowed energy level) and decreasing probability of finding atoms as one goes up the energy scale of allowed levels. If one supplies energy to the atom, then one can expect some of the lower energy state atoms to be excited and move into higher energy levels. Examples of a few of the allowed energy levels for the sodium atom are shown in Figure 4.2 1,2. It requires knowledge of quantum mechanics to predict the distribution of the atom population of these energy levels. While in reality electrons gain or lose energy transitioning to different levels, it manifests as the atom’s energy transitioning to different levels since an atom’s total energy is given by the sum of the electronic energies. 6 Sodium ionized (Na+) 5.12 eV 5 4.3 eV 4.1 eV 4 3.6 eV E(eV ) 3 3.2 eV 2.1 eV 2 1 0 Ground State 0 eV Energy levels in sodium atom Figure 4.2 Energy levels in a sodium atom. By supplying energy to the atom, one can excite them from a lower energy level to a higher energy level. The energy may be supplied by using electrical means such as applying a voltage or current through the medium, or using light (photons.) Conversely, when atoms occupy higher energy states (levels) they can transition down to lower energy states, sometimes giving out a photon whose energy equals the energy difference between the two levels. Similarly, one can conceive of energy level diagrams of other atoms or molecules, for example, helium, neon, argon, krypton, carbon dioxide, water vapor, etc. Notice that we have chosen atoms and molecules that are gaseous. You will see in the next section why we did so. 84 4.2 Spontaneous Emission In general, all bodies will strive to reach their lowest allowed energy levels since those are the most stable configurations. If we place marbles in a container and shake it, the marbles will move around and occupy the most stable configuration for the given setting. Thus, if we elevate the energy of an atom to higher energy levels, they may stay in the excited energy levels for a “short” period of time. The time it takes for one half of the number of atoms from an excited state to leave that state (to a lower energy state) is called the atomic life time or half life of that excited state1. The half life may vary from 10-14 s to 0.01 s, a wide range1,3. In comparison, for an atom not subjected to excitation energy, it can stay in the lowest energy state called the ground state for ever. Thus, normally, the ground state energy level of an atom will have the greatest number of atoms, and higher energy levels will have fewer and fewer number of atoms. When ground state atoms are excited using an external source of energy, such as electric current, they move into some of the higher energy levels, thus increasing the population of those energy states. When some of the atoms in the excited energy states lose their energies and move down to a lower energy state (which includes the ground state), the energy lost by the atom appears as the energy of a photon emitted. This process is called spontaneous emission where the higher energy atom “voluntarily” (i.e., without an external influence) gives up its excess energy and shoots out a photon. Knowing the energy of the photon emitted, one can calculate the wavelength of the photon. Conversely, if the conditions are right, an atom can absorb a photon of a distinct wavelength, and raise its own energy to a higher allowed energy level. This process known as absorption results in the loss of the photon. Thus, spontaneous emission and absorption are opposite and opposing processes. Looking at the energy levels diagram for sodium vapor, a high voltage applied can excite the ground state sodium atoms to an ionized state. While, theoretically, one could have 21 transitions possible for atoms occupying a higher energy state to move to a lower energy state, some transitions are forbidden or unlikely, based on selection rules of quantum mechanics theory. The most common transitions happen to be from the two closely spaced levels of about 2.1 eV to the ground state. In general, transitions that involve smaller energy differences are more likely than larger energy differences. Also, the atomic life times of the energy states dictate the likelihood and ease of transitions. Example 4.1 Calculate the wavelength of the photon emitted spontaneously when a sodium atom transitions from 2.1 eV energy state to ground state. λ = hc E(J) = 6.625 x 10-34 x 3 x 108 2.1 x 1.602 x 10-19 = 590 nm Answer This is the most common of the wavelength emitted by a sodium vapor lamp. The more 85 precise value of this wavelength corresponds to two closely spaced values of 589 nm and 589.7 nm. The mercury vapor lamp contains a gas discharge tube which is filled with mercury vapor through which an electric current is passed. The approximate energy levels for a mercury atom are shown below in Table 4.14. Table 4.1 Energy Level E1 (Ground State) E2 E3 E4 E5 E6 E7 E8 Energy (eV) 0.0 eV 4.8 eV 5.0 eV 5.5 eV 6.7 eV 7.7 eV 8.0 eV 11.2 eV Example 4.2 Calculate the wavelength of the photon emitted spontaneously when a mercury atom transitions from E6 energy state to E3 energy state. Energy of the photon would correspond to E6-E3 = 7.7-5.0 eV = 2.7 eV λ = hc E(J) = 6.625 x 10-34 x 3 x 108 2.7 x 1.602 x 10-19 = 459 nm Answer This is one of several possible wavelengths that are emitted from a mercury lamp. In fact, the lamp emits different wavelengths from UV to IR. Viewing UV radiation without eye protection is not wise. Thus, one should not look at the spectroscope directly when using a mercury lamp as a source. The mercury lamp emits strongly, among other UV wavelengths, at 365 nm wavelength. A UV filter must be used. Thus, when gas tubes, known as plasma tubes, are energized, atoms are excited to different higher energy levels and, during transitions to lower energy states, some photons are emitted. In other cases, the atom undergoes radiationless decays5 or transitions, i.e., they decay to a lower energy state without emitting a photon, or may undergo collisions with other atoms, losing their energy without emitting a photon. Consequently, photons of different wavelengths, traveling in different directions, and “marching to different drums” are emitted, resulting in incoherent light. The strong lines correspond to the most likely transitions arising from energy levels of short life times, whereas weak lines arise from energy level transitions arising from long life time states. In the case of solids, one does not see a clear line spectrum. Rather one sees a band spectrum, because of overlap of electronic and vibration levels. Additionally in solids, excitation of vibration levels manifests as heat. In order to control the characteristics of the photons to result in an intense, directional beam of light as in a laser, spontaneous 86 emission of light alone is not sufficient. We need to look at another phenomenon known as stimulated emission, which is presented in the next section. 4.3 Stimulated Emission Just as a leader needs to model his/her behavior so that the followers would follow suit, in order to create photons that are of the same wavelength, traveling in the same direction and in phase with the original photon, one needs to stimulate this process. Due to excitation, some atoms are elevated to higher energy levels. Let us say for the sake of argument that an atom reached the E2 level, as shown in Figure 4.3. Also assume that the E2 level happens to possess a relatively large half-life. It is therefore called a metastable state. It means that it has gained a net energy corresponding to E2-E1, since the atom was in the E1 (ground state) to begin with before excitation started. If, somehow, one hits the E2 level atom with a photon whose wavelength corresponds to E2-E1 energy level, what will happen? The photon induces or stimulates the E2 level atom to give up its energy and fall back to the ground state (E1) giving up a photon in the process3,5,6, as shown in Figure 4.3. This photon will be the same wavelength as the original stimulating photon. Further, the second photon will travel in the same direction as the original photon, as well as be in phase with it. The result is that we have now two photons of the same wavelength, in phase and traveling in the same direction – the essential characteristics of laser light. E3 Stimulating photon Ep=E2-E1 Stimulating photon Ep=E2-E1 x E2 Ep=E2-E1 E1 New photon Figure 4.3 Illustration of stimulated emission. A single stimulating photon has now created an additional photon. Thus, if we have many original stimulating photons, then we will have an equal number of photons arising from stimulated emission. Note that a photon with energy equal to E3-E2 hitting an E2 level atom will not cause stimulated emission. The essential conditions for stimulated emission3,6 to occur are (i) a stimulating photon with energy corresponding to a difference in two allowed energy levels of an atom is available, and (ii) it must strike an atom in the higher of the two allowed energy levels. Three questions arise. (1) Where did the stimulating photon come from? (2) What happens if a photon strikes an atom of lower energy level? (3) Is stimulated emission alone sufficient to produce intense laser light? The answers are as follows. 87 (1) When atoms are in an excited state, spontaneous emission will always happen, as implied by its name. Some of these spontaneously emitted photons happen to be traveling in the direction of the optical axis of the laser tube and thus serve as stimulating photons. (2) If a photon with energy = E2-E1 of an atom collides with an E1 level atom, it is conceivable that the photon transfers its energy to the E1 level atom, thereby increasing the atom’s energy level to E2. In the process, the original photon will cease to exist. This process is called absorption and is the opposite phenomenon to spontaneous emission. (3) Stimulated emission by itself will produce a few extra photons, however, not sufficient to sustain an intense laser beam. One needs feedback to take advantage of the stimulated emission process, which is the topic of discussion in the next section. 4.4 Light Amplification As mentioned briefly in the previous section, spontaneous emission, followed by stimulated emission start the lasing process; this needs to be amplified to obtain sustained laser output. This is achieved by using mirrors at either end of the laser cavity3,5, as shown in Figure 4.4. The mirrors provide the feedback mechanism necessary for light amplification. One of these mirrors is known as the total reflector or high reflectance mirror, sometimes abbreviated as HR, since its reflectivity is upwards of 99.9%. The second mirror is called the output mirror or output coupler, sometimes abbreviated as OC. Its reflectivity is somewhat lower since a portion of the light is transmitted by it. Total reflector Lasing Medium Output Mirror Laser Beam Beam undergoes multiple reflections Figure 4.4 Illustration of light amplification. When light travels through a medium, a portion of it is absorbed. The amount of light absorbed depends upon the thickness of the medium, as well as the absorption coefficient of the medium α. The light irradiance I at any distance x from the surface is given by1-6 I = I0e-αx where I0 is the irradiance of the incident light at the surface, as shown in Figure 4.5. For most materials of interest to us including glass, water, chemical solutions, etc., α is positive, i.e., for any distance x into the medium, the irradiance of light is always smaller than the light entering the medium. This is the exponential law of absorption (see Chapter 8) where e is the exponential number equal to 2.718.., the base of natural logarithm. 88 Air I0 Medium I x Figure 4.5 Illustration of light absorption. With a laser cavity, however, due to stimulated emission occurring inside the medium, the absorption coefficient is negative (see Chapter 9 also). In other words, irradiance, which is directly proportional to the number of photons, at any point inside the medium is higher, leading to light amplification. During each round trip of the laser beam, it bounces off the high reflectance once, goes through the medium, is reflected by the output mirror, goes through the medium and reaches the high reflectance mirror again. During a round trip, the light is amplified twice, once in its forward journey and the second time during its return journey6. This is called the round trip gain of the laser. If the round trip gain is more than one (1), the laser power will continue to increase; if it exactly one, laser power will be sustained; if it is less than one, the laser power will die out6. The OC transmits fraction of a per cent to 1% for He-Ne lasers3 and a few tens of per cent for solid state lasers. Population Inversion As mentioned previously when the laser is turned off, most atoms are in the ground state, with fewer numbers occupying higher energy states. In fact, the higher the energy level, fewer will be the number of atoms (also called the population) in the state. Thus, lower energy levels are highly populated and higher energy levels are sparsely populated. We know from a discussion of stimulated emission that we need greater number of atoms in a higher energy state so that a stimulating photon has an enhanced probability of collision with those, creating additional photons. Thus, one needs to create a situation whereby the higher energy metastable state is populated preferentially compared to the lower energy state between which the lasing occurs. This condition is called population inversion and is a prerequisite for successful spontaneous emission and subsequent lasing action3,6. How does one achieve population inversion? It is achieved by a process known as excitation or pumping. In essence, it involves supplying energy to the lower level energy atoms via electrical or optical means. It must be mentioned that the pumping process to create population inversion takes place between ground state and an upper energy state level. However, the laser photon is not necessarily emitted when the atom returns to its ground state from the higher energy state. In fact, this is seldom the case. Quite frequently, pumping elevates the atom to, say E3 level. If the atomic life time for this level is relatively short and the atom can transition to E2 level through a rapid radiationless decay5 or transition, then, if the atomic life time in E2 level is relatively long, population inversion is achieved for E2 level. E2 level is called a metastable level3. Thus, when the atom drops from E2 to E1 level, the 89 laser photon is emitted. This is the principle for a three-level laser as shown in Figure 4.6 3,5,6. E3 Radiationless transition E2 Pumping Lasing E1 Ground State Figure 4.6 Three level laser system. One can also envision a possible four level lasing system3,5,6 as shown in Figure 4.7. E4 Radiationless transition E3 Lasing E2 E1 Radiationless transition Ground State Figure 4.7 Four level laser system. Here the lasing occurs between E3 and E2 levels. Due to rapid radiationless decays5, population inversion is readily achieved between E3 and E2 levels. The details of lasing levels pertinent to each laser will be discussed in separate sections. The lasing phenomenon can be summarized thus. (i) one needs population inversion first which is accomplished by excitation or pumping process, (ii) a few higher energy atoms spontaneously emit photons, (iii) many of these photons are traveling in random directions by virtue of the characteristic of spontaneous emission, however, a few are traveling along the optical axis of the laser which (iv) stimulate higher energy atoms to emit stimulated photons that result in gain, (v) the stimulated photons are reflected between the two laser mirrors resulting in feedback and amplification, (vi) a small portion of the light leaves through the output mirror resulting in laser output. If one uses an output mirror (OC) with a reflectivity as high as the HR mirror, the laser output will be too small to be of practical use. 90 The properties of the laser light that emerges thus can now be summarized as follows: (a) monochromaticity (all photons of the same wavelength), (b) directionality (all photons are traveling in the same direction, and (c) coherence (waves in phase, both in space and time.) Regular light such as sunlight, candle light or incandescent lamp source lacks many of these properties. All sources are coherent to some extent, i.e., coherence is maintained up to a certain distance from the source. This is known as the coherence length. However, for many sources the coherence length is rather small. Lasers on the other hand have much longer coherence lengths, essentially infinite compared to other sources. 4.5 Gas and Ion Lasers Some examples of gas lasers include the helium-neon (He:Ne) laser, carbon dioxide (CO2) laser, nitrogen (N2) laser and excimer lasers; examples of ion lasers include argon ion (Ar+) and krypton ion (Kr+) lasers. The lasing medium (also called the active medium) in these cases involves a gas or a gas mixture. The pumping process uses an electrical current through the low pressure gas (mixture) in most cases and atmospheric or above atmospheric pressures in other cases. Helium-Neon (He:Ne) laser This is the most common laser used by the student for the laboratory experiments. As the name implies, it consists of a low pressure gas mixture of 90% helium and 10% neon7. The first step involves exciting the helium atom to an excited level, which we will call He*. This needs at least 21.8 eV of energy. The pumping process also creates excited neon atom states. When He* state atom collides with a ground state Ne atom, this creates a Ne impurity state band, which rapidly decays to a lower level Ne energy band. The lasing occurs when the higher level energy Ne atom drops to the ground state. A rough illustration of the process is shown below1,7. He excited states 21.8 eV (He*) +Ne Ne Impurity bands Ne excited states (via collision) Energy (eV) Pumping Figure 4.8 Simplified energy level diagram for the He:Ne laser.. Laser Output 91 The 632.8 nm wavelength is the strongest and most commonly used output of the He:Ne laser. It appears as a red beam of light. These lasers are also widely used as alignment lasers for other lasers such as the Nd:YAG laser. Argon ion Laser Group 8 elements of the periodic table have a complete outer shell of electrons, and therefore considered inert. Argon, krypton, xenon, etc. are some examples. Argon, at normal conditions, does not react with other types of atoms. However, given sufficient electrical energy, one could knock off an electron from the outermost shell. Losing an electron is tantamount to having a positive charge for the argon atom. This is called an argon ion having a positive charge (Ar+). In general, it requires a great deal of energy to ionize an atom than to elevate to a higher energy state without ionization. Once ionized, the argon laser acts on a four level laser principle, as shown in Figure 4.9 below8. Realize that the ground state for the lasing is the argon ion, not the argon atom. The same is true for other ion lasers as well. In fact, mixtures of argon and krypton are used in laser light shows due to the multitude of laser lines that are obtainable. The energy levels have been simplified significantly for clarity. In reality, the E3 and E2 levels shown between which lasing action takes place are by no means single valued. There are multiple energy levels within E2 and E3. Excited Ar+ E4 Pumping E3 ca. 35 eV Laser photon E2 Ionized argon (Ar ) ca. 19 eV (Ground state for Ar+) + E1 Argon atom ground state (Ar) Figure 4.9 Simplified energy level diagram for the argon-ion laser.. Thus, when lasing occurs, a photon is emitted when the atom loses its energy from any one of the sublevels in E3 and reaches any of the sublevels in E2. As a result, photons of wavelength in the range of 400 nm to 600 nm are emitted, with predominant lines at 488 nm and 514.5 nm8. Using a prism one can selectively obtain the required wavelength for the application. The details of the prism action will be discussed in Chapter 5. Notice that E1 is truly the ground state for lasing, even though the Ar+ atom has an energy of 19 eV compared to the ground state argon atom. 92 Carbon dioxide (CO2) laser These are typically composed of approximately 10% CO2, 10% nitrogen (N2) and 80% helium (He), maintained at a total pressure of 6-15 torr9 (1 torr = 1mm of mercury pressure). The most common wavelength emitted is in the IR at 10.6 µm. The essential steps involve excitation of nitrogen vibration level by electron impact. An 18 kV electrical discharge is needed to accomplish this. The vibration energy is then transferred via collision from the nitrogen to the CO2 molecule. The lasing occurs when the CO2 molecule transitions down to a lower energy level9. The helium gas is present to maintain the plasma discharge. These lasers are water cooled and are available in continuous wave (CW) or pulsed modes. They can heat the target material up very quickly and hence these lasers find applications in cutting, welding, surgery, skin resurfacing, etc. Excimer Laser These are pulsed lasers operating in the UV/deep UV regions. The laser has a noble gas such as argon (Ar), Krypton (Kr) or Xenon (Xe) and a halogen such as chlorine (Cl), fluorine (F) or bromine (Br.) The word excimer stands for excited dimer. Being a noble gas, Ar, Kr or Xe does not react with other atoms under normal conditions (i.e. when they are in the ground state.) Thus, the ground state consists of individual noble gas molecules and halogen molecules. However, when excited electrically, the molecules react to form a compound such as ArF, KrF, XeBr, KrCl, etc. These compound molecules exist only in the excited state and never in the ground state. Thus, a favorable population inversion is created whereby a large number of higher lasing level species are present and virtually zero number of the same species is present in the ground state. The life time of the excimer molecule in the excited state is of the order of a few nanoseconds. Some typical wavelengths emitted by excimer lasers are shown in Table 4.210. Table 4.2 Type of Excimer Laser Wavelength (nm) Argon Fluoride (ArF) 193 Krypton Chloride (KrCl) 222 Krypton Fluoride (KrF) 248 Xenon Chloride (XeCl) 308 Xenon Fluoride (XeF) 351 These pulsed lasers are capable of producing a few hundred millijoules per pulse, each pulse lasting a few to few tens of nanoseconds. Thus, the instantaneous power, as illustrated earlier, is very high, leading to photochemical ablation of the target materials that the laser beam is incident upon. By changing the composition of gases in the plasma tube, one can obtain various UV wavelengths. One can also obtain 157 nm wavelength radiation in the so-called vacuum-UV region by using a fluorine (F2) excimer laser. Excimer lasers have applications in eye surgery, angioplasty, photolithography in semiconductor manufacturing, etc. Laser pulsing is achieved by pulsing the power supply, for example, using a thyratron circuit consisting of higher energy capacitors. 93 Nitrogen Laser This emits light at 337.1 nm wavelength. The laser is pumped by a fast rise time electrical pulse. A transition from one of the vibration levels of a higher electronic state to a vibration level of a lower electronic state produces the lasing action. This transition from about 11 eV to about 7.3 eV results in release of a photon with approximately 3.7 eV, which is emitted as a UV photon. Typically these lasers have a flowing gas system with a pump. One of the main applications of the pulsed nitrogen laser is in pumping a dye laser. Dye Laser These lasers use an organic dye dissolved in a solvent, such as alcohol, as the active (lasing) medium. The main advantage of the dye laser is its tunability, meaning ability to obtain different wavelengths. Two of the most common dyes10 are Rhodamine 6G (Rh6G), a highly fluorescent dye, and Coumarin 630. The Rh6G is tunable in the 570650 nm range, while the Coumarin dye emits laser radiation at 504 nm. Dye lasers are pumped by a pulsed nitrogen laser, argon ion laser or using a flash lamp. The dye is continuously circulated by a pump. 4.6 Solid State Lasers In this section we will look at lasers where the active medium is a solid. It may be a solid crystal or a semiconductor material. The industry distinguishes the two types differently. Nd:YAG laser This crystal based laser is the successor to the ruby laser, which was the first optical laser. YAG stands for yttrium aluminum garnet, a crystal. Some of the aluminum ions (Al3+) in it are replaced by neodymium ions (Nd3+). The Nd+3 is the active species which is pumped optically, using a flash lamp or high pressure krypton or xenon lamp, or diode laser. The mirrors that reflect the 1064 nm wavelength IR light however are transparent to visible light11. The crystal rod and the gold-plated elliptical reflector that shine the light on the crystal are about the same length. An ellipse has such a property that if a ray of light passes from one focal point, after reflection, it will pass through the second focal point. Thus, the light source and the crystal are placed on either focal point to optimize light coupling from the lamp to the rod. Furthermore, the surface of the rod is roughened to couple the light diffusively into the rod. Krypton lamps have strong output between 730 nm and 760 nm, and also between 790 nm and 820 nm, the regions where the crystal rod absorbs the light. Xenon lamps, on the other hand, have only a small intensity in those wavelengths11. In any case, one must use the appropriate lamp suitable for the laser circuit. For example, a pulsed laser requires a pulsed lamp that draws relatively large currents. Therefore, a lamp designated for CW operation should not be used in a pulsed laser circuit since it could explode due to large currents11. The laser should be water cooled for high power operation. The cooling is not an issue with high efficiency diode laser pumped YAG lasers. The crystal rod itself is different for CW versus pulsed operation. The CW rod is doped with Nd3+ to a lesser 94 extent than the rod for pulsed laser operation. The former looks transparent to white light, while the latter has a distinct color. Semiconductor laser The semiconductor laser, also known as diode laser or injection laser, operates on the principle of a semiconductor junction. A brief description of the workings of a semiconductor diode is in order, prior to providing a description of the lasing action. Copper, silver, aluminum, gold, etc. conduct electricity well and, hence, are termed conductors. The energy required to excite electrons from the outermost orbit (called valence electrons) to the conduction band (i.e. an energy level that is adequate that electrons start conducting electricity by motion) is rather small for these conductors. On the other hand, the energy gap is large for insulators such as wood, plastic, rubber, etc. A class of materials such as silicon, germanium, gallium arsenide, etc. possess an energy gap that is somewhere in between the conductors and insulators, and hence their name, semiconductors. A pure semiconductor material such as silicon is not very useful as such since its conductivity is not large enough. Therefore, it needs to be doped with other types of atoms. If one adds boron atoms to silicon, then the conductivity is increased due to increased number of “positive type” carriers or holes. The resulting silicon is called ptype silicon. If one adds arsenic, phosphorus or antimony to silicon, again, the conductivity is enhanced due to increased number of “negative type” carriers, i.e. electrons. The resulting silicon is called n-type silicon. If one can fabricate a structure so that the silicon is doped p-type in one half and n-type in the other half, there will be a junction where the p and n type of silicon come in contact. This is called the pn junction. Silicon is one example of a semiconductor material from which a junction can be formed. Gallium aluminum arsenide, gallium indium phosphide and gallium aluminum indium phosphide are other examples. A semiconductor diode consists of a pn-junction. When the p-side of the semiconductor is connected to the positive of a DC source and the n-side connected to the negative of the same source, there will be current flow. On other hand, if it is reverse polarized, there will be very little current flow. This explains the diode action of allowing current flow in one direction and not the other. When the p-type carriers (holes) and the n-type carriers (electrons) recombine near the junction when the semiconductor diode is suitably biased, photons are emitted from the junction, as shown in the Figure 4.10. For laser diodes used for fiber optics applications, the actual layers are more complex than shown in Figure 4.10. In fact, there are several thin differently doped layers to accomplish the appropriate lasing wavelength and power. The mirrors are built in as part of the crystal12. Realize that the semiconductor crystal is of the order of 1 mm x 300 µm x 300 µm. The crystal is cleaved to have parallel faces forming the total reflectance and output mirrors respectively. The active region or 95 junction is only about 1 µm wide. To obtain pulsed laser output, one ideally needs a power supply that can produce ca. 20 A current pulses of about 20 ns pulse width. A lens may be used to collect and focus the original laser beam which may have large divergence12, up to 10o, or higher. + p Laser photon 1 µm High Reflectance Mirror n Active region Output Mirror Figure 4.10 Illustration of a semiconductor laser operation. Semiconductor lasers have applications as fiber optics transmitters. A discussion of pulse laser operation and their characterization is provided in Chapter 9. 4.7 Suggested Experiments: Power Meter Use A photo detector is used to measure laser beam power in the visible-near IR region. If the readout is calibrated to give the power of He:Ne beams directly, one needs to use a correction factor when dealing with wavelengths other than 632.8 nm. The calibration graph must be supplied by the detector manufacturer. Always follow safety precautions when using lasers. Determine the beam power at various distances from the laser to determine if there is a difference. Estimate the beam diameter using a piece of graph paper13 and calculate the area and then the beam irradiance. Make similar power measurements of reflections of a front surface mirror (specular reflection), reflections from a lens surface (Fresnel reflection) and reflection off the wall surface (diffuse reflections)13. Characterize expanded beams by making power measurements13. A beam can be expanded using a negative (concave) lens. So long as the beam falls completely inside the detector, the irradiance calculation should use the area. Otherwise use the active detector area supplied by the manufacturer to calculate the irradiance. • Chapter Summary A laser beam is monochromatic, directional, and temporally and spatially coherent. 96 • • • • • The main elements of a laser are the lasing medium, the pumping mechanism, the feedback mechanism between the mirrors and the output mirror itself. Spontaneous emission has to occur first to create a few photons; then stimulated emission ensures the directionality, coherence and monochromaticity of the output beam. Population inversion, along with amplification that results when photons are reflected back and forth between the mirrors, ensures that an intense beam is output from the laser. The lasing medium may be solid, liquid, gas or plasma. The laser may emit a continuous beam (CW mode) or pulses. End of Chapter Exercises 1. Referring to the atomic energy level table for the mercury atom, assuming that any transition from a higher energy state to a lower energy state is possible, calculate the wavelength of the spontaneously emitted photon. Remember there are 28 possibilities. Draw these wavelengths as a line spectrum on a LOGARITHMIC graph paper. 2. The important physical elements of a laser are ____________, _____________, _____________ and ________________ 3. Why is the reflectance of the output mirror less than that of the HR mirror? 4. A four level laser has energy levels in eV of E1=0, E2=1.3, E3=3.4 and E4=4.3. If lasing occurs between E3 and E2 levels, what is the wavelength of the laser photon? 5. An atom in a 2.1 eV energy state absorbs a 514 nm photon. What is the new energy level of the atom? 6. An energetic atom A collides with a less energetic atom B. Energy is transferred from A to B. Now B’s energy is increased and A’s energy is decreased. There is no loss in energy. What is this phenomenon called? 7. The name of the process in a laser where low energy atoms are pumped to high energy levels is known as _______________ . 8. The Nd: YAG laser emits 1.06 µm photons. What is the corresponding difference in energy levels that resulted in this photon being emitted? 9. A Nickel-like x-ray laser emits 14.7 nm wavelength radiation. What is the photon energy? 10. A He:Ne laser beam is expanded to 2 cm diameter and its power is measured by a power meter. If the meter reading shows 740 µW and the active area of the detector is 1 cm2, what is the irradiance? 97 11. A power meter with a silicon detector reads the beam power as 2.2 W. If the wavelength is 478 nm, what is the actual beam power? (Use Figure 3.7). 12. A power meter reads 80 µW when it is exposed to light from a sodium lamp at 589 nm? What is the actual power? (Use Figure 3.7). Group Discussion Questions These require the students to form small groups of three or four and discuss each question. Some questions may require you to do a search on the Internet and find the answers. 1. What is MASER? Who invented it? What principles does it use? 2. Wintergreen mint containing sugar, if chewed with your teeth, will result in bluishgreen light emanating from your mouth. (You have to watch yourself in the mirror in a darkened environment to observe this clearly.) What phenomenon causes this? 3. What is black body radiation? 4. What instrument is useful in looking at different wavelengths emitted by spontaneous emission from an ordinary light source? 5. Does one always require all the functional elements of a laser to obtain lasing? Are there any exceptions? References Sears, F.W, M. W. Zemanksy and H.D. Young. “College Physics.” Reading: AddisonWesley Publishing Company, 1991. 2 http://webexhibits.org/causesofcolor/4.html 3 Hecht, J. “Understanding Lasers.” Carmel: Howard H. Sams Company, 1988. 4 The Frank-Hertz Experiment and Quantized Energy Levels http://www.physics.sfsu.edu/~rrogers/Phys%20321/5_Frank.htm 5 Laurence, C. “Laser Book: A New Technology of Light.” New York: Prentice Hall (1986) 6 O’Shea, D.C., W.R. Callen and W.T. Rhodes. “Introduction to Lasers and their Applications.” Reading: Addison-Wesley Publishing Company, 1978. 7 Physics Education Research Group, Kansas State University Department of Physics, Visual Quantum Mechanics, http://phys.educ.ksu.edu/vqm/html/henelaser.html 8 University of Florida, http://itl.chem.ufl.edu/4411L_f00/i2_lif/ar_laser.html 9 Davidson College Department of Physics http://www.phy.davidson.edu/StuHome/sethvc/Laser-Final/co2.htm 10 University of Waterloo, http://www.adm.uwaterloo.ca/infohs/lasermanual/documents/section5.html 11 Niagara College, http://www.technology.niagarac.on.ca/people/mcsele/lasers/LasersYag.htm 12 . RIEGL General Information, 1994. http://www.riegl.co.at/principles/e_gi003.htm 13 Gottlieb, H. “Experiments Using a Helium Neon Laser.” Bellmawr: Metrology Instruments, Inc. (1981) 1 98 Chapter 5 Rectilinear Propagation of Light 5.1 Fundamentals of geometry and trigonometry 5.2 Laws of Reflection 5.3 Refraction and Snell’s Law 5.4 Total Internal Reflection 5.5 Dispersion 5.6 Lenses 5.7 Laboratory Experiments – Reflection, Refraction Chapter Summary End of Chapter Exercises Group Discussion Questions Chapter Objectives: After completing this chapter, the student should be able to (i) Perform ray tracing based on laws of reflection (ii) Perform ray tracing based on laws of refraction (iii) Define index of refraction, Snell’s law, critical angle and total internal reflection (iv) Explain wavelength dependence of refractive index (v) Distinguish among different types of lenses Key words: reflection, refraction, Snell’s Law, dispersion, critical angle, total internal reflection, converging and diverging lenses The term rectilinear propagation of light refers to light traveling in a straight line. This is a good approximation for many practical cases. During early morning or late evening, the shadow of a tall building conforms to its shape. In other words, a straight feature of the building does not appear curved or bent when you look at its shadow. In order to make such predictions, one should be conversant with principles of plane geometry and trigonometry. 5.1 Fundamentals of geometry and trigonometry Some principles of plane geometry are discussed first. Principle 1 If a straight line stands on another straight line, the sum of the adjacent angles is equal to 180o. D C A Figure 5.1 Illustration of Geometry Principle 1. B 99 In Figure 5.1, line CD stands on line AB. /DCA + /DCB = 180o Principle 2 If two straight lines intersect, the vertically opposite angles are equal. D A P C B Figure 5.2 Illustration of Geometry Principle 2. In Figure 5.2, AB and CD intersect at point P. Then, /APC = /DPB and /DPA = /CPB Principle 3. If a straight line intercepts a pair of parallel lines, the alternate angles are equal and the corresponding angles are equal. E A K C B D L F Figure 5.3 Illustration of Geometry Principle 3. In Figure 5.3, AB and CD are parallel to each other. EF intercepts AB and CD at points K and L, respectively. Then, /EKB = /KLD; /BKL = /DLF; /EKA = /KLC; /AKL = /CLF (corresponding angles are equal) Also, /BKL = /KLC; /AKL = /KLD (alternate angles are equal) Also, /BKL + /KLD = 180o and /AKL + /KLC = 180o Principle 4. The sum of the three angles of any triangle equals 180o. In the following principles, when two triangles are congruent, they are exactly identical in all respects, i.e. the three sides and three angles of one triangle are respectively equal to the three sides and three angles of the second triangle. In other words, one can draw out one triangle on a piece of paper cut out its outline and align it on the second triangle. Then the triangles will match up perfectly. 100 Principle 5 Two triangles are considered congruent, if two sides and the included angle in one triangle are equal to two sides and the included angle in the second triangle. Principle 6 Two triangles are considered congruent, if any side and any two angles in one triangle are equal to any side and any two angles of the second triangle. Principle 7 Two triangles are considered congruent if each side in one triangle is equal to each side in the second triangle. Principle 8 Two right triangles are considered congruent if the hypotenuse and any other side of one triangle equal the hypotenuse and any other side of the second triangle. Principle 9 Two triangles are considered to be similar to each other, if each angle in one triangle equals each angle in the second triangle. Note that while congruent triangles will be similar to each other, similar triangles are not necessarily congruent with each other. Similar triangles have proportionate sides. Principle 10 Pythagorean Theorem B c a A C b Figure 5.4 Illustration of Geometry Principle 10. Refer to Figure 5.4. Pythagorean Theorem states that c2 = a2 + b2 c = √a2 + b2 or _______ Some principles of trigonometry are discussed next. Definitions based on right triangle Refer to Figure 5.5. sin A = a/c sin B = b/c cos A = b/c cos B = a/c tan A = a/b tan B = b/a 101 csc θ = 1/sinθ sin A = cos B sec θ = 1/cos θ cos A = sin B cot θ = 1/tan θ tan A = cot B cot A = tan B (Note: The Greek letter θ is used to designate an angle. It is pronounced “theta.") B c a A C b Figure 5.5 Illustration of trigonometric relationships. In general, sin θ = cos (90o - θ ) cos θ = sin (90o -θ ) tan θ = cot (90o-θ ) cot θ = tan (90o-θ ) The values of some trigonometric functions for select angles are presented in Table 5.1. Table 5.1 o o Function 0 30 45o 60o 90o θ= 0 ½ 1 sin θ √2/2 (0.707) √3/2 (0.866) 1 ½ 0 cos θ √3/2 (0.866) √2/2 (0.707) 0 1 tan θ √3/3 (0.577) √3 (1.73) ∞ Some Trigonometric Identities: sin2θ + cos2θ = 1 1 + tan2θ = sec2 θ 1 + cot2θ = csc2 θ 5.2 Laws of Reflection Now that we have reviewed some essentials of geometry and trigonometry, we can apply some of these concepts to locate the path of light which, in this chapter, is assumed to travel in a straight line. Since it is impossible to track all the light from place to place, we will look at representative rays that originate from the object. We will figure out what happens when rays of light, which travel in straight lines, encounter objects such as a mirror or a glass block. Reflection from plane surfaces The best examples are flat mirrors which we encounter in every day life, in bath rooms, in cars, etc. We use mirrors because they reflect practically all the light back enabling objects to be seen clearly. Mirrors are coated with a thin layer of silvery material which is responsible for the reflective characteristics. A plate of glass can be coated and used as a front surface mirror or back surface mirror as shown in Figure 5.6. 102 Light Front surface mirror Back surface mirror refers to mirror coating (exaggerated) Figure 5.6 Reflection from mirrors. The same mirror can be used with its silvered surface facing the light first (left figure) or the glass facing the light first (right figure.) In each case, an arbitrary ray from the light source is shown hitting each mirror and is reflected. The following laws of reflection shown below are always obeyed1,2: (a) The angle of incidence must always equal the angle of reflection. (b) The incident ray, the reflected ray and the normal to the surface at the point of incidence must lie in the same plane. These principles are illustrated using Figure 5.7 for a plane mirror. The same rules also hold for curved surfaces which will be discussed later. O Plane mirror θi θr P N Q Figure 5.7 Laws of reflection. The incident ray PO strikes the mirror at an angle θi from the normal ON. ON is perpendicular to the mirror at O, the point of incidence. After striking the mirror, the ray is reflected and it is shown by OQ. Thus, OQ is the reflected ray. The angle of incidence is /PON = θi and the angle of reflection is /NOQ = θr. • • The first law of reflection states that θi = θr. The second law states that PO, ON and OQ lie in the same plane; in this case, the plane of the paper on which the figure is drawn. In other words, if PO and ON lie 103 on the plane of the paper, the reflected ray ON cannot, all of a sudden, come out of the plane of the paper and strike the reader! That is good to know especially with laser beams! Note that the angles of incidence and reflection are measured always from the normal, not from the mirror surface. We will maintain this rule when we get into refraction in the next section. Mirrors cause lateral inversion, i.e. your left hand appears as the right hand in the image and vice versa. Let us look at some examples which use the laws of reflection. In all the cases, the incident laser beam lies in the plane of the paper. Example 5.1 Trace the reflected laser beam off the mirror in the following Figure. reference line parallel to laser beam 60o O Incident laser beam Mirror Figure 5.8 Illustration of Example 5.1. Solution: Draw the normal ON at O, the point of incidence of the laser beam, as shown below A B reference line parallel to laser beam o 60 60o P Incident laser beam 30o N 30o O C Mirror Q Figure 5.9 Solution for Example 5.1. Note that the reference line AB is parallel to the incident beam PO. Thus, using Principle 3 of Section 5.1, we have /BAO = /AOP = 60o Since NO is perpendicular to the mirror AC, then /PON = 90o – 60o = 30o, the angle of incidence, which, per the first law should equal the angle of reflection /NOQ = 30o. 104 Answer. The reflected beam undergoes a 60o turn compared to the original beam. Example 5.2 Trace the reflected laser beam off the mirrors in the following Figure. A B reference line parallel to original laser beam 30o O Incident laser beam D C Mirror #1 45o E reference line parallel to original laser beam Mirror #2 Figure 5.10 Illustration of Example 5.2. Construct the normal to mirror #1 at point O. A B reference line parallel to laser beam 30o O P Incident laser beam 60o D Mirror #2 C Mirror #1 60o 75o N Q M 75o o 45 E reference line parallel to laser beam Figure 5.11 Solution for Example 5.2. Extend the normal ON, past mirror #2 to intersect the bottom reference line at point M. Using arguments similar to the previous example, we obtain /AOP = 30o, leading to /PON = 60o, as also the angle of reflection /NOQ. Also /NME = 60o, leading to /ONQ=105o. This leads to /OQN=180-165=15o. Thus the angle of incidence at Q for mirror #2 is 75o, resulting in the beam reflected off at 75o with respect to the normal at Q. The image obtained using a plane mirror is always virtual, i.e., it cannot be caught on a screen since it appears “behind” the mirror. For a plane mirror, the image is of the same size as the object and is as far behind the mirror as the object is in front of it. 105 Reflection at curved surfaces (spherical mirrors) If the surface is not flat then it must be curved, either convex or concave. Figure 5.12 illustrates the differences between the two. Either surface is part of a sphere. X---------------C--------F-------------------------------------------------------F--------------C-- Y Concave surface Convex surface Figure 5.12 Illustration of curved surfaces. For either mirror, the laws of reflection still hold. How do we establish the normal? The point of incidence is joined to the center of curvature C to provide the surface normal. Unlike plane surfaces where the normal to the surface at the point of incidence can quite simply be drawn using a perpendicular line, in the case of a curved surface, one has to know where the center of curvature C is of the circle (or sphere) that the mirror is part of. Once C is located, joining C with the point of incidence establishes the normal to the surface1. Then, using the law of reflection, one can trace the reflected ray. Referring to Figure 5.13, PO is the incident ray that strikes the mirror at the point O. The normal to the surface at point O is obtained by joining it to C, the center of curvature. Then, using the first law of reflection, one draws the reflected ray OQ such that /POC =/COQ in each case. The second law of reflection states that the PO, CO and OQ lay in the same plane, in this case, the plane of the paper on which the figure is drawn, as shown in Figure 5.13. incident ray P -------------C---------------------------------normal O Q reflected ray P incident ray O normal ---------------------------C-------- optical axis Q reflected ray Figure 5.13 Reflection from curved spherical surfaces. In either case, referring to Figure 5.12, there exists an optical axis XY that passes through C, the center of curvature. Any object to be imaged has to be placed standing on the axis and very close to it for the sake of mathematical simplicity of the problem. For the concave surface, C is to the left of the surface and for the convex surface, C is to the right. This is obvious since in order to draw a circle of which the curved mirror is a part, the center of the circle must be to the left for the concave surface and to the right for convex surface. Also there exists a point called F, the focal point which is approximately half way between C and the mirror. Again, this approximation holds only for short 106 objects standing on the optical axis that need to be imaged. The approximation is called paraxial approximation that basically considers rays emanating from the object that are parallel and close to the axis. In the following, light is traveling from left to right, i.e. the object is located to the left of the mirror, either concave or convex. Image formation in spherical mirrors is a direct result of the laws of reflection. These manifest as the following three rules which are mentioned without proof1,2 (for either concave or convex surface): 1. If a ray of light passes through C or is headed toward it, after reflecting off the curved surface, it will pass through C again. 2. A ray of light passing through F or is headed toward it, after reflecting off the curved surface, will become parallel to the optical axis XY. 3. A ray of light that was parallel to the optical axis will, after reflecting off the curved surface will pass through F or will appear to be coming from F. These are the three basic rules of ray tracing for curved surface mirrors. Any two of the three will enable determination of image location using a scaled graphical procedure, as shown in the example below: Example 5.3 A concave mirror has a radius of curvature 16 units. Using two of three properties above and a graphical procedure, determine the location of the image when the object is (a) 10 units, and (b) 4 units to the left of the mirror. (a) The letter C is used to denote the center of curvature and F the focal point. We use Rules 2 and 3 here. 60 Image 50 C 40 Object F 30 20 10 0 0 10 20 30 40 50 60 Figure 5.14a Illustration for Example 5.3a. An enlarged, inverted image is formed to the left of the mirror, 40 units from it. If a screen is placed there, you can capture the image on the screen. For example, if the object were a candle flame, the image will be an inverted candle flame on the screen. The image is considered real. 107 (b) We use Rules 2 and 3 here. Image 60 Object 50 F 40 30 20 10 0 0 5 10 15 20 Figure 5.14b Illustration for Example 5.3b. In this case, the image is virtual since it is formed to the right side of the mirror, 8 units from it. In other words, one can look from the left of the mirror and see the image of the object (for example a candle flame) “inside” the mirror. The image is enlarged and of the same orientation as the object. For a concave mirror, if the object is located within the focal point (denoted by F in the above figures), the image is enlarged and virtual. If the object is located outside the focal point, the image is real. It may be enlarged, diminished or of the same size as the object depending on the object location from the lens. Example 5.4 A convex mirror has a radius of curvature 16 units. Using two of three properties above and a graphical procedure, determine the location of the image when the object is (a) 10 units, and (b) 4 units to the left of the lens. 1 0.8 0.6 Object 0.4 Image 0.2 0 -0.2 0 5 10 15 20 -0.4 -0.6 -0.8 -1 Figure 5.15a Illustration for Example 5.4a. Here the image is virtual and oriented in the same way as the object and to the right of the mirror “inside” it, about 4.4 units from the surface. Again, the image is virtual and located “inside” the mirror, about 0.44 times the height of the object. 108 60 Object 50 Image F 40 30 20 10 0 0 5 10 15 20 Figure 5.15b Illustration for Example 5.4b. Convex mirrors always produce a virtual, diminished image. This principle is used in the passenger’s side mirror of the automobile. The mirror can capture diminished images of objects located quite a distance away, i.e., many cars in the right lane. Such mirrors are also used in supermarkets, as well as warehouses, and located strategically in corners so that persons/forklifts in a perpendicular aisle may be viewed and collisions averted. Concave mirrors, by virtue of their magnification property, find applications in telescopes, vanity mirrors, some dental mirrors, solar energy research, car headlights, etc. Concave mirrors can produce real or virtual images that are magnified, of the same size or diminished in size with respect to the object. While the three simple rules provided a method to obtain the image graphically, one can use any arbitrary ray at an oblique angle striking the mirror. Plane mirrors always result in a virtual image, with image size same as the object size. The applications include household, reflecting laser beams, driver’s side mirror, rear view mirror, trick photography/cinematography, kaleidoscope, periscope, etc. A few further principles of plane mirrors will be discussed next. Multiple images with plane mirrors With a single object and a single plane mirror, one obtains a single, virtual image, whose dimensions we know from experience are the same as the dimensions of the object, i.e. there is no lateral magnification. By virtue of this, the image is located as far behind the mirror as the object is in front of it, as shown in Figure 5.16. In this figure, an object “x” sits at a distance xA from a plane mirror. Use the ray xA which is perpendicular to the mirror and extend it behind the mirror. This is because the angle of incidence and angle of reflection are both zero and the reflected ray retraces the path of the incident ray. Draw an oblique ray xB (i.e. striking the mirror at an angle other than 90o.) At point B draw a normal NB. Draw the reflected ray BY such that /xBN = /NBY, obeying the first law of reflection. As far as ray BY is concerned, it appears to be coming from behind the mirror. Thus, extend YB backwards. The point, where this line intersects the extension of line xA is denoted by x’, the location of the image. By looking at the congruence of the two right triangles xAB and x’AB (use Principle 6 presented at 109 the beginning of the chapter), we can conclude that xA=Ax’, i.e. the object distance from mirror = image distance from the mirror. This holds for a plane mirror1. x A x’ | N Y B Plane Mirror Figure 5.16 Locating virtual image in a plane mirror. A single object provides a single virtual image with a single mirror, with the image distance and object distance being the same from the mirror. How about multiple mirrors? Let us look at two plane mirrors inclined at 90o with respect to each other and a single object in between them, as shown in Figure 5.17. M1 x x M2 x x Figure 5.17 Multiple virtual images in a plane mirror. Two mirrors M1 and M2 are separated by 90o and an object shown in bolded x is in between. One obtains three images, shown by italicized letter x for clarity. What if the two mirrors are separated by 60o? One would obtain 5 images. In general if θ is the angle between two plane mirrors, one would obtain (360o/θ) -1 number of images from a single object. If the two mirrors are placed parallel to each other, theoretically one would obtain infinite number of images of a single object placed in between them. Realize however that after as few passes, the intensity of the image will diminish and only a finite number will be visible. This is the theory behind some trick photography/cinematography shots. An attractive marketing strategy used in restaurants or in displays of jewelry or tempting desserts or vegetables is to place two or more mirrors at right angles to obtain multiple images creating the illusion of more, attractive product displayed! 110 We looked at two examples of diverting a laser beam using two plane mirrors placed at different angles and performed ray tracing. Let us look at other specific applications, where one can turn a laser beam by 90o or 180o. A single mirror inclined at 45o with respect to a laser beam can deflect the beam by 90o, following the laws of reflection, as shown in Figure 5.18. Incident laser beam reference line parallel to original laser beam 45o 45o Normal 45o Mirror Reflected laser beam Figure 5.18 Use of a plane mirror to deflect a laser beam by 90o. Two mirrors inclined at 45o with respect to a laser beam can deflect the beam by 180o, following the laws of reflection, as shown in Figure 5.19. Incident laser beam reference line parallel to original laser beam 45o 45o Normal 45o Final beam Mirror #1 Reflected laser beam Mirror #2 45o reference line parallel to original laser beam Figure 5.19 Use of a plane mirror to deflect a laser beam by 180o. Two mirrors inclined at 45o with respect to a laser beam can deflect the beam by a net of 0o, following the laws of reflection, however, displace the beam, as shown in Figure 5.20. Figure 5.19 and Figure 5.20 illustrate the principle of a periscope used in submarines, where the sailor can unobtrusively watch objects above the surface of the water while the submarine stays mostly submerged under water. Does one always need a silvered surface for reflection to occur? The answer is clearly no. First, from the chapter on laser safety, we discussed that reflection can occur at any surface. Fresnel reflection is the case when a laser beam strikes a lens surface and the reflections off a concave surface can converge to be hazardous. Secondly, when you look at the quiet surface a pool or pond, you notice reflections of the sky, clouds, trees, etc. Thirdly, during night time, if it is dark outside and the room light is on, you can see reflections of objects in the room clearly in window glass. While a mirrored surface has 111 a high reflectance (approaching 100%), other surfaces, such as still water and a sheet of glass, also reflect light. It is sufficient to know that for light incident more or less perpendicularly to a typical glass surface, about 4% of the light is reflected. Additional details will be presented in Chapter 7. Incident laser beam reference line parallel to original laser beam 45o 45o Normal 45o Mirror #2 Mirror #1 Reflected laser beam 45o Final beam reference line parallel to original laser beam Figure 5.20 Use of a plane mirror to displace a laser beam. Surfaces such as paper also reflect light. Since the surface is rough compared to the scale of the wavelength of light used, and since the paper absorbs some of the light, the balance light is reflected in all directions. This is known as diffuse reflection. Note that the two laws of reflection are still obeyed. The normal to the surface at the point of incidence is highly variable since the surface is rough and highly irregular. Thus, the reflected light is bounced off in all directions and the power of the reflected light measured at any location is small compared to the incident light. In contrast to diffuse reflectors and Fresnel reflectors, the power of the beam reflected off specular reflectors such as mirrors is high. Thus the reflectance is also high. Example 5.5 An object is 20 inches (50 cm) in front of a plane mirror 3’ (90 cm) x 2’ (60 cm) in dimension. Where is the image located? Answer The image is located 20 inches (50 cm) behind the mirror and is virtual. Example 5.6 Two plane mirrors are 30o apart. If a candle is placed between the mirrors, how many candle images are visible? # of images = (360o/θ) -1 = (360o/30o) -1 = 12 – 1 = 11 Answer Example 5.6 Trace the reflected ray in the following figures. Angle of incidence for the first case is 25o and for the second case is 32o. 112 incident ray P 25o -------------C---------------------------------O O incident ray o 32 ---------------------------C-------- optical axis Figure 5.21 Illustration for Example 5.6. Solution: Notice that the surface normal is OC. Draw a line that makes 25o with OC in the first case and 32o with OC in the second case. These are the reflected rays as shown below. incident ray P 25o -------------C---------------------------------normal O Q 25o reflected ray O incident ray o 32 normal ---------------------------C-------- optical axis 32o Q reflected ray Figure 5.22 Solution for Example 5.6. Example 5.7 A 2 mW laser beam is reflected off a plane mirror. The power of the reflected beam is measured to be 1.98 mW. What is the per cent reflectance of the mirror? % Reflectance = Reflected Beam Power x 100 Incident Beam Power = 1.98 mW x 100 = 99% Answer 2.00 mW Exercise 5.2A 1. Trace the reflected ray in the following figure: 22o Mirror Figure 5.23 Illustration for Exercise 5.2A, problem 1. 113 2. What are the angles of incidence and reflection in the following figure? Mark them. Mirror 112o Figure 5.24 Illustration for Exercise 5.2A, problem 2. 3. Trace the reflected laser beam off the mirror in the Figure 5.25. reference line parallel to laser beam 37o O Incident laser beam Mirror Figure 5.25 Illustration for Exercise 5.2A, problem 3. 4. Trace the reflected laser beam off the mirror in the following Figure 5.26. Incident laser beam Mirror 72o reference line parallel to laser beam Figure 5.26 Illustration for Exercise 5.2A, problem 4. 5. Trace the reflected laser beam off the mirrors in the following Figure 5.27. A B reference line parallel to laser beam 40o O Incident laser beam D C Mirror #1 55o E Mirror #2 reference line parallel to laser beam Figure 5.27 Illustration for Exercise 5.2A, problem 5. 114 6. Trace the reflected laser beam off the mirrors in the following Figure 5.28. A B reference line parallel to laser beam o 70 O Incident laser beam D C Mirror #1 55o E Mirror #2 reference line parallel to laser beam Figure 5.28 Illustration for Exercise 5.2A, problem 6. 7. A concave mirror has a radius of curvature 16 units. Using two of three properties above and a graphical procedure, determine the location of the image when the object is (a) 16 units, and (b) 6 units to the left of the lens. 8. A concave mirror has a radius of curvature 20 units. Using two of three properties above and a graphical procedure, determine the location of the image when the object is (a) 10 units, and (b) 4 units to the left of the lens. 9. A convex mirror has a radius of curvature 16 units. Using two of three properties above and a graphical procedure, determine the location of the image when the object is (a) 16 units, and (b) 6 units to the left of the lens. 10. A convex mirror has a radius of curvature 20 units. Using two of three properties above and a graphical procedure, determine the location of the image when the object is (a) 10 units, and (b) 4 units to the left of the lens. 11. Two plane mirrors are 45o apart. If a candle is placed between the mirrors, how many candle images are visible? 12. Two plane mirrors are 120o apart. If a candle is placed between the mirrors, how many candle images are visible? 13. Trace the reflected ray in Figures 5.29. Angle of incidence for the first case is 35o and for the second case is 37o. 115 incident ray P 35o -------------C---------------------------------O O incident ray o 37 ---------------------------C-------- optical axis Figure 5.29 Illustration for Exercise 5.2A, problem 13. 14. Mark the incident and reflected rays, as well as the angles of incidence and reflection for the following cases. incident ray P O -------------C---------------------------------72o Qo reflected ray incident ray 58o ---------------------------C-------- optical axis Q reflected ray Figure 5.30 Illustration for Exercise 5.2A, problem 14. 15. Prove by ray tracing that if you are a six foot tall person, you would need only a three foot tall plane mirror to view yourself completely. 16. A 5 mW laser beam is reflected off a plane mirror whose reflectance is 97.5%. What is the power of the reflected beam? 5.3 Refraction and Snell’s Law In the previous section we looked at a phenomenon that occurs when light interacts with a surface. We considered mirrors whose reflectance is usually high. We also know that even a plate of glass reflects some light back. What happens to the balance? Of the light incident on a surface, part of it is reflected, part absorbed by the material (or sometimes scattered) and the balance “refracts” into the material. Figure 5.31 illustrates this concept. A slab of glass is shown as an example above. Light practically travels at the same speed in air as in vacuum at 3 x 108 m/s. Therefore, in this section we will consider air or vacuum to be one and the same. Why is the refracted ray bent? To answer this question, analyze the phenomena of why a straw inside a glass filled with water appears bent, or why a rock at the bottom of a clear, shallow stream appears closer than it really is to the person outside. As mentioned earlier, light has its highest speed in vacuum (or air.) Thus, any medium (including air) will only slow down light. When a light ray strikes the 116 glass surface, while a portion of it is reflected and it travels at the same speed as the incident ray, the portion that travels into the glass (i.e. refracted) slows down. Incident ray Vacuum (or air) Refracted ray Some absorption Glass Transmitted ray Figure 5.31 Refraction of light in a glass slab. The degree to which light slows down is dictated by the index of refraction of the medium. Using Fermat’s principle2, the details of which are beyond the scope of this elementary book, the optical path length of light inside the glass is optimized so that the refracted ray has to bend closer to the normal (shown by a dashed line.) Clearly, the bending phenomenon minimized the distance that the light ray travels inside the medium (i.e. glass in this case.) compared to what it would have been had the ray proceeded straight into the glass without bending. The index of refraction, also known as refractive index, of the medium is crucial in tracing the ray inside the medium. Denoted by the symbol n, it is defined as the ratio of the velocity of light in vacuum (c) to the velocity of light in the medium1,2 (v.) n = c v Since light is always slowed down by any medium including air, the refractive index of all media must be greater than one (1.) In this text, we will use this definition of (absolute) index of refraction, where we use vacuum as the reference. Other books define a relative index of refraction as the ratio of the speed of light in one medium with respect to another medium; neither of the media has be a vacuum in that case. What does the index of refraction depend on? It depends upon the wavelength of light, temperature, pressure and concentration of the medium, to name a few. In general, for most materials, index of refraction decreases slightly with increasing wavelength, i.e. it is slightly higher for violet light compared to red light. A graph that depicts the variation of refractive index with wavelength is known as a dispersion curve. If you heat up most materials, its mass density decreases due to an increase in volume. If the refractive index is an indication of a medium’s ability to slow down light, as the medium is heated up (i.e. its temperature increased), the molecules move farther apart, decreasing the medium’s ability to slow down light. Thus, generally, for most media, an increase in temperature results in a decrease of the refractive index. Using similar arguments, one can state that, 117 in general, for most media, an increase in pressure results in increase in refractive index. These concepts are illustrated as follows: λ↑ n ↓ T↑ n ↓ P↑ n↑ The effect of concentration is not quite straight forward and is pertinent more to a liquid mixture of two or more components. Depending upon the concentration of the components, the refractive index varies. This technique is used in some chemical industries to measure refractive index of liquid mixtures which in turn can be translated into concentration. For this chapter, if the refractive index of a material is given, it is assumed that it is valid for that particular wavelength of incident radiation. If the wavelength is not mentioned in a problem, it is assumed that white light is used and an average refractive index (roughly corresponding to yellow part of the spectrum) is given and will be used for calculations. The wavelength dependence of the refractive index is discussed later. Example 5.8 Light from a sodium lamp (λ=589 nm) enters a certain type of glass of refractive index 1.55 at that wavelength. (a)What is the speed of light inside the glass? (b) What is the speed of light as it exits the glass into air? (a) Starting from n = c v we have v = c/n or v = 3 x 108 m/s = 1.94 x 108 m/s Answer 1.55 (b) Answer: So long as the medium of interest is air the speed of light will always be 3 x 108 m/s in air. From this example, we see that light slowed down in the glass and its speed was restored when it exited glass into air. This explains why the direction of the transmitted light is the same as the incident light. Light unbent itself to accomplish this, again following Fermat’s principle. Overall, the light ray is displaced by a certain extent which depends on the refractive index of the medium and the angle at which the ray strikes the glass, i.e., the angle of incidence. How do we predict the location of the refracted ray and the ray displacement? We need Snell’s law. Laws of Refraction: Refer to Figure 5.32. Light is traveling from medium 1 of refractive index n1 to medium 2 of refractive index n2. Then, the laws of refraction are (1) Snell’s Law1,2 118 n1sin θ1 n2sin θ2 = (2) The incident ray, the refracted ray and the normal to the surface at the point of incidence lie in the same plane. The laws of refraction hold for plane or curved surfaces and hold for any interface between two media of different refractive index. P Incident ray θ1 N1 Medium 1 θ2 N2 Medium 1 Q Refracted ray θ2 R Medium 2 N3 N4 θ1 S Transmitted ray (Note that n2 > n1 in this illustration.) Figure 5.32Illustration of laws of refraction of light. In Figure 5.32, a ray of light originally traveling in medium 1, strikes the interface to medium 2 at point Q. PQ is thus the incident ray. N1N2 is the normal drawn at the interface at the point of incidence Q. The angle between the incident ray PQ and the normal N1N2 is the angle of incidence θ1, which is in medium 1. Once light enters medium 2, following Fermat’s principle, it bends closer to the normal, since we have considered medium 2 to have a higher refractive index than medium 1. The angle that the refracted ray QR makes with the normal N1N2 is the angle of refraction. While in medium 2, the light ray proceeds in a straight line until it hits the second interface at point R of the bottom face. A normal N3N4 is drawn at point R. Note the angle of incidence that the refracted ray QR makes with the surface normal N3N4 is the same as θ2, by virtue of Principle 3 discussed in section 5.1 earlier. The ray bends away from the normal N3N4 at point R such that the angle of refraction in air is the same as θ1, the original angle of incidence. The ray RS is called the transmitted ray since it is transmitted from the block of medium 2. The second law of refraction states that PQ, N1N2 and QR lie in the same plane, i.e. the plane of the paper in this case. Applying this to the bottom face, it also states that QR, N3N4 and RS lie in the same plane, i.e. the plane of the paper in this case. Combining these principles, we can boldly state that the incident, the refracted, and the transmitted rays, and the two surface normal lie in the same plane, i.e. the plane of the paper. Principle of reversibility1,2 119 This principle merely states that if the direction of the refracted ray is reversed, the direction of the incident ray will also be reversed. In other words, what we consider as incident ray in medium 1 is also the refracted ray in medium 1 if we reverse the direction of light and say that light is now traveling from medium 2 to medium 1, as shown in Figure 5.33. Transmitted ray P θ1 N1 θ2 N2 Medium 1 Medium 1 Q Refracted ray θ2 R Medium 2 N3 N4 θ1 S Incident ray (Note that n2 > n1 in this illustration.) Figure 5.33 Illustration of the principle of reversibility of light. Let us apply Snell’s law in solving problems involving plane surfaces such as a block of glass, a crystal gem or quiet surface of a lake. Example 5.9 Light of a certain wavelength is incident upon a window glass material at 30o from air. The angle of refraction is measured to be 19o. Calculate the index of refraction of the glass. The starting equation is n1sin θ1 n2sin θ2 = Denote medium 1 for air and medium 2 for glass. Thus n1=1 (air), θ1=30o, and θ2=19o. Rearranging Snell’s law, n2 = n1sin θ1 sin θ2 = 1 sin 30o sin 19o = 1.54 Answer Example 5.10 Light traveling in air hits a quiet surface of a pond at an angle of incidence of 42o. If the index of refraction of water is 1.333, what is the angle of refraction in water? Medium 1 is air and medium 2 is water. n1=1, θ1=42o and n2=1.333, θ2=? Rearranging Snell’s law, we have 120 sin θ2 = n1sin θ1 n2 To solve for θ2, take inverse on both sides, i.e. sin-1 [sin θ2] = sin-1 n1sin θ1 n2 where sin-1 is known as the inverse sine or arc sine. Taking the inverse sine of a sine function gives back the angle itself. Thus, θ2 = sin-1 n1sin θ1 n2 Substituting the known values, θ2 = sin-1 1sin 42o 1.333 o = 30.1 Answer Example 5.11 Light enters from air to water (n=1.333). The angle of refraction in water is seen to be 22o. What is the angle of incidence? n1=1, n2=1.333 and θ2=22o Rearranging Snell’s law, we have sin θ1 = n2sin θ2 n1 or θ1 = sin-1 n2sin θ2 n1 θ1 = sin-1 1.333sin 22o 1 30o Answer = Exercise 5.3A 1. Calculate the speed of light in the following materials with the given refractive indices: (a) 1.44 (b) 1.67 (c) 2.00 and (d) 2.72 2. What is the refractive index of each material if the corresponding speed of light inside each respective material is (a) 1.44 x 108 m/s (b) 2.1 x 108 m/s, and (c) 1.75 x 108 m/s? 3. Light is traveling from air into glass of refractive index 1.50 at the given wavelength. Calculate the angle of refraction corresponding to an angle of incidence of (a) 20o, (b) 35o, (c) 45o and (d) 67o. 121 4. Light is traveling from water (n=1.333) to glass (n=1.555). Calculate the angle of refraction corresponding to an angle of incidence of (a) 20o, (b) 35o, (c) 45o and (d) 67o. 5. Light is traveling from glass (n=1.57) into air. Calculate the angle of refraction corresponding to an angle of incidence of (a) 20o, (b) 35o, (c) 45o and (d) 67o. What happens in cases (c) and (d)? 6. Light is traveling from air to water (n=1.333.) Calculate the angle of incidence corresponding to an angle of refraction of (a) 10o, (b) 25o, (c) 35o and (d) 67o. What happens in case (d)? 7. The following observations were noted in a laboratory experiment when He:Ne laser beam (λ=632.8nm) was traveling from air to synthetic glass material Angle of incidence Angle of refraction 15o 20o 25o 30o 35o 40o 9o 12o 15o 18o 21o 23.5o Draw a graph of sin θ1 on the x-axis and sin θ2 on the y-axis. Draw a smooth line connecting the data points. Determine the index of refraction from the reciprocal of the slope. Beam displacement2 Let us calculate the beam displacement for light traveling from a medium of lower refractive index to one of higher refractive index and finally transmitting into the original medium of lower refractive index. As we discussed, the transmitted beam is parallel to the incident beam, however, is displaced by a certain distance called displacement, d. The thickness of the slab is dented by t. It is further assumed that the slab is long enough that light will exit from the opposite face. Later we will discuss other cases where the beam may exit from the side faces. In Figure 5.34, t is the thickness of slab of medium 2. Light is traveling from medium 1, which in most cases is air. d is the beam displacement, i.e. the shortest distance between the transmitted ray and the extension of the incident ray, shown in dashed line, as though it did not undergo refraction. The displacement we need is the same as the distance N3T, the line drawn from N3 and perpendicular to the dashed line QW. From triangle QN3T, /N3QT = θ1 - θ2 Therefore d = N3T = QN3sin(θ1 - θ2 ) From triangle QN2N3, we have 122 P Incident ray θ1 θ2 t N1 Medium 1 Q Refracted ray θ2 T R N2 N3 V Medium 1 d N4 θ1 Transmitted ray Medium 2 S W Figure 5.34 Illustration for beam displacement calculation. QN3 = QN2 = cos θ2 t . cos θ2 Substituting for QN3 in the equation for d, we obtain, d = t sin(θ1- θ2) ------------cos θ2 The angle of refraction first needs to be calculated from Snell’s law, given the angle of incidence. Then, knowing the angles of incidence and refraction and the slab thickness t, one can calculate the beam displacement easily using this equation. Example 5.12 A 20 mm thick paperweight is placed on a typed piece of paper. The paperweight is made of a glass material whose refractive index is 1.57. If the typist views the paperweight at a 35o angle from the vertical, how far is the image of the print displaced compared to its actual location? Given n1=1, θ1=35o , and n2= 1.57, we first calculate θ2 using θ2 = sin-1 n1sin θ1 n2 o = 21.4 Now we use 123 d = with t=20 mm d = t sin(θ1 - θ2 ) cos θ2 20 sin(35o – 21.4o ) cos 21.4o = 5.04 mm Answer As the thickness of the slab is increased the beam displacement is also increased proportionately. The print appears to be about 5 mm off from the original location caused by beam displacement. This phenomenon is mistakenly called the lens effect. It is clearly not so. There is no magnification involved here. The size of the print remains the same; it is only displaced. Example 5.13 A thin, transparent bag is laid flat. It contains water. When viewed at a 45o angle from the vertical, a pattern on the tablecloth on which the bag is placed appears 1 cm off. What is the thickness of the bag? Neglect the effects of the thin bag material itself. The refractive index of water is 1.333. As before, Given n1=1, θ1=45o , and n2= 1.333, we first calculate θ2 using θ2 = = Now we use d 1 sin-1 n1sin θ1 n2 32.0o = t sin(θ1 - θ2 ) cos θ2 = t sin(45o – 32o ) cos 32o t= 3.77 cm Answer Exercise 5.3B 1. Calculate the beam displacements for the following angles of incidence: (a) 10o, (b) 25o, (c) 35o and (d) 67o. Light is entering a long glass slab 15 mm thick. The refractive index of the glass is 1.63. 2. Calculate the slab thickness for the following angles of incidence: (a) 10o, (b) 25o, (c) 35o and (d) 67o. Light is entering a long glass slab and is displaced by 15 mm in each case. The refractive index of the glass is 1.56. 3. At what angle should light strike a glass slab from air so that the beam displacement is one half of the slab thickness. The refractive index of glass is 1.5. 124 4. At what angle should light strike a glass slab from air so that the beam displacement is equal to the slab thickness. The refractive index of glass is 1.5. Small angle approximation How small is small as far as an angle is concerned? To accomplish this let us calculate the three basic trigonometric functions, sine, cosine and tangent of various angles expressed in degrees and radians. The radian measure is derived from a circle. Since the circumference of a circle is given by 2πr and a circle has 360o of angular measure at the center, the radian measure was devised to denote 360o to equal 2π radians. From this, we get the following relationships: Angle in degrees = Angle in radians x 180o π and Angle in radians = Angle in degrees x π 180o Table 5.2 presents the value of angle in degrees and radians, as well as sine, cosine and tangent of those angles. Angle (deg) Table 5.2 Angle Sine Cosine Tangent (radians) 1 0.017453 0.017452 0.999848 0.017455 2 0.034907 0.034899 0.999391 0.034921 3 0.052360 0.052336 0.99863 0.052408 4 0.069813 0.069756 0.997564 0.069927 5 0.087266 0.087156 0.996195 0.087489 7.5 0.130900 0.130526 0.991445 0.131652 10 0.174533 0.173648 0.984808 0.176327 12 0.209440 0.207912 0.978148 0.212557 14 0.244346 0.241922 0.970296 0.249328 16 0.279253 0.275637 0.961262 0.286745 18 0.314159 0.309017 0.951057 0.32492 20 0.349066 0.34202 0.939693 0.36397 25 0.436332 0.422618 0.906308 0.466308 30 0.523599 0.5 0.866025 0.57735 40 0.698132 0.642788 0.766044 0.8391 60 1.047198 0.866025 0.5 1.732051 80 1.396263 0.984808 0.173648 5.671282 90 1.570796 1 0 ∞ As we can see, the sine and the tangent approximate the value of the angle itself in radians very well for small angles. Even for an angle of 20o (0.349 radians), sine of that angle is 0.342 and tangent of the angle is 0.364; both are within about 4% of 0.349. 125 Value of trigonometric function Graphical illustration is shown in Figure 5.35. 1.5 1 0.5 0 -0.1 0.1 0.3 0.5 0.7 0.9 1.1 1.3 1.5 Angle, radians Angle sine cos tan Figure 5.35 Small angle approximation. Also, cosine of the angle has decreased only 6% at 20o angle from its starting value of 1.0. Thus, the small angle approximation states sin θ ~ tan θ ~ θ for small angles so long as θ is expressed in radians and cos θ ~ 1 This is a useful approximation in geometric optics since it simplifies the mathematics tremendously. Consider the following example. Example 5.14 An angler looks down into a clear stream more or less vertically. (S)he sees a trout, which appears to be 2 feet (60 cm) from the surface. What is the actual distance of the trout from the surface? We know that objects inside a medium of higher refractive index appear bent to an observer in air. A straw in a clear beverage cup filled with liquid appears bent to the person consuming the beverage. The object also appears to be closer than it really is. In Figure 5.36, the point P denotes the trout. The image of the trout as it appears to the angler is located at P’. Clearly P’ is closer to the surface compared to P. Compared to the normal, the point P subtends an angle θ1 and the point P’ subtends an angle θ2. Thus, if x is the horizontal distance of the observer from the trout1, x = s’tan θ2 Also x = s tan θ1 giving us the relationship 126 s tan θ1 = s’ tan θ2 θ2 x OBSERVER AIR θ1 s s’ P’ θ2 P WATER Figure 5.36 Use of small angle approximation. or s’ s = tan θ1 -------tan θ2 For small distances of x, i.e. the observer looking at the object nearly vertically as stated in the problem, we can approximate the sine of an angle to the tangent of the angle (small angle approximation.) Thus sin θ1 = and sin θ2 = giving us s’ = s tan θ1 tan θ2 sin θ1 sin θ2 By Snell’s law the right hand side equals 1/n, the index of refraction. Thus or s’/s s’ = = 1/n s/n or s = s’n Given s’=2 feet and n= 1.333 for water, we obtain s = 2 x 1.333 = 2.67 feet (2’8”) (80 cm) Answer Thus, the trout is actually 2 ft 8 in below the surface, 8 inches (20 cm) lower than what it appeared to be. Again, some people mistakenly attribute the closeness of the object under water to the lens effect. The fish under water as viewed from air has the same size 127 as the fish that is caught and is in air. Refraction at plane boundaries causes no magnification. Refraction through several media placed in series Example 5.15 (a) Trace the light ray, mark all the angles and calculate the speed of light in each of the material for the following case. (b) What is the velocity of light in each medium? 42o 1 Air 2 Glass (n=1.485) 3 Glass (n=1.667) 4 Gem (n=2.105) 1 Air Figure 5.37 Illustration for Example 5.15. The media are marked 1,2,3,4, and 1 respectively. Refraction will occur at each of the four boundaries. At each boundary, Snell’s law will apply. For example at the 1-2 boundary, we have n1sin θ1 = n2sin θ2 with n1=1, =42o and n2=1.485. Solving we obtain, θ2 = 26.8o At the 2-3 boundary, we have n2sin θ2 = n3sin θ3 Substituting the known values, we obtain θ3 = 23.7o At the 3-4 boundary, we have n3sin θ3 = θ4 = 18.6o At the 4-1 boundary, we have n4sin θ4 128 n4sin θ4 θ1 n1sin θ1 = = 42.2o Thus the formulations are correct since we obtain the original angle to the accuracy of the round offs. 42o 1 2 Air 26.8o 26.8o Glass (n=1.485) 3 4 1 23.7o Glass (n=1.667) 18.6o Gem (n=2.105) 42o Air Figure 5.38 Solution for Example 5.15. Please note that you will get exactly the same results for the angles if you pretended that light was coming directly from air at an angle of incidence of 42o into each medium directly and apply Snell’s Law. After you have convinced yourselves that you obtain the same answer, ponder why. (b) To calculate the velocity of light in each medium simply use the relation v = c n For medium 2, we have v For medium 3, we have v For medium 4, we have v = 3 x 108 / 1.485 = 2.02 x 108 m/s Answer = 3 x 108 / 1.667 = 1.80 x 108 m/s Answer = 3 x 108 / 2.105 = 1.43 x 108 m/s Answer The velocity of light in any medium does not depend on its history, i.e. it does not matter where the light traveled from. So long as light is in a medium, all that matters for speed calculations is the refractive index of that medium. Example 5.16 What is the (a) frequency of light and (b) wavelength of light in each of the media in Example 5.15 above, if the original incident beam had 633 nm wavelength photons? 129 While some photons are lost due to reflection and absorption when light enters from vacuum/air into a medium, the photons that successfully leave the medium have the same energy as the incident photons. Since the photon energy is given by hν, and if the photon energy is the same no matter which medium the photon is present in, this implies that the frequency ν remains unaltered no matter where the photon is. (a) ν = c/λ = 3 x 108/(633 x 10-9) = 474 THz Answer in all the media (b) λ = v/ ν where v is the velocity in the medium given by v = c/n Substituting for v and simplifying we obtain λmedium = λvacuum/n (n is the absolute index of refraction). For medium 2, λmedium = 633 nm/1.485 = 426 nm Answer For medium 3 λmedium = 633 nm/1.667= 380 nm Answer For medium 4 λmedium = 633 nm/2.105= 301 nm Answer The wavelength of 633 nm is restored in the final transmitted beam. Thus the red colored beam, turns violet colored in medium 2 and changes to UV wavelengths in media 3 and 4! How is it that beam inside these media still appears red to an outside observer? The key word here is “outside.” The observer is in the air. Therefore, when the light which was inside media 2, 3 or 4 reaches the observer, it has already traveled in air. Next time, you are completely inside a swimming pool observe a red colored light which is located in air above the pool and note what color it looks to you. Caution: Do not use a laser! Exercise 5.3C 1. An air bubble is 5 mm below the surface of a glass paperweight (n=1.66.) Where does it appear to be when viewed more or less vertically? 2. A fish appears to be 42 cm from the surface when an angler is viewing it more or less perpendicularly. What is the actual location of the fish? 3. (a) Trace the light ray, mark all the angles and calculate the speed of light in each of the material for the following case. (b) What is the velocity of light in each medium? (c) What is the wavelength of light in each medium if the original wavelength was 710 nm? 130 52o 1 Air 2 Glass (n=1.585) 3 Glass (n=1.767) 4 Gem (n=2.105) 1 Air Figure 5.39 Illustration for Exercise 5.3C, problem 3. 4. (a) Trace the light ray, mark all the angles and calculate the speed of light in each of the material for the following case. (b) What is the velocity of light in each medium? (c) What is the wavelength of light in each medium if the original wavelength was 589 nm? 48o 1 Air 2 Glass (n=1.685) 3 Glass (n=1.467) 4 Gem (n=2.105) 1 Air Figure 5.40 Illustration for Exercise 5.3C, problem 4. 5.4 Total Internal Reflection As we have seen, when light travels from a medium of lower refractive index to one of higher refractive index, it bends towards the normal. The opposite is true when light travels from a medium of higher index to one of lower index. Figure 5.41shows what happens when the angle of incidence is progressively increased. 131 Case A Case B Case C Case D 90o θc Lower n Higher n θ> θc (Condition of TIR) Figure 5.41 Total Internal Reflection (TIR). In cases A and B the light ray gets refracted and passes into the medium of lower refractive index. As the angle of incidence is progressively increased, in case C, it reaches a point at which the angle of refraction reaches 90o, i.e. the refracted beam just grazes the interface between the two media. The angle of incidence that corresponds to case C is known as the critical angle θc. At θc, the maximum possible value of the angle of refraction (90o) occurs because if the angle of incidence is increased any further, light cannot pass into the medium of lower refractive index. Rather, it will be reflected back into the first medium following the laws of reflection, as in Case D. This phenomenon is known as total internal reflection1,2 (TIR) and is the basis for light propagation in optical fibers. The phenomenon also results in the fish eye effect where a fish (or a person) inside a pool of water (refractive index=1.333) can only see objects in the air that lie within a circle of fixed diameter. Objects in air outside this circle are not visible to the fish. Outside this circle, the fish sees only reflections of the pool bottom. Birds of prey capitalize on this phenomenon by swooping at a shallow angle into the water and catching the fish by surprise! TIR can also explain the phenomenon of mirage that occurs on hot days when one sees a reflection of the sky on the pavement or sand resembling water. Example 5.17 Calculate the critical angle for light traveling from a glass (n=1.567) to air. Denote medium 1 as glass and medium 2 as air. Given n1=1.567, θ1 =?, n2=1, θ2 = 90o. Apply Snell’s law, replacing θ1 with θc 1.567sin θc = 1sin 90o gives θc = 39.7o Answer Angles of incidence higher than the critical angle will result in TIR and no refraction. Example 5.18 Calculate the critical angle for light traveling from a glass (n=1.567) to water. Denote medium 1 as glass and medium 2 as water. Given n1=1.567, θ1 =?, n2=1.333, θ2 = 90o. 132 Apply Snell’s law, replacing θ1 with θc 1.567sin θc = gives θc 1.333sin 90o = 58.3o Answer Example 5.19 A fish is inside a pool of water at a depth of 3 feet (90 cm) from the surface. Calculate the critical angle and the diameter of the cone within which the fish can see objects in the air. Refractive index of water is 1.333. r 90o Air 3 feet θc Water θc FISH Figure 5.42 Illustration for Example 5.19. θc = sin-1[n2sin θ2] [ n1 ] where subscript 1 refers to water and subscript 2 refers to air. We know that θ2 = 90o and thus sin 90o = 1. Further, n2=1 and n1=1.333. Substituting these values, we obtain θc = sin-1[1x1/1.333] = sin-1[0.75] = 48.6o Answer The fish can see objects that are within a cone of half-angle 48.6o and height 3 feet. The problem asks for the diameter of the cone. or tan θc = r/3 r = = 3 tan θc 3 tan (48.6o ) = 3.4 feet Diameter of the cone = 2 r = 2 x 3.4 = 6.8 feet (2.07 m)Answer Example 5.20 Trace the beam in the glass block shown in Figure 5.43. Show all dimensions where the ray strikes different surfaces until (if) it exits the block. The refractive index of the block is 1.56. (Please note that the diagram is not to scale) 133 3 cm 40o 7 cm 20 cm Figure 5.43 Illustration for Example 5.20. Unlike ray tracing is some previous examples where the ray always exited the opposite side undergoing refraction only, in this case because of limiting dimensions, we should always check for possibility of total internal reflection at interfaces. For this, first calculate the critical angle for light traveling from glass to air as θc= sin-1 (1/1.56) = 39.9o Now we know that if light strikes any of the faces from inside the glass at an angle exceeding 39.9o, it will undergo TIR. Let us call calculate the angle of refraction at the left face given θ1 = 40o, n2=1.56 and n1=1 using Snell’s law as θ2 sin-1 n1sin θ1 n2 o 24.3 (Diagram not to scale) = = P 3 cm 40o Q 24.3o T 24.3o 7 cm R 20 cm Figure 5.44 Solution for Example 5.20, step 1. Let us denote the progression of the ray inside the block denoted by QR. We assumed that the ray hits the bottom face first. We need to verify this. If, by calculations, the length of TR is less than 20 cm, then the assumption is correct. Examine triangle QTR where /QRT = 24.3o using Principle 3 for section 5.1. tan /QRT = QT/TR 134 = 7 cm/tan 24.3o = 15.5 cm Yes, the beam will strike the bottom face at 15.5 cm from the bottom left edge at an angle of incidence of 90o-24.3o = 65.7o or TR P 3 cm 40o Q 7 cm T 24.3o 65.7o 65.7o 24.3o R 15.5 cm S U 4.5 cm 20 cm Figure 5.45 Solution for Example 5.20, step 2. The angle of incidence at the bottom face of 65.7o exceeds the critical angle of 39.9o, and therefore TIR will occur at the bottom face. In other words, the beam cannot exit the block into air through the bottom face! Laws of reflection will be followed such that the reflected ray RS makes 65.7o with the normal to the bottom face at R. Again we made an assumption. Let us verify if the ray will hit the right face next rather than the top face. Look at triangle RSU where /SRU = 90o-65.7o = 24.3o. or tan /SRU = SU/RU SU = 4.5 cm.tan24.3o = 2.03 cm Now we can be confident that the beam indeed strikes the right face since 2.03 cm is less than 10 cm, the total. Will the beam again undergo TIR? We need to verify this. The angle of incidence at the right face is 24.3o which is less than the critical angle of 39.9o. Thus, no TIR will occur at the right face (see Figure 5.46).. Rather, the beam will exit from the block into air at the right face. Following the principle of reversibility, the angle of refraction in air will be 40o (compare with the right face) and the beam will exit at a point about 2 cm from the bottom right corner through the right face of the block. Answer It is important to perform these systematic calculations. Otherwise, one might think that the beam may exit the right face inclined downward. Had it been a laser beam with the student holding the block against the beam as shown in the figure, the student might be surprised when the exiting laser beam hits her/his eye! The beam has undergone a total 135 angular deflection of 80o from its original path. Notice also that consequently the transmitted beam is no longer parallel to the incident beam as in the case of refraction through thin transparent slabs. P 3 cm Transmitted beam o 40 24.3o 65.7o 65.7o 24.3o R Q 7 cm T 15.5 cm 40o 2.03 cm S U Angle of incidence=24.3o 4.5 cm 20 cm Figure 5.46 Solution for Example 5.20, step 3. Similar principles are used in periscopes. We looked at periscopes in the previous section where a beam can be deflected by 90o or 180o. TIR can accomplish the same effect. In fact, in many applications, glass prisms that bring about TIR are preferred over mirrors whose silvery finishes are likely to degrade with time and need replacement. Since typical glass has a refractive index around 1.5, the critical angle for light traveling from glass to air can be easily calculated to be sin-1(1/1.5) = 41.8o. If a prism can be fabricated so that light will be incident on one of the interior faces at 45o, then TIR will occur since 45o > 41.8o. Thus, a 45o-45o-90o prism can be used in two configurations to deflect light as shown in Figure 5.47. These are also called Porro prisms. 45o 45o 45o 45o 45o 45o 90o 45o 90o o 45 (180o rotation) (90o rotation) Figure 5.47 Total reflecting prisms. The 180o rotation principle can also be used to invert images and finds applications in binoculars as shown above. Also, one can use it in reflecting telescopes (Newtonian telescope), shown in Figure 5.48. 136 OBSERVER Lens Distant object Concave Mirror Prism Figure 5.48 Reflecting telescope. Two such 45o-45o-90o prisms can also be used to construct a periscope as shown in Figure 5.49. Sea level Enemy Ship Periscope Submarine Observer Figure 5.49 Periscope. Other physical phenomena associated with TIR The phenomenon of mirage is due to TIR. On a hot day, surfaces such as roads and sand are readily heated by the sun. The air closest to the surface is heated rapidly while air farther away from the surface less rapidly. This results in a temperature gradient of the air, with the highest air temperature closest to the road or sand surface while the air farthest away is relatively unaffected, as shown in Figure 5.50. As we learned in the beginning of section 5.3, refractive index decreases with increasing temperature. For most practical purposes, we have considered the refractive index of air as being one, without much error. In reality, if we express it accurately it is 1.0003 at 20oC. As air is heated from 20oC (68oF), its refractive index decreases reaching closer to one but never really reaching it. Thus, as shown in the figure below, one can, for mathematical purposes divide the air into segments of different temperature zones, and for convenience label them “Extremely Hot”, “Very Hot”, “Hot”, “Warm”, and “Cool.” Light traveling from the “sky” (or clouds or tall palm trees) originates from relatively cool air with relatively high refractive index. As light travels towards the ground, it 137 passes through air layers of increasing temperature or decreasing refractive index. As we discussed earlier, when light travels from a medium of higher refractive index to lower refractive index, it bends away from the normal. As shown in the figure above, it bends continuously since the refractive index is continuously decreasing as light is approaching the pavement/sand surface. At some point, the light will undergo total internal reflection. When this happens, it will appear to the observer close to the surface that light is actually originating from the sand/pavement. SKY...CLOUD…….. AIR Air Temperature cool warm hot OBSERVER very hot extremely hot Image of Sky, Cloud, etc... SAND/PAVEMENT Figure 5.50 Mirage. Thus, reflections of objects in the “sky” including clouds, tall tree tops or mountains are seen in the sand/pavement, giving the illusion of reflection from water. No wonder the thirsty desert traveler is frequently fooled by the appearance of water. This phenomenon is called mirage. When traveling on highways on hot days, again one can notice mirage by way of a shimmering road surface. It is not due to the tar on the road melting. It is the reflection of the sky. The slight movement is due to convective air currents near the surface, causing local temperature variations. The last point can also be observed when you look at the hot hood of a car where “vapor like” refractive index gradient patterns can be observed. The same explanation holds to distinguish far away planets from stars. Stars twinkle, planets do not. This is because the temperature of stars (for example, our own sun) is much higher than the planets, which have no sources of energy of their own. The temperature gradients around stars result in refractive index changes which manifest as “twinkling.” Light from planets does not twinkle. Seeing the sun before it has actually risen is another example of TIR. As a result of refractive index differences, light bends so much so that while the sun itself has not “officially” risen, it is visible from the earth. This is sometimes termed false dawn. Conditions have to be just right to observe and recognize this phenomenon. Exercise 5.4A 1. Calculate the critical angle for each of the fictitious materials with the given refractive indices: (a) 1.44 (b) 1.62 (c) 1.77 (d) 2.25 (e) 3.00 (f) 5.00. 138 Draw a smooth graph showing refractive index on the horizontal axis and the critical angle on the vertical axis. 2. Calculate the critical angle for each of the fictitious materials with the given refractive indices: (a) 1.33 (b) 1.52 (c) 1.87 (d) 2.55 (e) 4.00 (f) 6.00. Draw a smooth graph showing refractive index on the horizontal axis and the critical angle on the vertical axis. 3. Trace the beam in the glass block shown below. Show all dimensions where the ray strikes different surfaces until (if) it exits the block. The refractive index of the block is 1.58. (Diagram not to scale) 3 cm 30o 7 cm 10 cm Figure 5.51 Illustration for Exercise 5.4A, problem 3. 4. Trace the beam in the glass block shown below. Show all dimensions where the ray strikes different surfaces until (if) it exits the block. The refractive index of the block is 1.414. (Diagram not to scale) 5 cm 45o 5 cm 15 cm Figure 5.52 Illustration for Exercise 5.4A, problem 4. 5. Trace the beam in the glass block shown below. Show all dimensions where the ray strikes different surfaces until (if) it exits the block. The refractive index of the block is 1.89. 139 (Diagram not to scale) 5 cm 75o 5 cm 15 cm Figure 5.53 Illustration for Exercise 5.4A, problem 5. 6. Trace the beam in the glass block shown below. Show all dimensions where the ray strikes different surfaces until (if) it exits the block. The refractive index of the block is 1.89. Light is normally incident on the prism first. 4” 45o 2” 45o Figure 5.54 Illustration for Exercise 5.4A, problem 6. 5.5 Dispersion Thus far, we have discussed very little concerning the refractive index dependence on the wavelength except to say that, in general, refractive index of most material decreases slightly with increasing wavelength, especially in the 300 nm-10 µm range where we typically operate lasers. Figure 5.55 shows the wavelength dependence of selected materials3-6. White light is composed of wavelengths roughly in the 400 nm – 700 nm range. When white light impinges on, say a glass material, it will disperse, i.e. each wavelength will be refracted differently. An exaggerated view of the phenomenon is shown in Figure 5.56. Violet light (λ=400 nm) has a higher refractive index than red (λ=700 nm). As a result, for the same angle of incidence, violet light has a relatively smaller angle of refraction compared to red light. This implies that violet light undergoes greater deviation than red light. Due to dispersion, white light is broken up into its individual components, namely violet, blue, green, yellow, orange and red. Only three colors are shown in the diagram above for clarity. Also, the spread of the colored rays is exaggerated for clarity. 140 Refractive Index Refractive Index vs. Wavelength for select materials 1.9 1.85 1.8 1.75 1.7 1.65 1.6 1.55 1.5 1.45 1.4 100 Sapphire Quartz CaF2 Crown Glass 1000 10000 Wavelength (nm) Figure 5.55 Dependence of refractive index on wavelength of select materials. WHITE LIGHT VIOLET AIR AIR GREEN GLASS RED Figure 5.56 Dispersion of white light inside glass (exaggerated). If white light can be dispersed using glass or water, how is it that one does not see the individual colors as light exits a flat slab? The operand word is “flat.” For a flat slab, the dispersed colors reconstitute upon exiting, and, thus, the transmitted beam is white. In order to keep the colors separate and obtain a rainbow spectrum, one needs to have nonparallel faces, such as in a prism. With a prism, the dispersed light inside the prism is never allowed to reconstitute, as shown in Figure 5.57. Rather, the differently colored rays are separated farther at the second prism face leading to a well defined rainbow pattern which can be captured on a screen1,2. 141 Red Red White light Green Violet Figure 5.57 Dispersion of white light inside a prism (exaggerated). Rainbow formation follows this principle. Substitute the glass prism with a water drop, sunlight for white light, the sky for the screen, and the result is the rainbow which is always formed in the direction opposite to the sun. Sometimes reflection produces a secondary rainbow which is not as bright as the primary rainbow. In any instance where white light is broken down into its constituents, the phenomenon of dispersion takes place. The case with the rainbow or prism is dispersion due to refraction. Dispersion can also happen due to interference (gasoline film on water) and diffraction (looking at a compact disk and tilting it from side to side.) 5.6 Lenses Lens materials are so chosen for optimum performance at definite wavelengths of interest. They need to transmit a high percentage of the incident light of whatever wavelength. In a lens, refraction takes place at two curved surfaces. The surfaces may have some of the following configurations1,2. More detailed discussion of lenses is provided in Chapters 10 and 11. Positive lenses (or converging lenses or convex lenses:: thicker at the center) Figure 5.58 Some examples of positive lenses. Figure 5.58 and Figure 5.59 provide examples of both types of lenses. The laws of refraction are still obeyed. The reader is referred to other books in optics to understand the lens maker’s equation and imaging using lenses. It is sufficient to know in this chapter the following preliminary rules1,2. More details are presented in Chapters 10 and 11. 142 Negative lenses (or diverging lenses or concave lenss : thinner at the center) Figure 5.59 Some examples of negative lenses. Rules: 1. A real image can be obtained only with a positive lens. With a single positive lens, the real image is always inverted. 2. Virtual images can be obtained using either type of lenses. 3. A negative lens always produces only a virtual image whose size is smaller than the object. 4. A positive lens is thickest in the middle; a negative lens is thinnest in the middle. 5. A parallel beam of light such as from the sun or from a laser (which is directional and coherent) will be focused to a spot at the focal point F by a positive lens. F‘ Figure 5.60 Parallel beam of light focused by a positive lens. (Never focus a laser beam on to a power meter detector. It will damage the coating.) 6. One obtains a divergent beam when a parallel beam is incident on a negative lens. Negative lenses are also called diverging or concave lenses. F’ ‘ Figure 5.61 Parallel beam of light expanded by a negative lens. 143 7. Multiple lenses can be arranged to obtain a high degree of magnification with applications in telescopes, microscopes and binoculars. 8. When using high energy pulsed excimer lasers operating at UV wavelengths, due to repeated exposure to laser pulses, some lens compaction may take place, i.e. the center of the lens is flattened somewhat by the pulses. 5.7 Suggested Laboratory Experiments – Reflection, Refraction7 Reflection of light: 1. Set up a He:Ne laser. Observe safety precautions. Obtain an optical axis by ensuring that the laser beam is centered on a cross hair target at both ends of the meter rail7. 2. Set up a rotating table with a circular graph paper affixed to it and graduated in degrees. 3. Mount a plane mirror with the silvered side facing the laser beam (i.e. front surface mirror) on the table. 4. Rotate the table such that the zero degree mark of the graph paper lines up with the laser beam. Adjust the mirror such that the reflected beam retraces the path of the incident beam. This establishes the surface normal at the point of incidence. 5. Determine the power of the incident beam. 6. Rotate the table by an additional 10o. 7. Locate the reflected beam and determine the angle of reflection. Measure the power of the reflected beam. 8. Record in Table 5.3 below. 9. Repeat steps 6-8 until 80o. 10 Complete the calculations in Table 5.3. Note: Percent reflectance = Reflected Beam Power x 100 Incident Beam Power 11. Replace the plane mirror with a concave mirror and repeat steps 4-10 recording answers only for the selected angles in Table 5.4. 12. Comment on your observations and conclusions from Table 5.4. 144 13. Replace the cocave mirror with a convex mirror and repeat steps 4-10 recording answers only for the selected angles in Table 5.5. 14. Comment on your observations and conclusions from Table 5.5. Angle of incidence Table 5.3 (Plane Mirror) Angle of reflection Power of reflected beam % Reflectance Angle of incidence Table 5.4 (Concave Mirror) Angle of reflection Power of reflected beam % Reflectance Table 5.5 (Convex Mirror) Angle of reflection Power of reflected beam % Reflectance 10o 20o 30o 40o 50o Angle of incidence 10o 20o 30o 40o 50o To verify Snell’s Law7 1. Set up a He:Ne laser. Observe safety precautions. 2. Set up a rotating table with a circular graph paper affixed to it and graduated in degrees. 3. Place a flat glass block on the table. 4. Rotate the table such that the zero degree mark of the graph paper lines up with the laser beam. Adjust the block such that the incident beam is perpendicular to the leading edge of the block. This establishes the surface normal at the point of incidence. 5. Determine the power of the incident beam. 6. Rotate the table by an additional 10o. 145 7. Locate the refracted beam and determine the angle of refraction. Locate the transmitted beam. Measure the power of the transmitted beam and the displacement from the incident beam. 8. Record in Table 5.6 below. 9. Repeat steps 6-8 until 50o. 10. Complete the calculations in Table 5.6 (Note: Percent transmittance = Transmitted Beam Power x 100 Incident Beam Power Index of refraction of block, n = sin (θ1 )/ sin (θ2) 11. Measure the thickness of the block with a ruler. Verify using formula if the measured beam displacement agrees with the calculated values. Table 5.6 (Refraction) Angle of Angle of Calculated Power of % Measured Calculated incidence refraction n transmitted Transmittance beam beam beam displacement displacement 10o 20o 30o 40o 50o 12. Comment on your observations and conclusions from Table 5.6. 13. Besides calculating n individually in Table 5.6, can you, using a simple graphical procedure, determine n that takes into account all the data points at once? • • • • • Chapter Summary Rectilinear propagation of light theory posits that light rays travel in straight lines and will continue doing so until they encounter a medium with different optical properties. When light encounters a surface, plane or curved, part or most of the light may be reflected following the laws of reflection. The laws of reflection state that (i) the angle of incidence and angle of reflection must be equal, and (ii) the incident ray, the reflected ray and the surface normal at the point of incidence lie in the same plane. Plane mirrors, while causing lateral inversion, do not magnify or invert the image. Concave mirrors can magnify or diminish the image and produce real or virtual images, while convex mirrors always produce a virtual, diminished image of the object. The index of refraction of a medium is the ratio of the velocity of light in vacuum to that in the medium. Also called the refractive index, denoted by n, its value is always greater than one (1.) Thus, light always slows down, even in air. 146 • • • • • • • • • • Snell’s law relates the sine of the angle of incidence to the sine of the angle of refraction through the index of refraction of the two media in contact at an interface. The second law of refraction states that the incident ray, the refracted ray and the normal to the surface at the point of incidence lie in the same plane. Be it reflection or refraction, all angles are always measured with respect to the surface normal at the point of incidence. When light travels from a medium of lower refractive index to one of higher refractive index, it bends towards the normal and vice versa. Refractive index of a medium depends on wavelength and decreases slightly with increasing wavelength. When light enters a medium from vacuum, its velocity and wavelength decrease and the frequency remains constant. When light travels through a flat block of transparent material, the transmitted ray is parallel to the incident ray, however is displaced from it. When light travels from a medium of higher refractive index to one of lower refractive index, there is a maximum angle beyond which light is not refracted into the second medium. This angle is called the critical angle. If the angle of incidence exceeds the critical angle, total internal reflection (TIR) occurs, following the laws of reflection. Natural phenomena such as mirage can be explained by TIR. White light is composed of different colors with different refractive indices, however small the difference may be. Due to refraction, different wavelength rays are refracted to different degrees leading to dispersion. This explains the phenomenon of rainbow formation. End of Chapter Exercises 1. Calculate the remaining seven marked angles in the following figure. AB and CD are parallel to each other. E A 1 75o 2 3 4 5 C B D 6 7 F Figure 5.62 Illustration for End of Chapter 5 Exercises, problem 1. 2. Trace the reflected laser beam off the mirrors in Figure 5.63. 147 3. A concave mirror has a radius of curvature 25 units. Using two of three properties above and a graphical procedure, determine the location of the image when the object is (a) 16 units, and (b) 6 units to the left of the mirror. 4. A convex mirror has a radius of curvature 10 units. Using two of three properties above and a graphical procedure, determine the location of the image when the object is (a) 10 units, and (b) 4 units to the left of the mirror. A B reference line parallel to laser beam 30o O Incident laser beam D C Mirror #1 50o E Mirror #2 reference line parallel to laser beam Figure 5.63 Illustration for End of Chapter 5 Exercises, problem 2. 5. A 3.3 mW laser beam is reflected off a plane mirror whose reflectance is 94.5%. What is the power of the reflected beam? 6. Light is traveling from air into glass of refractive index 1.60 at the given wavelength. Calculate the angle of refraction corresponding to an angle of incidence of (a) 20o, (b) 35o, (c) 45o and (d) 67o. 7. Light is traveling from water (n=1.333) to glass (n=1.655). Calculate the angle of refraction corresponding to an angle of incidence of (a) 20o, (b) 35o, (c) 45o and (d) 67o. 8. Light is traveling from glass (n=1.50) into air. Calculate the angle of refraction corresponding to an angle of incidence of (a) 20o, (b) 35o, (c) 45o and (d) 67o. What happens in cases (c) and (d)? 9. Calculate the beam displacement for light incident on a glass aquarium tank from air at an angle of 45o. The height of water in the tank is 18”. The refractive index of water is 1.333. Neglect effects of the glass. 10. (a) Trace the light ray, mark all the angles and calculate the speed of light in each of the material for the case in Figure 5.64. (b) What is the velocity of light in each medium? 148 (c) What is the wavelength of light in each medium if the original wavelength was 589 nm? 48o Air 1 2 Glass (n=1.885) 3 Glass (n=1.657) 4 Glass (n=1.500) 1 Air Figure 5.64 Illustration for End of Chapter 5 Exercises, problem 10. 11. Calculate the critical angle for each of the materials with the given refractive indices with respect to air: (a) 1.793 (sapphire @ 589 nm)) (b) 1.434 (calcium fluoride @589 nm)) (c) 3.436 (silicon at 3000 nm) (d) 1.531 (crown glass @ 400 nm) 12. Trace the beam in the glass block shown below. Show all dimensions where the ray strikes different surfaces until (if) it exits the block. The refractive index of the block is 1.500. (Diagram not to scale) 4 cm 40o 6 cm 15 cm Figure 5.65 Illustration for End of Chapter 5 Exercises, problem 12. Group Discussion Questions These require the students to form small groups of three or four and discuss each question. Some questions may require you to do a search on the Internet and find the answers. 149 1. How does a kaleidoscope work? 2. Find additional details (from the web) regarding the operation of a reflecting telescope. 3. What are deformable mirrors? Where are they used? 4. Is there an algebraic relation to predict image location for a spherical mirror, given the object distance from the mirror and its focal length? (Hint: Refer to the Thin Lens Equation in the chapter on Geometric Optics: Part I). 5. Refer to any one of the problems where we had layers of different materials in series and a light ray refracted through it. Rework it assuming that there is an air gap between each solid layer. How do your answers compare with the original situation? 6. Besides the refractive index being so high, what are the other (optical) properties of diamond that makes it so valuable and thus expensive? References Sears, F.W, M. W. Zemanksy and H.D. Young. “College Physics.” Reading: AddisonWesley Publishing Company, 1991. 2 Jenkins,F. and H. White. “Fundamentals of Optics.” Singapore: McGraw Hill, 1981. 3 Saphikon Product Catalog, http://www.saphikon.com/ 4 Tydex Product Catalog, http://tydex.ru/materials/ 5 Oriel Corporation Product Catalog, www.oriel.com 6 Sciner Product Catalog, http://sciner.com/ 7 Gottlieb, H. “Experiments Using a Helium Neon Laser.” Bellmawr: Metrology Instruments, Inc., 1981 1 150 Chapter 6 Basic Principles of Fiber Optics 6.1 History of Fiber Optics 6.2 Elements of a Fiber Optics Cable 6.3 Modes of Propagation in a Fiber 6.4 Acceptance Angle of a Fiber 6.5 Power Losses in a Fiber 6.6 Properties of Light Sources 6.7 Fiber Optics Connectors 6.8 Suggested Laboratory Experiments Chapter Summary End of Chapter Exercises Group Discussion Questions Chapter Objectives: After completing this chapter, the student will be able to (i) Identify elements of a fiber optic cable (ii) Calculate critical and acceptance angles of a fiber (iii) Explain modes of transmission in a fiber and its consequences (iv) Calculate power loss in a fiber (v) Explain characteristics of light sources used in fiber optics communication (vi) Describe the role of fiber optic connectors Key terms: core, cladding, loss, dB, critical angle, acceptance angle, mode, transmitter, detector, dispersion, pulse broadening, bandwidth. 6.1 History of Fiber Optics (From “Technician’s Guide to Fiber Optics” 3rd edition by STERLING © 2000. Reprinted with permission of Delmar Learning, a division of Thomson Learning: www.thomsonrights.com FAX 800-730-2215). The principle of using light to transmit signals has been known for a long time. Lanterns were used to communicate in the days of Paul Revere. Lamps were used by Navy personnel to communicate from ship to ship or shore. Short and long bursts of light conveyed the dots and dashes of the Morse code. Also lamps were used in the Navy for semaphore signaling from ship to ship. The relative locations of two lamps held in outstretched arms of a sailor conveyed different letters of the alphabet. In the late part of the 18th century, the French built the first optical telegraph. Towers stretching 230 km relayed signals from one to the next using movable signal arms, enabling message transmission in 15 minutes. A similar system was operational between Boston and Martha’s Vineyards. The optical telegraph was ultimately replaced by electrical telegraphs. 151 The most compelling demonstration of guiding light occurred in 1870, when John Tyndall used a stream of water. The source of light was placed in a container to which was attached a spout. Even though the stream of water that issued out of the spout was curved, it was able to guide the light through the bend. The Royal Society members who saw the zigzag path of light inside the “water fiber” were stunned. The concept is illustrated in Figure 6.1. Spout Light Source Light confined inside the “water fiber” Figure 6.1 Illustration of John Tyndall’s experiment with a “water” fiber. Alexander Graham Bell, the inventor of the telephone, also invented the photophone in the later part of the 19th century. The principle involved carrying speech using unguided light over nearly a 200 meter distance. A mirror vibrated with sound causing modulation of light reflected off it. At the other end, the receiver demodulated the light and reproduced the sound. Use of fibers to look inside a human body started over fifty years ago. The term “fiber optic” was coined by Narinder Kapany in 1956. A glass rod with a glass coating was invented and it became the first optical fiber. The invention of laser in the 1960s evoked further interest in the field of communications. Over the next few decades, losses in fiber optic cables were reduced from ca. 20 dB/km in 1966, to ca. 0.2 dB/km or less these days. For the sake of comparison, a 20 dB/km loss fiber would lose 99% of its input power in a 1 km distance, whereas by 1982, the losses were cut down to such an extent that the same power loss would occur now in 125 km distance of the fiber cable. The details of loss characterization are provided in a later section. In the 1970’s, the military started replacing conventional communication methods with fiber optics due to the light weight offered by fiber cables. Communication companies started replacing existing cabling with fiber optic systems in the 1970s and 1980s. In the 152 1990s computer manufacturing companies started using fiber optic systems for rapid communications transfer with rates of up to 40 billion bytes per second by the late 1990s. (Byte per second [bps] is a measure of signal transmission speed). Fiber optic systems are extensively used in communications today, as well as in computers, LANs (local area networks), medicine, defense, etc. Their advantages over copper coaxial cables, which the fiber cable has been fast replacing, are summarized below1(From “Technician’s Guide to Fiber Optics” 3rd edition by STERLING © 2000. Reprinted with permission of Delmar Learning, a division of Thomson Learning: www.thomsonrights.com FAX 800-730-2215). 1. Wide Bandwidth: Bandwidth refers to the frequency range of signal transmission. As we studied in Chapter 2, time period and frequency are inversely related to each other. Short duration pulses with a short duration between pulses result in a high pulse repetition rate or frequency. Such high frequency pulses require a high frequency carrier wave to transport them. While radio waves have frequencies from 100 kHz to near 10 GHz, light waves typically used in fiber optics as carrier waves have frequencies approaching hundreds of THz, many orders of magnitude higher! Lasers capable of generating these frequencies have found their place as light sources. The information carrying capacity of which the bandwidth is a measure is indeed very high. This enables simultaneous transmission of voice, data and video signals through the fiber system. For digital telephone communication, while a conventional copper coaxial cable can provide nearly 700 voice channels at less than 50 million bytes per second, the fiber cable can transmit over 100,000 voice channels at 10 billion bytes per second and using fewer repeaters along the way. 2. Low loss: Attenuation is another name to denote loss. In conventional copper cable, while losses are in the acceptable regime with low frequency signals, the attenuation is so large to be unacceptable at higher frequencies. Thus, copper cables are limited in the range of frequencies with considerable loss of signal strength at higher frequencies. Fibers on the other hand have low losses over a wide range of frequencies. For comparison, while copper coaxial cables were adequate for transmission frequencies up to 10 MHz, the losses increased three-fold at 100 MHz to nearly 30 dB/km, which was unacceptable for long distance communications. On the other hand, multimode fiber optic cables extended the frequency of low loss transmission to the GHz regime and single mode fibers cable pushed this even further. (Refer to other sections of this chapter for definition and calculation of power loss in a fiber). 3. Light Weight: The densities of materials used in the construction of fiber cables are much less compared to the copper cable counterparts. As a result, a fiber optic cable weighs about nine times less compared to the same length of a copper cable, an important feature in such applications as aircraft and automobiles, where weight is a consideration. 4. Small size: As a result of lower density and smaller size of fiber optic cables, their sizes are correspondingly smaller than copper cables. Where space limitations exist, such as aircraft, submarines and communication networks, small size is important. As a 153 comparison, a 4.5 inch (113 mm) diameter copper cable can carry over 40,000 two-way conversations over short distances, whereas a fiber bundle with 144 fibers and one-half inch diameter can carry a total of nearly two million calls on all the fibers combined. Space savings is also an advantage when laying underground conduits in urban areas, and also in expanding existing computer rooms with limited sub-floor space. 5. Safety: The components of fiber optic cables are all insulators. The typical hazards associated with copper cables such as arcing, fires and lightning attraction are absent with fibers. Since there are no fire hazards, fiber cables may go straight through a fuel tank, if necessary! 6. Security: It is difficult to tap into fiber networks. Copper cables can generate electric fields which can be tapped into using antennas, and thus security can be compromised. 7. Electromagnetic immunity: A fiber cable is not susceptible to external electric or magnetic fields unlike a copper cable. Several problems have been reported due to EMI emanating from conductors (such as copper cables and devices) including erasure of vital records from computer memory, video games interfering with police communications, resetting of gasoline pumps from CB radio transmissions, to name a few. Cables can act as antenna for EMI; adjacent copper cables can result in cross talk and spurious pulses can result from bursts of EMI. None of these are concerns with fiber optic cable since the fiber material is a dielectric (nonconductor) enabling it to be placed next to high voltage equipment, motors and in a high EMI environment with no detrimental effect. 8. Cost: With improvement in fiber technology, the cost has become competitive to a point that sometimes it is cheaper to use fiber cabling rather than copper. Again the final analysis is made based on the advantages that fiber provides over copper for the same dollar spent toward building a communication network. 6.2 Elements of a Fiber Optics Cable1,2,2a,2b,2c,2d,2e The two most essential elements in a fiber optic cable, without which light transmission will not occur, are the core and the cladding. The core is the innermost element and the cladding surrounds the core. Outside the cladding are several layers as shown in Figure 6.2. They function mostly as strength members or shock absorbers or provide structural integrity and protection to the glasses that they surround. The buffer coat acts like a shock absorber and has no optical properties. The buffer coat may be absent in plastic clad fibers, i.e. fibers with a plastic cladding. The Kevlar® is a synthetic aramid yarn. When one cuts into a fiber cable, it is seen as a yellow, fibrous material. Fibers may be classified according to material of core and cladding. Typically, ultra pure glass (known as fused quartz) and polymethylmethacrylate (PMMA) based plastic are common materials of construction for the core and the cladding. Accordingly, one could have (i) glass core/glass cladding combination, simply known as glass fiber, or (ii) glass core/plastic cladding combination, known as plastic clad silica (PCS), [silica is another name for glass], and (iii) plastic core/plastic cladding. 154 Outer Jacket Kevlar® Jacket Buffer coat Cladding Core Figure 6.2 Illustration of the elements of a fiber optic cable. (i) Glass fibers are the heaviest in terms of weight and the most expensive. However, they offer the most superior performance. For applications such as LAN, or if the transmission length is greater than eight feet, glass fibers are to be used. (ii) PCS fibers are a compromise between glass fibers and plastic fibers in terms of cost and performance. (iii) Plastic fibers are the cheapest and lightest in weight. Typically used for lamps, they have low bandwidths and high losses. These inexpensive fibers can be used in short distance (i.e. a few feet) applications if their bandwidth and power loss problems are tolerable. (Bandwidth refers to the range of signal frequencies that can be successfully transmitted through a fiber. Further discussion is presented in later sections.) As you might have guessed from John Tyndall’s experiment and proceeding discussions, light propagates through the core and it does so even if the core goes through turns. The mechanism responsible for this is Total Internal Reflection (TIR) discussed in Chapter 5. In the case of Tyndall’s experiment, he had a “water fiber” of refractive index 1.333 surrounded by air (n=1.) Light was injected into the water at one end. Since the refractive index of water is higher than air for light inside the water fiber, TIR occurs so long as the angle of incidence exceeds the critical angle. Those rays for which this criterion is continually met will propagate successfully down the fiber and reach the exit end. In the case of fiber optics, theoretically one could have a bare glass fiber and it would transmit the light downstream. However, it is more efficient to enclose the fiber 155 core in a cladding whose refractive index is slightly lower than the core (typically ncore= 1.48, ncladding=1.47) so that TIR will occur as shown in Figure 6.3. cladding core Figure 6.3 Illustration of total internal reflection in a fiber optic cable. Only one ray is shown for clarity. Each time the ray strikes the core/cladding interface, provided that the angle of incidence in the core exceeds the critical angle, the ray undergoes TIR. The next section presents calculations. Fibers can also be classified based on their size. In particular, fiber cables are specified by the diameters of the core and the cladding. For example a fiber designation of 62.5/125 indicates that the core diameter is 62.5 µm and the cladding diameter is 125 µm. Similarly, 50/100 designation means a core diameter of 50 µm and a cladding diameter of 100 µm. A fiber cable is also available in 8/125 format. Clearly, in this case the core diameter is much smaller than the cladding diameter. Designations discussed thus far are common for glass and PCS fibers. Cladding diameter of 125 µm has become the unwritten “standard.” Fiber strippers are easily available to strip such fibers. Plastic fibers are usually much larger, for example 600/1000 and 980/1000. Common sense tells us that larger the diameter of the core, more the light that it can transmit. What other consequences exist when large size cores are used? We will examine this next. Fibers can be classified by refractive index profile and modes of propagation. As discussed, the core refractive index should always be slightly higher than the cladding refractive index. One could have two different possibilities, typical values shown in Figure 6.4. Step Index (62.5/125 fiber) Graded Index (62.5/125) 1.482 1.482 1.48 1.48 1.478 1.478 1.476 1.476 1.474 1.474 1.472 1.472 1.47 1.47 1.468 0 50 100 150 1.468 0 50 100 150 Figure 6.4 Refractive index profile of (a) step index, and (b) graded index fiber. 156 The figure on the right is for a step index (SI) profile. The refractive index jumps abruptly at the core/cladding interface. As we learned from Chapter 5, higher the refractive index of a medium, lower the speed of light in it. Thus, the speed of light in the core is somewhat less than that in the cladding. Light traveling in the core would take longer time to reach the exit end of the fiber compared to light traveling near the cladding interface. Thus, while two rays of light started at the entrance end of the fiber at the same time, the one that traveled through the center of the core would take longer time to reach the exit end than the one that traveled at the outer edges even though the speed of light is the same within the core. This results in time delay known as dispersion, which depends on many factors including the frequency of light, refractive indices of core and cladding, fiber length, etc. Further discussion is provided in the next section. The figure on the left is for a graded index profile. As the name implies, the refractive index varies continuously in the core, from the highest value at the core center to the lowest value corresponding to the cladding index at the core/cladding interface. The advantage of such a profile is that the time delays among different rays of light traveling at different locations in the core are not as drastic as is with the case of SI fiber, resulting in less dispersion. The refractive index profile is not to be confused with the mechanical profile. Mechanically, the physical length of the core and the cladding are the same and the entrance and exit ends of a fiber are ground to be perfectly flat during connectorization process, i.e., the process of attaching connectors to terminate the cable at either end. Fibers such as with 62.5/125 and 50/100 designations are available in either SI or graded index formats. The application dictates which one is chosen. With such “large” core diameters, there are multiple pathways for the light to travel in the core. This number of different pathways is referred to as modes of propagation. Mathematically speaking, the modes arise from the solution of Maxwell’s equations, a topic which is beyond the scope of this book. It is sufficient for the student to know that greater the number of modes in a fiber, greater the number of pathways for light to travel and more severe the modal dispersion (i.e. dispersion caused by the presence of multiple modes.) The fibers whose refractive index profiles are shown in the figures above are called multimode fibers. Clearly, the larger core diameter is partly to be blamed for it. Why then would one use multimode fibers? The answer is obvious when one realizes that it is relatively easier to allow light to enter a 62.5 µm core diameter fiber than an 8 µm fiber. Light sources used with small core diameter fibers are expensive. The advantages of using small core diameters are obvious. One can decrease the number of pathways available for light travel and minimize modal dispersion. Such fibers are called “single” mode fibers. Typically, the core diameters are at least ten times smaller than the cladding diameter. As mentioned earlier, the 125 µm cladding outer diameter has become a “standard,” so that tools and connectors are readily available. Thus, while many “single” mode fibers have 125 µm for the cladding diameter, the core diameter can be 6, 7, 8, 9, 10, 11 or 12 µm. The refractive index profile for a typical 8/125 “single” mode fiber is shown in Figure 6.5. 157 Step Index 8/125 Fiber 1.482 1.48 1.478 1.476 1.474 1.472 1.47 1.468 0 50 100 150 Figure 6.5 Refractive index profile of a “single” mode, step index fiber. By virtue of the small core diameter, modal dispersion is considerably less, compared to a multimode step index fiber. Do we have “single” mode graded index fiber? The answer is: since the dispersion is low enough, and since it is difficult to fabricate a graded index small core diameter fiber, “single” mode fibers are available in step index format only. You may be wondering why we used double quotes “” to denote the “single” mode fiber. The answer is provided in the next section. Fibers may be simplex which allow one-way communication or duplex which is a combination to two simplex cables. Duplex cable allows two-way transmission of signals, each simplex allowing light to pass one way. 6.3 Modes of Propagation in a Fiber1,2,2b,2c,2d As mentioned in the previous section, a mode refers to a pathway for light to travel from one end of the fiber to the other. Clearly, a multimode fiber has thousands and tens of thousands of pathways or modes. A select number of rays are shown in Figure 6.6 for a multimode fiber with step index. cladding 3 2 1 1 2 core cladding Figure 6.6 Modes of propagation in a multimode, step index fiber. 3 158 Here ray #1, practically goes through the center of the fiber. It is an ideal ray and is designated as a low mode ray. Rays 2 and 3 are incident at various angles and clearly go through longer optical path lengths. These are the higher mode rays. The arrival times are different for rays 1, 2 and 3, although all of them would have originated at the same time at the entrance. This is called modal dispersion. While only three rays are shown for clarity in this figure, in reality there are hundreds of thousands of modes. Calculation of the number of modes is shown later in this section. When one uses multimode graded index fiber, the differences in arrival times of the rays is less pronounced, as shown in Figure 6.7: cladding 3 2 1 1 2 core 3 cladding Figure 6.7 Modes of propagation in a multimode, graded index fiber. By virtue of gradual variation of the index which limits the number of modes, the difference in arrival times of the rays is less drastic. Finally, for a “single” mode fiber, despite the step index profile, since the diameter of the core is so small, it clearly permits one, if not a minimum number of modes, leading to the least modal dispersion among the three types of fibers discussed here. See Figure 6.8. cladding 1 1 core cladding Figure 6.8 Modes of propagation in a “single” mode fiber. What is the implication of modal dispersion on the quality of the output at the exit end of the fiber? For the sake of illustration, if, one starts with an ideal square pulse of light as input at the entrance end of the fiber, if life were ideal, light would travel through the center of the fiber and the output pulse will mimic the input pulse in its shape, as shown below. Its amplitude (or height) may be smaller due to fiber losses discussed in a later section. In Figure 6.9-6.13, the horizontal axis represents time, and the vertical axis is the signal strength which the light intensity or power is a measure of. In other words, in Figure 6.9, the pulse is not elongated in the horizontal direction or time axis. Let us consider a “single” mode step index fiber. The result is shown in Figure 6.10. 159 Ideal input Ideal output Figure 6.9 “Ideal” output from an “ideal” input. Ideal input Real output (“single” mode) Figure 6.10 Output from an “ideal” input for a “single” mode fiber. For a multimode, graded index fiber, the result is shown in Figure 6.11. Ideal input Real output (multimode, graded index) Figure 6.11 Output from an “ideal” input for a multimode, graded index fiber. For a multimode, step index fiber, the result is shown in Figure 6.12. Ideal input Real output (multimode, step index) Figure 6.12 Output from an “ideal” input for a multimode, step index fiber. By comparison of the three figures (6.10-6.12), it is evident that the shape of the pulse is distorted more and more as one proceeds progressively from the “single” mode to the multimode graded to the multimode step index fibers. This phenomenon is called pulse broadening when the output pulse is “broader” compared to the input pulse. In each of the cases above, we had a single square pulse as input. What if we had a train of pulses? For three sequential pulses the input and output of a multimode step index fiber are shown in Figure 6.13. 160 1 2 3 1 (input: 3 square pulses) 2 3 (output: 3 overlapping broadened pulses) Figure 6.13 Pulse broadening due to modal dispersion. Clearly, it is difficult to distinguish the output pulses from each other due to pulse broadening and overlap. This limits the frequency of input pulses and hence the bandwidth. As mentioned earlier, bandwidth relates to the frequency range of signal pulses, and is an important parameter in communication. Clearly, the graded index will have a higher bandwidth compared to step index multimode fibers. A “single” mode fiber has one of the highest bandwidths of around 50-100 GHz-km. While modal dispersion is all but eliminated in “single” mode fibers, other types of dispersion still exist explaining why one still does not obtain the ideal output. Let us look at calculating the number of modes in a fiber. The following equations are used. V = πdcore (NA) ------------- λ where V refers to the V-number, a dimensionless, normalized frequency number, dcore refers to the core diameter in meters, NA refers to the numerical aperture (a measure of a fiber’s light gathering ability -further details are provided in the next section), and λ refers to the wavelength of light expressed in meters. If the calculated V-number is less than or equal to 2.405, the fiber will support only a single mode. Otherwise, we can calculate the number of modes that the fiber will support, Nm, using V2 2 for a step index fiber, and Nm = V2 4 for a graded index fiber. Nm = Clearly by comparing these last two formulas, it is seen that a fiber will support half the number of modes for a graded index compared to a step index, all other parameters remaining the same. 161 Example 6.1 For a step index fiber with designation 62.5/125, using an 850 nm source and a 0.283 numerical aperture fiber, calculate how many modes the fiber will support. Repeat calculations for a graded index fiber. We need to calculate the V-number first, as π x 62.5 x 10-6 x 0.283 V = 850 x 10-9 = 65.4 Clearly, it is not a single mode fiber since V>2.405. Calculate the number of modes as Nm = 65.42 2 = 2137 modes Answer For a graded index fiber, we have Nm = 65.42 4 = 1068 modes Answer As advertised, the graded index fiber will support one-half the number of modes as the step index fiber. Example 6.2 For a step index fiber with designation 9/125, using a 1550 nm source and a 0.222 numerical aperture fiber, calculate how many modes the fiber will support. This is the so-called “single” mode fiber due to the core diameter << cladding diameter. We need to calculate the V-number first, as π x 9 x 10-6 x 0.222 V = 1550 x 10-9 = 4.05 V-number exceeds 2.405! Therefore, it is not truly a “single” mode fiber. Now you realize why we had denoted the word single using double quotes all along! We will omit the double quotes from now on so as not to offend experts in the fiber field! Since a single mode fiber is always step index, calculate the number of modes using Nm = 4.052 2 = 8 modes Answer Eight modes are much better than over a thousand modes that we had for a multimode graded index fiber. Modal dispersion will be minimal for the single mode fiber despite the eight modes it can support! Can we ever have a single mode fiber to behave so that it supports only one mode? Look at the next example. 162 Example 6.3 For a step index fiber with designation 9/125, using a 0.222 numerical aperture fiber, calculate what wavelength of light must be used so that the fiber will support only one mode. To support only one mode, we know that the V-number must not exceed 2.405. Using this as the cut-off point, we obtain 2.405 = π x 9 x 10-6 x 0.222 λ Rearranging, we have = λ π x 9 x 10-6 x 0.222 2.405 = 2.61 x 10-6 m or 2.61 µm Answer It does not mean that a 2.61 µm wavelength source is readily available. These calculations were performed to illustrate a point that there is a definite cut-off wavelength above which a fiber can behave truly in a single mode. Clearly lower the V-number, the lower the number of modes, i.e. Nm ↓ V↓ To decrease V, we note that V↓ as d↓ V↓ as NA↓ and V↓ as λ↑ While theoretically one can reduce the number of modes (and modal dispersion) by decreasing the core diameter, decreasing the numerical aperture and increasing the wavelength, there are practical limitations, some of which are described in the next section. In this section, we considered only modal dispersion which is present only in multimode fibers. Let us denote this by (∆tm). Fiber cables suffer from other kinds of dispersions such as material dispersion (∆tmat), waveguide dispersion – for single mode fibers only (∆twg) and polarization mode dispersion (∆tPM). Material dispersion is inherent even if there are no impurities in the glass. It is caused by the light source not being entirely monochromatic even if a laser is used as a source. Needless to say other light sources such as light emitting diodes (LEDs) would exacerbate the problem. The differences in speeds of different wavelength light leads to dispersion. Waveguide dispersion occurs in single mode fibers due to light exiting the core into the cladding and reentering the core. Since the core and the cladding have different indices of refraction, the speed of light is different leading to dispersion. Polarization dispersion is caused by reflections leading to light being partially polarized inside the fiber. Details of the polarization phenomenon are presented in Chapter 7. Empirical formulas exist for calculating the different dispersions, except the polarization mode. The total dispersion (∆ttotal) is now given by 163 _______________________________ (∆ttotal) = √ (∆tm )2 + [(∆tmat )+ (∆twg )]2 With empirical formulas available to calculate each term within the square root sign, the total dispersion can be evaluated. The terms in the square bracket together contribute to chromatic dispersion. Refer to the Appendix V at the end of the book for relevant empirical formulas. While this formula allows for calculation of dispersion, measurement of total dispersion can be performed as follows. Shown in Figure 6.14 are input and output pulses. Input (source) Output (receiver) t1(3dB) t2(3dB) Figure 6.14 Pulse broadening due to dispersion. In each case the dashed line represents half the maximum power of the pulse. The corresponding time duration in each case is known as the full width half maximum pulse width (FWHM). It is sometimes also called the 3 dB pulse width. The pulse width corresponding to the source is t1(3dB) and for the receiver it is t2(3dB). Then the (∆ttotal) is given by ___________ (∆ttotal) = √t2(3dB)2-t1(3dB)2 As an example if the FWHM pulse width at the source is 25 ns and that at the receiver is 60 ns, then using this formula, one would obtain a (∆ttotal) of 54.5 ns. ∆ttotal causes signal degradation to different degrees depending on its magnitude and the application. Dispersion by itself is not a loss mechanism. It degrades the quality of the pulse in terms of broadening while not necessarily contributing to reduced amplitude of the output signal. (See also Appendix V for additional dispersion formulas). A final point before we conclude this section. While we have been concentrating on the core as being responsible for light transmission, there is some evidence that a small portion of light travels through the cladding very close to the core-cladding interface. It has been demonstrated that with single mode fibers, the diameter of the exiting beam is slightly larger than the diameter of the cladding. The former diameter is called the mode field diameter since it pertains to the diameter of light energy transmitted. This makes the optical transmission properties of the cladding somewhat more critical for single mode fibers. With multimode fibers, it is not as significant since the core is large and can transmit much of the light. 164 Exercise 6.3A 1. In each of the following cases, indicate if it is (a) multimode or single mode fiber, (b) step index or graded index and (c) glass or plastic core. (i) 480/500, (ii) 50/125, (iii) 10/125, (iv) 100/140, (v) 735/750, (vi) 6/125 2. In each of the following cases, indicate if it is (a) multimode or single mode fiber, (b) step index or graded index and (c) glass or plastic core. (i) 980/1000, (ii) 85/125, (iii) 200/380, (iv) 300/440, (v) 5/85, (vi) 485/500 3. For a step index fiber with designation 7/125, using a 1550 nm source and a 0.222 numerical aperture fiber, calculate how many modes the fiber will support. 4. For a step index fiber with designation 7/125, using a 1300 nm source and a 0.222 numerical aperture fiber, calculate how many modes the fiber will support. 5. For a graded index fiber with designation 62.5/125, using an 850 nm source and a 0.222 numerical aperture fiber, calculate how many modes the fiber will support. 6. For a step index fiber with designation 62.5/125, using an 850 nm source and a 0.222 numerical aperture fiber, calculate how many modes the fiber will support. Redo this problem for a graded index fiber. 7. A typical telecommunications optical fiber is made of ________core and ________ cladding. 8. A typical material used in the construction of plastic fibers is ____________. 6.4 Acceptance Angle of a Fiber Thus far, we have looked closely at the internals of a fiber and transmission. Let us pause for a moment and look at the entrance end of a fiber. The question that arises is, “Can light from all directions illuminate the entrance end of the fiber?” The answer is yes, however, the important consideration is, of all the light rays incident upon the entrance end of a fiber, which of them can successfully propagate down the fiber and reach the exit end. Let us go back to the core/cladding interface again. Earlier, we looked at the “ideal” ray which travels down more or less at the center of the fiber. Clearly one can conceive of an ideal ray if we have a straight, short piece of single mode fiber. Real life is such that the fiber needs to go through several 90o bends between the entrance and exit ends. Thus, a mechanism has to exist which will allow light to travel through some other means. This mechanism is total internal reflection (TIR) which was discussed in detail in Chapter 5. The refractive index of cladding is slightly smaller than the core (by about 1%) allowing TIR at the core/cladding interface. The following example illustrates calculation of critical angle for TIR. 165 Example 6.4 A fiber cable has a core refractive index of 1.500 and a cladding refractive index of 1.485. What is the critical angle at the core/cladding interface? 90o Cladding Core θc Figure 6.15 Illustration for Example 6.4. If the angle of incidence in the core exceeds the critical angle, then TIR will occur. Otherwise, light will be refracted into the cladding and more than likely permanently lost. Thus, for light to propagate down the fiber via TIR, it should always be incident at the core/cladding interface at an incident angle exceeding θc. To calculate θc, apply Snell’s law (from Chapter 5) as n1 sinθ1 = n2sinθ2 where the subscript 1 stands for the core and subscript 2 for the cladding. We know that θ1=θc, θ2=90o, n1=1.500 and n2=1.485. Substituting and simplifying, we obtain θc θc = sin-1(ncladding/ncore) = sin-1(1.485/1.500) = sin-1(0.99) = 81.9o Answer The useful formula for the critical angle that we derived in the above example is θc = sin-1(ncladding/ncore) In order to continually satisfy the critical angle criterion, one must also have a restriction on the angle of incidence at the entrance face of the core. Let us look at another example. Example 6.5 Two of the most essential elements in an optical fiber cable are the core and the cladding. The goal is to propagate light from one end of the fiber through the core via TIR (total internal reflection). The refractive index of the core is 1.52 and that of the cladding is 1.48. (a) Find the critical angle for light propagation. (b) To what angle should rays of light entering the fiber from air be restricted to so that propagation will occur along the fiber core? 166 (a) Solution TIR Cladding AIR Core Fiber Axis TIR Cladding Figure 6.16 Illustration for Example 6.5. In Figure 6.16, a ray of light is shown that exceeds the critical angle when it strikes the core-cladding interface. Thus, it undergoes TIR. It undergoes TIR a second time at the opposite end and so on. To calculate the critical angle, we use θc = sin-1(ncladding/ncore) where ncladding=1.48, ncore=1.52 θc = sin-1[1.48] [1.52] = 76.8o Answer (b) Solution Cladding AIR θa θc 90o- θc Core Fiber Axis Cladding Figure 6.17 Solution for Example 6.5. To answer the second question, sketch the situation where the light ray just grazes the surface of the core-cladding interface. This would correspond to an angle of incidence of θc = 76.8o as shown above. Examining the geometry of the system, the angle of refraction at the air/core interface should be 90o-76.8o = 13.2o. Applying Snell’s law at the air-core interface, the angle of incidence can be calculated as 167 θa sin-1[n2sin θ2] [ n1 ] = where n2=1.52, n1=1 and θ2= 13.2o Substituting the values, we obtain θa = sin-1[1.52sin 13.2o] [ 1 ] 20.3o Answer = This is called the (half-) angle of acceptance. Thus, if light rays within a cone half-angle of 20.3o strike the entrance of the core, they will propagate successfully down the fiber. Angles of incidence larger than this will result in light lost into the cladding more or less permanently. The above example required us to calculate the critical angle first and then work from there to the half angle of acceptance. Following Figure 6.18 is a derivation that results in a much simpler and direct formula for the acceptance angle. Cladding 90o A θa AIR B 90- θc Core θc C X Cladding Figure 6.18 Relationship between critical angle and (half-)angle of acceptance. Consider a ray of light incident at the core/cladding interface at point A with the angle of incidence θc. Normal AC is drawn at point A. The point C lies on the optical axis of the fiber. Retracing the ray back to the entrance face, it originates from the point B on the entrance face and in line with the optical axis. The ray incident on the entrance face is XB at point B, with an angle of incidence of θa in air. Apply Snell’s law at point B at the air/core interface. 1sin θa =ncore sin (90- θc) From the principles of trigonometry discussed in the beginning of Chapter 5, we can simplify the right hand side as sin θa =ncore cos θc 168 From the principles of trigonometry discussed in the beginning of Chapter 5, we can further simplify the right hand side as ________ sin θa =ncore √1- sin2 θc We have already derived that θc = sin-1(ncladding/ncore) or sin θc = ncladding/ncore Substituting for sin θc in the equation for sin θa, we obtain _______________ sin θa =ncore √1- (ncladding/ncore) 2 Simplification of the right hand side yields ____________ sin θa = √ncore2 - ncladding2 or ____________ θa = sin √ncore2 - ncladding2 -1 We can rework the last example using the new formula and expect to obtain the same result for the angle of acceptance. ____________ = θa = sin-1√1.522 – 1.482 20.3o If light is incident on the entrance side of the fiber at a cone half-angle of 20.3o or less then it can successfully propagate down the fiber via TIR. This is such a narrow cone angle that much of the light that is incident on the entrance face is lost right away. Clearly the half-angle of acceptance is related to the fiber’s light gathering ability known as the numerical aperture (NA) which we briefly alluded to in the previous section. The numerical aperture for a fiber is defined as1,2 NA = sin θa or ____________ NA = √ncore2 - ncladding2 Numerical aperture varies from 0 to 1. We see many conflicts here. (a) We need a large NA to allow most light into the fiber. 169 (b) However, large NA results in a large V-number and higher modes, as discussed in the previous section. (c) We need the core and cladding refractive indices to be as close as possible (obviously with ncore>ncladding) to minimize the number of modes, at the same time compromising on the amount of light that is let into the fiber to begin with. Example 6.6 Calculate the numerical aperture for the fiber in Example 6.5. ____________ NA = √ncore2 - ncladding2 ____________ NA = √1.522 – 1.482 = 0.346 Answer Presence of residues such as fingerprints or condensed moisture (dew) on the exit end of the fiber can alter the amount of light that can exit from a fiber core. Exercise 6.4A Calculate (a) critical angle in the core, (b) acceptance angle and (c) numerical aperture for the following fiber cables. 1. ncore=1.49 ncladding= 1.47 2. ncore=1.49 ncladding= 1.46 3. ncore=1.50 ncladding= 1.48 4. ncore=1.49 ncladding= 1.45 5. ncore=1.52 ncladding= 1.50 6. ncore=1.55 ncladding= 1.00, bare fiber core in air (no jacket) 7. ncore=1.33 ncladding= 1.00, water fiber in air 8. ncore=1.54 ncladding= 1.49 6.5 Power Losses in a Fiber Even for an ideal pulse propagating through an ideal fiber through its middle, signal power loss occurs. This would have been evident from the refraction experiments in Chapter 5 when the transmitted beam from a glass slab always has lower power than the incident beam. Losses occur through absorption, scattering, etc. For a fiber devoid of impurities that could potentially absorb light, losses still occur due to scattering (see the chapter on Absorption and Scattering for additional details.) Power losses are expressed in units of decibels (dB). dB itself may be expressed based on current, voltage or power. Recalling that in electronics, Watt's law relates power to current, voltage and resistance. The three forms of Watt's law3 are P = I2R 170 P = V2 R P = VI and where P is the power in Watts (W), I the current in amperes (A), V the voltage in volts (V) and R the resistance in ohms. Looking at the first two formulas, we see that the power is proportional to the square of current or voltage. dB relates an output variable to an input variable. The variable can be current, voltage or power. Accordingly, one can have three formulas for dB calculation as dB = 20 log (Iin/Iout) dB = 20 log (Vin/Vout) dB = 10 log (Pin/Pout) and where the subscript “in” indicates the variable calculated at the input end and the subscript “out” refers to the output end. By virtue of the square dependence of voltage or current on power, the multiplier is 20 for the first two equations and is 10 for the third form. We will use the third form of the equation, namely dB = 10 log (Pin/Pout) exclusively. “log” represents logarithm to the base 10 and you will see this button on your calculator. To understand a little further about the logarithmic function, consider the function y = 10x This is an exponential function. Substituting various values for x, positive or negative, whole or decimal one will obtain various values of y. This results in an exponential function of the exponent 10. A table of corresponding x-y values is shown below. x y -2 0.01 -1 0.1 0 1 1 10 1.5 31.6 2 100 2.2 158.5 The corresponding graph is shown in Figure 6.19. As is evident from the graph a regular (linear) scale does not do justice to represent small y-values4. Therefore, a logarithmic scale for y is shown in Figure 6.20 for the same function. 171 180 160 140 120 100 80 60 40 20 0 -3 -2 -1 0 1 2 3 2 3 Figure 6.19 Graph of an exponential function. 1000 100 10 1 -3 -2 -1 0 1 0.1 0.01 Figure 6.20 Graph of an exponential function drawn using logarithmic scale. You can immediately see the advantage of using the logarithmic (log, for short) scale. The above two graphs are exponential functions of the base 10, namely y=10x. One can define logarithm as follows, log10y = x where the base 10 appears as a subscript immediately following “log” which stands for logarithm. The equation is read as logarithm of y to the base 10 equals x. When the base is 10, as shown below, it is often omitted. If the base is anything other than ten, it must be mentioned. Base 10 is referred to as the common logarithm, base e (value 2.71828…) 172 is referred to as the natural logarithm. Logarithm to the base e is often abbreviated as ln. Further discussion of the natural logarithm is provided in later chapters. The graph of y = log x is shown in Figure 6.21. The base of 10 is understood. You can generate log values using your calculator. Some calculators such as the TI-502Xa® require you to punch in the value first, and then press the “LOG” button, while others require you to press the “LOG” key followed by the number whose logarithm to the base 10 is needed. y = log x 2.5 2 1.5 1 0.5 0 -0.5 0 -1 -1.5 -2 -2.5 20 40 60 80 100 120 x Figure 6.21 Graph of a logarithmic function. Using a log scale for the x-axis, the graph is redrawn as shown in Figure 6.22. y = log x 0.01 0.1 2.5 2 1.5 1 0.5 0 -0.5 1 -1 -1.5 -2 -2.5 x 10 100 Figure 6.22 Graph of a logarithmic function drawn using logarithmic scale. 173 Now that you have a clearer understanding of the logarithm function, let us revisit the dB formula dB = 10 log (Pin/Pout) Some books report the formula with output power in the numerator and the input power in the denominator. This will result in a negative value for dB, which is acceptable since it is an indication of a “loss.” However, in this book, we will define dB as shown above so that dB losses will always turn out to be positive. There are several mechanisms leading to power loss in a fiber. Even before the light has an opportunity to enter the fiber some of it is reflected through Fresnel reflection, which was discussed in Chapter 3 under “Laser Safety.” For near normal incidence of light on the entrance end of the fiber from air, the fraction of light reflected, known as reflectance R, is given by2,5 R = (ncore - 1)2 (ncore+1)2 For typical core refractive index of around 1.5, the reflectance calculated using this equation works out to be 0.04 or 4%. Simply put, for each reflection on a connector face containing the fiber, 4% of the light is lost due to reflection, i.e. only 96% of the incident light enters the fiber for propagation. The dB power loss for this case is given by2,2e,5a,5b dBFresnel = -10log(1-R) = 10 log 0.96 = 0.177 dB Mechanical splices introduce dBFresnel losses at each connection. As an example, consider the cable system shown in Figure 6.23, where light enters the connector at the left end and there ate two mechanical splices in between before it exists. entrance splice#1 splice#2 exit of fiber Figure 6.23 Illustration of losses due to splices in a fiber. In this case Fresnel losses occur at the entrance, at the splice #1 and splice #2 interfaces, a total of three places. Total Fresnel dB loss = 3 x 0.18 = 0.54 dB. This is just based on reflective losses only. There are additional losses inside the fiber, as discussed below. Impurities in the fiber absorb power leading to attenuation. Additionally, some light undergoes scattering and is ejected out of the core even if the glass is pure and devoid of impurities. Very small bends called microbends, nicks and imperfections also add to the attenuation. Environmental factors also affect attenuation. Presence of moisture and/or debris/residue on connector faces contributes to both power loss and dispersion. 174 Additionally, with connector junctions, one could have an end gap or an angular misalignment of connectors or an offset in the alignment of connectors – all leading to additional light loss. Fiber cabling will necessarily have to take 90o bends and a certain minimum bend radius of at least ten times the diameter of the outer cable jacketing has to be followed so that stresses and losses are minimized1,2,2c,2e. Fiber optic losses depend on the wavelength of light used. Of the three wavelengths commonly used, 850 nm, 1300 nm and 1550 nm, the loss is the highest at 850 nm and the lowest at 1550 nm. Example 6.7 Twenty mW of power are injected into a fiber. At the other end, which is 2 km away, 150µW of power are measured. What is the dB loss? How would you convert the loss so that you can compare it to loss in a specification sheet? dB = 10 log (Pin/Pout) dB = 10 log (20 x 10-3/150 x 10-6) = 21.2 dB Answer Specification (spec) sheets always denote loss as dB per km of fiber length. In order to compare what we have to a spec value, we need dB/km which we obtain by dividing the total dB by the total distance. dB/km = 21.2/2 = 10.6 dB/km Answer This number may now be compared to the spec sheet that is available from fiber vendors to assess how much extra loss occurs due to fiber cable connections. Example 6.8 A graded index fiber has a loss specification of 0.7 dB/km at 1300 nm wavelength of light. After 5 km of length, what would be the measured output power if the input power is 1 mW? dB = = (dB/km) x km 0.7 x 5 = 3.5 dB This is the value to be used in dB = 10 log (Pin/Pout) with Pin=1mW to back calculate Pout. If we transform and rearrange the dB equation to solve for Pout, we obtain Pout = Pin / 10[dB/10] where the logarithmic function was transformed into an exponential function of base 10 upon inversion. Substituting dB=3.5, Pin=1 x 10-3 W, we obtain, 175 Pout = 1 x 10-3 W / 10[3.5/10] = 447 µW Answer Example 6.8 If a detector requires at least 2 mW to detect a signal at the exit end of a 2 km fiber cable of loss specification 3.5 dB/km at 850 nm, what should be the minimum input power at the entrance end of the fiber? dB = (dB/km) x km = 3.5 x 2 = 7 dB This is the value to be used in dB = 10 log (Pin/Pout) with Pout=2mW to back calculate Pin. If we transform and rearrange the dB equation to solve for Pin, we obtain Pin = Pout x10[dB/10] Pin = 2 x 10-3 x10[7/10] = 10.0 mW Answer With these three examples, we have covered all the possibilities of problem solving given the equation dB = 10 log (Pin/Pout) A diagram of typical fiber loss specifications is shown in Figure 6.24. The data is obtained from a variety of manufacturers’ data and graphed for convenience. Top:8/125 fiber; middle: graded 62.5/125: bottom: graded 100/140 10 1 0 0.1 Loss spec, dB/km 500 1000 1500 2000 Wavelength, nm Figure 6.24 Typical fiber losses in a fiber. In terms of bandwidths, the single mode fiber provides around a few GHz.km while the multimode fibers range below 2 GHz.km. To summarize this chapter, (Fresnel) reflection losses, power losses and dispersion degrade the signal in terms of pulse broadening as well as decreased pulse amplitude. 176 Exercise 6.5A 1. For a fiber with five mechanical splices, what is the total Fresnel reflection loss? 2. For a fiber with one mechanical splice, what is the total Fresnel reflection loss? Complete the following table for a 3.3 km fiber cable Problem# 3 4 5 6 7 8 9 10 Pin 5 mW 20 mW Pout 300 µW 1 mW 0.2 mW dB 7.5 dB/km 1 5 25 mW 0.3 0.5 0.8 mW 10 mW 15 mW 2.1 1.5 mW 6.6 Properties of Light Sources1,2,2c,2e,5a,5c Electrical signals are converted into optical signals which are then modulated by the carrier waves generated by the light source. The fiber optic cable transmits this signal to the detector. Conversion of the signal from optical form into electrical form takes place at the receiving end. This is the crux of the entire communication process as illustrated by a simplified block diagram shown in Figure 6.25. Complex electronic circuits are present inside each of the boxes. Electrical Electrical Fiber Optical Signal Signal in Transmitter Signal Out Receiver Figure 6.25 Block diagram of signal transmission. The electrical signal is converted into an optical signal which is then modulated by the carrier waves. Communication transmittal is rapid since optical signals travel at the speed of light through the fiber optic cable. At the receiver end, the optical signal is converted back into an electrical signal and the carrier waves are no longer needed since they have served their purpose. The electrical signals thus transmitted might have been analog or digital, or audio, video or computer related signals from the modem. The possibilities are limitless. The source of light can be infra red (as in communications and medical fields), visible (as in electronic scoreboards and message signs) or occasionally ultraviolet (as in curing dental epoxies.) The power of light source typically varies from tens of µW to tens of mW. Typically, analog signals are used in plastic fibers, employing visible light and 177 short distances. Communication signals are typically digital in nature and use infrared light. Light sources must be of suitable wavelength, must carry the signal (i.e. modulation), and must send the light through the fiber. They must be of sufficient power so that despite attenuation that is bound to occur in fibers, connectors, and splices, the signal received at the exit end is strong enough to be decoded back to an electrical signal. Laser diodes and light emitting diodes (LED) are common sources. LED’s 850 nm wavelength is commonly used. Diode lasers operating at 1300 nm and 1550 nm wavelengths are also popular. One encounters LEDs in every day life. Many power on/off buttons in computer monitors, drives, etc. are equipped with a small light which can emit sometimes three distinct colors of light for off, standby or on. Prior to extensive use of liquid crystal display (LCD) as in calculators or screens, LED segments were commonly used. LEDs are inexpensive and have long lives. Diode lasers on the other hand require extensive maintenance and are more expensive. In either case, the output power can be varied by varying the input current into the source. Why then do they use diode lasers? Figure 6.26 gives us the answer. Laser diode Light output LED Wavelength Figure 6.26 Relative comparison of spectral output of a laser versus LED . From this figure, it is seen that the light output from the LEDs is much less compared to the laser diode1,2. Secondly and more importantly, the output wavelength of the diode laser is only a few nm wide in range, whereas light from an LED is distributed across a wider range of wavelengths (tens of nm), an undesirable feature, because it adversely affects attenuation as well as dispersion. Whichever the source, the light emitted has to be effectively coupled into the fiber. From a discussion of the acceptance angle of a fiber in a previous section, it is imperative that the source and fiber be properly aligned and the angle of light emission from the source carefully controlled. Secondly, the numerical apertures (NA) discussed earlier, a measure 178 of the light gathering ability of any optical component, of the source and the fiber will have to be carefully matched. If not, this leads to attenuation of light even prior to light entering the fiber. The equation relating to the dB loss due to NA mismatch is dBNA = -10log NAfiber 2 NAsource Similarly, the spatial extent of the source and the diameter of the core need to be matched to minimize entrance losses given by dBsize = -10log dcore 2 dsource In these equations, NAfiber = numerical aperture of the fiber calculated using ncore and ncladding NAsource= numerical aperture of the source including any optics present dcore = diameter of the fiber core dsource = diameter of the light source If the NA of the source is larger than that of the fiber or if the diameter of the source is larger than the core diameter of the fiber, losses occur. Otherwise, there are no losses due to these mechanisms. Thus, it is desirable to keep both the NA and the diameter of the source as small as possible. However, real life is such that light from typical sources, such as an LED, tends to spread out prior to incidence on the entrance end of the fiber, leading to losses. (see also Appendix V for additional formulas). Example 6.9 A source has an NA of 0.25 and an output diameter of 80 µm. Light is incident on a 50/125 fiber with core refractive index of 1.500 and cladding index of 1.488. Calculate dB losses due to NA mismatch and diameter mismatch. We first need to calculate the NA of the fiber using ____________ NAfiber = √ncore2 - ncladding2 ____________ NAfiber = √1.5002 – 1.4882 dBNA = 0.189 = -10log NAfiber 2 NAsource dBNA = = -10log 0.189 2 0.25 2.43 dB Answer 179 dBsize = -10log dcore 2 dsource dBsize = -10log 50 2 80 4.08 dB Answer = Total dB loss due to NA and diameter mismatch is 6.51 dB. Let us say that a 2 W source is available. Of the 2 W, only a smaller portion will actually enter the fiber, given by 6.51 = 10 log (2 W/Pout) resulting in a power of only 0.447 W (447 mW) actually entering the fiber. A discussion of semiconductor diodes was presented in Chapter 4. The semiconductor laser, also known as diode laser or injection laser, operates on the principle of a semiconductor junction. Examples of materials used as active medium in diode lasers include gallium aluminum arsenide, gallium indium phosphide, and gallium aluminum indium phosphide. Exercise 6.6A For each of the problems below calculate the dB loss due to NA mismatch. Problem NAsource NAfiber 1 0.32 0.30 2 0.32 0.28 3 0.30 0.27 4 0.29 0.27 5 0.31 0.28 6 0.33 0.25 For each of the problems below calculate the dB loss due to diameter mismatch. Problem dsource (µm) dcore (µm) 7 100 62.5 8 100 50 9 120 100 10 120 50 11 120 62.5 12 80 50 6.7 Fiber Optics Connectors1,2,2e,5a,6,7 Fibers need to be terminated using connectors. These connectors interface with other devices or other connectors with matching interfaces. The fiber optics technician is often required to terminate fibers, which is both an art and a science. The fiber needs to be carefully stripped first to remove the outer jacket and the buffer and to cut off the excess Kevlar®. One needs to follow the safety rules discussed in Chapter 3. These include 180 using a dark surface so that fiber debris can be easily seen, avoiding puncture wounds, stripping and cleaving the fiber below eye level and cleaning debris carefully by pressing a masking tape on the dark work surface and disposing of the tape properly. Safety, cleanliness, and careful handling are important practices that a fiber optics technician should value and cultivate. Once a sufficient length of the cable has been stripped, the outside of the cladding needs to be wiped with a lint free tissue soaked in isopropyl alcohol (IPA). This will remove any debris, as well as oil residues and fingerprints. Then, depending on the type of connector, one follows one of two procedures. If one uses hot melt connectors, the connector needs to be heated in a hot melt oven following the directions for a sufficiently long enough time to melt the adhesive inside the connector. There is not much of a danger of prolonged heating or problems associated with the adhesive oozing out of the connector. At this point, the connector is removed from the oven and the fiber end quickly poked through the connector. If the fiber experiences resistance, it is an indication that the connector needs to stay in the oven longer. Once the fiber has been pushed all the way through the connector, allow the connector to cool. If one uses an epoxy, it must be dispensed into the connector, the fiber inserted, and the assembly cured in an oven, and then the connector be allowed to cool. When the connector is cool enough to handle with bare fingers, the fiber needs to be cleaved using a special scribe. Great care must be exercised with this step. If the fiber end falls over upon contact with the scribe, it is a sign that the fiber has cracked inside the connector, and, hence, the connector is not good. The proper procedure is to scribe the fiber gently, and then using two fingers, pull the fiber up. If the fiber is removed in this fashion then it has been cleaved cleanly at the base. The next steps involve polishing the connector using a series of abrasive pads of successively finer grit size, using IPA during the process. Care must be exercised not to over-polish the fiber to such an extent that a portion of the fiber is gouged out below the level of the connector ferrule. (A ferrule is a member inside the connector that holds the fiber in place so that fiber alignment is ensured.) Using a power meter, the dB loss of a connector can be measured. If the dB loss is slightly higher than that allowed, some polishing will enable the loss to be brought down to within specifications. The procedure for attaching the second type of connector, the one that uses anaerobic glue, is very similar with the following exceptions. There is no need to heat up the connector. After stripping and cleaning the outside of the fiber with IPA soaked lint free tissue, one applies a liquid primer over the fiber, and then inserts the fiber through the glue-filled connector. No heat curing is necessary, since the glue sets instantly. The cleaving and polishing procedures are the same as those for the hot melt connector mentioned above. Prior to cleaving, tug on the strength members of the fiber to ensure that the fiber has been set in place by the hot melt glue or the anaerobic glue or the thermal cure epoxy. 181 In addition to safety and cleanliness, proper handling of fiber cables is important. As mentioned previously, debris and residue on connector faces can cause signal attenuation. Care must be exercised not to nick or scratch the cladding during connectorization as this can lead to microbends, another source of attenuation in fibers. Additionally, loss in fibers in connectors and couplers may be due to misalignments or gaps as shown in Figure 6.27. cladding cladding core core cladding End offset cladding Angular offset End Gap Figure 6.27 A few possible sources of power losses in couplers and connectors(highly exaggerated) . The situations are exaggerated for clarity in the figures above. Nevertheless, these mechanisms lead to signal attenuation with connectors. Connectors and splices should introduce no more than 0.2 dB loss for telecommunication purposes. Systems using plastic fibers for short distance signal transmission can tolerate up to 3 dB loss for connectors. A further requirement for connectiorization, in addition to ease of installation and low cost, is reproducibility and repeatability of dB loss in a connector so that the losses do not vary with time, location, and the number of times they are disconnected and reconnected. Discussion of losses associated with fiber insertion into a connector, mismatches in NA and diameter of two fibers brought together, reflection losses and misalignment losses shown in the diagrams above are beyond the scope of this introductory book. Connectors are available in many formats. In the last decade, connectors were standardized. The classification of connectors as SC type, ST-type, D4-type, FC-type, etc. is based on the coupling mechanism and ferrule design and dimensions. A ferrule, as mentioned earlier, is a member of the connector that enables the alignment of the fiber inside the connector. Among the connectors, the ST type is the most popular. It uses a ceramic ferrule and a quick release coupling and is used extensively in LAN, premise wiring, etc. Splicing, on the other hand, requires two fiber ends to be brought together and joined via mechanical means or melting (called fusion splicing.) While expensive, fusion splicing results in very low loss compared to mechanical splicing. 182 To conclude this chapter, let us summarize some good practices when handling fiber cables, which are (i) never wear a fiber around your neck, (ii) avoid twisting or pinching cables, (iii) never bend the cable to a radius less than ten times the outer jacket diameter, (iv) avoid pulling on a fiber or connector – pull on strength members only, and (v) never place objects on top of fibers. 6.8 Suggested Laboratory Experiments – Identification of Fiber Cable Elements Follow safety precautions of fiber handling! Wear safety gloves and safety glasses! Required materials and tools: -A simplex cable with 62.5/125 designation. (If a simplex cable is not available, use a duplex, pull the two sections apart; each will then be a simplex fiber) -Kevlar® cutter -Micrometer screw (metric) -Cable stripper -Dark surface workbench • Cut about two feet (60 cm) of simplex cable from the cable spool using the Kevlar® cutter. The cutter should cut through all the elements of the cable, namely outer jacket, Kevlar®, inner jacket, buffer, cladding and core, in that order. • Using the stripper, cut about 3 inches (75 mm) of the jacket off from either end of the cable. At this point, one should see the Kevlar® fiber, which resembles fiber glass in appearance. • Carefully move the Kevlar® sideways and cut off using the Kevlar® cutter. At this point, you should see the inner jacket of the fiber. • Determine what a main scale division measures and what a vernier scale division on the micrometer measures and record it in Table I • Measure the outer diameter of the inner jacket using a micrometer screw. Record in Table I. • Use a 125 µm fiber stripper. Keep the fiber below eye level and as close to the dark surface workbench as possible. Keeping the fiber horizontal to the work surface, angle the stripper as horizontal as possible to the surface, strip ¼ inch (6 mm) off each time. If the stripper is held close to vertical, it may cut through the core and the cladding! Using a shallow angle from the horizontal will ensure that the stripper cuts through the inner jacket and the buffer coating and not the cladding/core inner layers. Moving the ¼” cut piece off gently to the free end will enable the buffer coating also to be cleanly removed simultaneously. If the buffer coating is not removed completely you will see pieces of it left behind on the otherwise smooth cladding outer surface. • Once 3” of outer jacket has been cleanly stripped of either end, gently wipe the outer surface of the cladding with an alcohol (IPA) soaked lint free wipe. • Measure the outer diameter of the cladding with a micrometer screw. The measurement should be close to 125 µm • Record it in Table 6.1. 183 • • If your laboratory has connectors follow the connector attachment process described in the manufacturer’s manual and/or video. Otherwise turn in your report to your instructor/lab aide. Using a masking type wrapped around two fingers, with the sticky side of the tape out, gently dab the work surface to remove any fiber debris. Dispose of the masking tape in suitable disposal containers labeled appropriately. Dispose of alcohol soaked wipes in appropriate chemical disposal containers. Seek medical help if you believe that glass fiber might have caused puncture wound. Table 6.1 Each division on the main scale corresponds to -> How many revolutions of the vernier scale are required to travel one main scale division -> Distance traveled on the main scale for one vernier scale rotation -> Number of graduations on the vernier scale -> Each division on the vernier scale corresponds to -> Outer diameter of the inner jacket is -> Outer diameter of the cladding is -> • • • • • • • • • Chapter Summary Fiber optics has been in use for many years. The major difference is that, in recent years, systems have been developed with minimum losses so that communication over long distances is now feasible. The essential elements of a fiber are the core and the cladding which possess optical properties; the other elements of buffer coating, jacketing and Kevlar® provide structural integrity and mechanical protection to the delicate fiber inside. Fibers are available with different core diameters and cladding diameters. However, the cladding diameter is often available in a standard 125 µm outer diameter format. Fibers are classified as single mode or multimode to denote the number of pathways that light is allowed to travel in. Multimode fibers result in modal dispersion and pulse broadening. Fiber is classified based on the refractive index profile as being step index or graded index. For multimode fibers, modal dispersion is less with a graded index fiber than with a step index fiber. Numerical aperture (NA) of a fiber refers to a fiber’s light gathering ability. Given the NA, core diameter and wavelength of light, one can determine the number of modes a fiber can support through the V-number. In addition to modal dispersion, signal degradation through dispersion can occur due to waveguide dispersion, polarization mode dispersion and material dispersion. Not all types of dispersion exist in single mode or multimode fibers. For any core/cladding combination, there is a critical angle that needs to be exceeded in order for successful light propagation through TIR to occur. To meet this criterion, there exist certain values of acceptance angle at the fiber entrance. 184 • • • • • • Light that is incident on the entrance end of the fiber exceeding the acceptance angle will not propagate down the fiber through TIR. Power loss occurs due to various mechanisms. Fresnel reflection losses occur at entrance end and at splices. Losses occur due to NA and diameter mismatches between the source and the fiber core. Once light enters the fiber, further losses occur due impurities in the fiber, micro- and macrobends, end gap, end and angular misalignment with connectors and residues/debris/moisture on connector faces. Power loss is characterized using dB which relates the input and output powers using a logarithmic function. Fiber cable losses are specified in dB/km of fiber length. Typical light sources used in fiber optic transmitters operate in the IR region predominantly at 850nm, 1300 nm and 1550 nm using either a laser diode or a light emitting diode (LED). LEDs, while inexpensive and long lasting, suffer from a wide spectral range of wavelengths, whereas laser diodes have 1-3 nm typical spectral range with very high output power. dB/km losses decrease with increasing wavelength irrespective of fiber size or mode. Connectors are used to terminate fibers. Connectors are available in hot melt or anaerobic adhesive types and in many formats such as ST, SC, FC, etc. Fibers need to be stripped, cleaned, connector attached and polished to result in a low loss connection. Safe practices must be followed when handling fibers and fiber debris must be disposed of properly. Safety, handling and cleanliness are important when handling fibers so that a technician can deliver a quality product to the customer. End of Chapter Exercises 1. In each of the following cases, indicate if it is (a) multimode or single mode fiber, (b) step index or graded index and (c) glass or plastic core. (i) 980/1000, (ii) 85/125, (iii) 12/125, (iv) 50/80, (v) 5/85, (vi) 485/500 2. For a step index fiber with designation 10/125, using a 1550 nm source and a 0.222 numerical aperture fiber, calculate how many modes the fiber will support. 3. For a graded index fiber with designation 62.5/125, using an 850 nm source and a 0.235 numerical aperture fiber, calculate how many modes the fiber will support. Calculate (a) critical angle in the core, (b) acceptance angle and (c) numerical aperture for the following fiber cables. 4. ncore=1.48 ncladding= 1.45 5. ncore=1.50 ncladding= 1.46 6. For a fiber with six mechanical splices, what is the total Fresnel reflection loss? Complete the following table for a 3.3 km fiber cable 185 Problem# 7 8 9 10 Pin 15 mW 25 mW Pout 400 µW 1 mW 0.3 mW dB 7.5 dB/km 1 4 For each of the problems below calculate the dB loss due to NA mismatches. Problem NAsource NAfiber 11 0.310 0.285 12 0.300 0.265 For each of the problems below calculate the dB loss due to NA mismatches. Problem dsource (µm) dcore (µm) 13 100 12 14 75 62.5 Group Discussion Questions These require the students to form small groups of three or four and discuss each question. Some questions may require you to do a search on the Internet and find the answers. 1. How do they make the core and the cladding in the factory? 2. Explore the existence of a dual core, single cladding fiber. 3. Explore the existence of a multiple cladding fiber. 4. Can you build a fiber optic cable with an air core? 5. How can fiber optics be used in medicine and surgery? 6. Compare and contrast a laser and an LED as light sources for fiber optics. 7. What is a fiber laser? 8. What is an OTDR? How does it work? 9. Investigate methods for splicing fiber cables together, example fusion splicing vs. mechanical splicing. 10. How do connectors differ from couplers? References Sterling, D. Jr., “Technician’s Guide to Fiber Optics.” Albany: Delmar Publishers, 2000. 2 Meardon, S.L.W. “The Elements of Fiber Optics.” Englewood Cliffs: Prentice Hall, 1993. 2a Robinson, D., Hallam. A., and Hapanowicz, R., “Key Parameters for Testing Multimode Fiber Optic Cables and Transmitters” http://www.ardenphotonics.com/datasheets/WhitePaper-Key-MultimodeParameters01.pdf 2b U of Tenn Phusics Department Optics website: http://electron9.phys.utk.edu/optics 2c Digital Library Network for Engineering and Technology www.dlnet.vt.edu/.../EE000000/EE007000/ EE007002/DISK1/DLNET-09-03-20020237/resources/Topic%204.pdf . 2d Bell College School of Science and Technology website http://floti.bell.ac.uk/MathsPhysics/system.htm 1 186 2e Corning Cable Systems web site http://www.corningcablesystems.com/web/college/fibertutorial.nsf/ofpara?OpenForm 3 Floyd, T.L. “Electronics Circuits Fundamentals.” Englewood Cliffs: Prentice Hall, 2002. 4 Vasan, Srini. “Technical Mathematics (with Applications in Electronics and Photonics.” Victoria: Trafford Publishing, 2003. 5 Jenkins, F. and H. White. “Fundamentals of Optics.” Singapore: McGraw Hill, 1981. 5a Lanshack Technical Information on Fiber Optic Connectors http://www.lanshack.com/fiber-optic-tutorial-termination.asp 5b Society of Cable Telecommunication Engineers, “Calculating fiber losses and distances- A Technology Overview.” http://www.scte.org/chapters/cascade/Calculating%20Fiber%20Loss.pdf (Fresnel loss) 5c SEL Fiber Optics Products and Applications data sheets http://www.selcom.com/lit/FiberOpticsDS.pdf 6 Hayes, J. “Fiber Optics Technician’s Manual.” Delmar: Albany, 2001. 7 3M Corporation, Hot Melt and Epoxy Fiber Optic Connector Attachment Procedures, video and documentation. 187 Chapter 7 Polarization 7.1 Electric and Magnetic Fields 7.2 Polarization, Types of Polarization 7.3 Methods of Polarizing Light 7.4 Law of Malus 7.5 Birefringence 7.6 Laboratory Experiments Chapter Summary End of Chapter Exercises Group Discussion Questions Chapter Objectives After completing the chapter, the student should be able to (i) distinguish between electric and magnetic field (or vector) (ii) characterize different types of polarization (iii) distinguish the p-ray from the s-ray (iv) explain the different methods of producing polarized light. (v) derive an expression for Brewster angle and perform calculations (vi) apply the Law of Malus Key words: polarization, Poynting vector, E-field, B-field, Brewster angle, Extinction Ratio, Law of Malus, birefringence 7.1 Electric and Magnetic Fields In Figure 7.1, the entire electromagnetic spectrum is repeated below from Chapter 2 with approximate values of wavelength and frequency. Notice that the shorter wavelengths are towards the left, and the longer wavelengths are to the right, whereas the shorter frequencies are to the right and larger frequencies to the left. Frequency in Hz. 1019--1018--1017--1016--1015--1014--1013--1012--1011--1010--109--108—107—106—105--104--Gamma rays-X-rays—UV-vis-IR------microwave---------- FM radio-AM radio-TV— Wavelength in m. 10-11--10-10-10-9--10-8--10-7-- 10-6--10-5-- 10-4-- 10-3--10-2-- 10-1-100---101---102---103--104 Figure 7.1 The electromagnetic spectrum. No matter which portion of the electromagnetic spectrum we are looking at, the photon, which has been characterized in Chapter 2, has the same velocity in vacuum of 3 x 108 m/s denoted by the letter c. Furthermore, irrespective of the wavelength or frequency, the light wave exhibits both electrical and magnetic properties, and, hence, the name electromagnetic radiation. This concept may be difficult to comprehend at first since we know that we do not get “shocked” by the electrical component of the light coming from 188 the sun or from a lamp. Nevertheless, electrical and magnetic fields make up electromagnetic radiation. While traveling light waves are in reality represented by more complex equations, for the sake of ease of understanding, we can characterize a wave by a simple sine function. Its equation is y = A sin(2πft) where y is the height of the wave or vibration from the reference plane at time t. The wave has an amplitude (i.e. peak height) of A and frequency f. This represents a stationary wave. A light wave, on the other hand, is traveling. In reality, an electromagnetic wave consists of two types of vibrations, one representing the electrical component, the other the magnetic component1,2. The two waves lie in perpendicular planes with respect to each other, as shown by the simplified diagram in Figure 7.2. y y λ x x λ z z Figure 7.2 Electric and magnetic fields. The figure on the left hand side represents the electrical wave corresponding to one of millions of light waves. The electrical wave was purposefully chosen so that it lies on the plane of the paper on which it is drawn, i.e., the xy plane. The arrows are drawn further to clarify that all the arrows that make up the vertical heights of the wave location lie in the xy plane. For the corresponding case, the magnetic wave is shown on the right hand side. Here the wave oscillates in the xz plane, i.e., in a plane perpendicular to the paper. The arrows are shown as dashed to distinguish from the arrows of the electrical field. The dashed arrows lie in the xz plane. The two waves, the electrical wave on the left and the magnetic wave on the right, lie in planes perpendicular to each other and the light wave is a composite of both. The figures shown above are one of several possibilities, where we confined the electrical wave to the plane of the paper. This does not necessarily have to be. The light we receive from typical sources have the electrical waves existing in many different orientations and planes. In each case, the magnetic wave will be in a plane perpendicular to the electrical wave. Thus, light from such sources are called unpolarized, i.e. the electric wave (or field) is not preferentially confined to any one plane. In fact it is present 189 in all planes. The same is true of the corresponding magnetic waves (or fields). Since the magnetic wave is always perpendicular to the electrical wave, in future, we will denote the location of the electrical field only as a convention. A vector is a quantity which has a magnitude and a direction. Looking at Figure 7.2, left, we see arrows lined up vertically which are of different lengths, however, they are all confined to the vertical direction. We also use the terminology E-vector to denote the electrical wave. The E-vector lies in the xy plane. The corresponding magnetic wave, called the B-vector (see figure on the right) is confined to the xz plane as shown by the dashed arrows. In unpolarized light the E-vector is oriented in all directions (the same being true for the B-vector.) The Poynting vector2 provides the direction of travel of the light wave. If one uses the left hand thumb rule, the direction of travel can be determined. Stretch the thumb, index finger and the middle finger of your left hand so that they are perpendicular to one another. If you orient the index finger in the direction of the E-vector (for example, up), the thumb in the direction of the B-vector (for example, out of the plane of the paper), the direction to which the middle finger points denotes the direction of light travel, which in this case is from left to right of the page. This direction is denoted by the Poynting vector, a vector that points in the direction of light travel. The left hand thumb rule use is illustrated in Figure 7.3. The principle works equally well with the right hand also which is used in physics. However, we chose to use the left hand rule only to maintain the convention throughout the text that light travels from left to right. The wavelength of the electrical, magnetic, and light wave are all the same, equal to λ. Index finger E-vector Middle finger Thumb B-vector Direction of light wave (Poynting vector) Figure 7.3 Determining the direction of the Poynting vector. 7.2 Polarization, Types of Polarization Unpolarized light can be polarized. Methods of polarization are presented in the next section. This section looks at polarized light of different forms. Unpolarized light for which the E-vector is oriented every which way is designated as shown in Figure 7.42. An alternate way of representing unpolarized light is shown on the right. The arrows represent the plane of the E-vector, also called the plane of polarization. The circular dots represent the E-vector in a direction perpendicular to the paper. In essence, while looking head on when only the tip of the arrow is visible. The figure in the right hand side suggests that the E-vector is present in two extreme planes, 90o with respect to each other, and, therefore, all other planes in between imply unpolarized nature. 190 Figure 7.4 Representation of unpolarized light. (Reproduced with permission of The McGraw Hill Companies from “Fundamentals of Optics” JENKINS and WHITE, 4th ed. © 1981). Light can be polarized so that the E-vector remains in one plane only. The resulting light is called plane-polarized or linearly polarized. Both terminologies are equivalent, the first implies that the electrical wave is confined to a plane; the second implies that the direction of the E-vector is confined to a fixed direction. Two examples of plane polarized light are illustrated in Figure 7.52. Figure 7.5 Representation of plane-polarized light. (Reproduced with permission of The McGraw Hill Companies from “Fundamentals of Optics” JENKINS and WHITE, 4th ed. © 1981). In the left figure, light is polarized so that the E-vector lies in the plane of the paper. This is called vertically polarized light, or plane polarized in the plane of the paper. The figure on the right hand side is horizontally polarized light, or light that is plane polarized in a direction perpendicular to the plane of the paper. The notation of the direction of polarization has meaning only with respect to a reference plane of incidence (on a target). A ray of light that follows the first type of polarization is also referred to as p-ray (p meaning E-vector polarized in a direction parallel to the plane of incidence). A ray of light that follows the second type of polarization is also referred to as an s-ray (meaning E-vector polarized in a direction perpendicular to the plane of incidence, “s” representing the German word senkrecht, meaning perpendicular2.) One can break down a three dimensional space only into two planes, vertical and horizontal. While light can be preferentially polarized in any direction, it can be conveniently resolved into mutually orthogonal directions. The p- and the s- ray together represent the two possibilities of plane (or linearly) polarized light with reference to the plane of incidence of light2. 191 Let us look at other types of polarization that are not linear or plane polarized, namely circular and elliptical polarization1. The Poynting vector gives us the direction of light wave travel. If during one cycle of the sine wave, the E-vector undergoes clockwise rotation (with respect to an observer “looking” at the wave head on as it approaches that person) of one cycle while its magnitude remains the same, the light is said to be circularly polarized clockwise, or right circularly polarized. If during one cycle of the sine wave, the E-vector undergoes counterclockwise rotation of one cycle while its magnitude remains the same, the light is said to be circularly polarized counterclockwise, or left circularly polarized. In either of the cases, however, if the magnitude of the E-vector changes from a maximum to a minimum, to a maximum, to a minimum and back to a maximum, for example, in one cycle, the light is said to be either elliptically polarized, either clockwise (right) or counterclockwise (left)1, as shown in Figure 7.6-7.7. λ Circularly polarized Clockwise Counterclockwise denotes the E-vector Figure 7.6 Representation of circularly polarized light. λ Elliptically polarized Clockwise Counterclockwise Emin Emax Emax Figure 7.7 Representation of elliptically polarized light. Thus, with circularly and elliptically polarized light, the E-vector goes though a revolution when the light wave traverses a cycle. 7.3 Methods of Polarizing Light Lasers can be fabricated so that they produce polarized light. By using Brewster windows, the resultant output of such lasers is polarized3. Discussion of this process is 192 provided in a later section. Polarized light also results from accelerating electrons subjected to high energy fields. Polarization by reflection off dielectric materials: A dielectric material does not conduct electricity. An example is glass. When unpolarized light is incident on, say, a glass surface, Fresnel reflection takes place. While the majority of light is refracted into the glass a portion is reflected. The s-ray is reflected to a greater extent than the p-ray at all angles of incidence except at the grazing angle (90o incidence), when 100% of both types of rays are reflected. The Figure 7.82 shows the relationship among fraction of light reflected as a function of the angle of incidence for the p-ray, the s-ray and the composite for a typical glass material of refractive index 1.5. Reflectance Glass (n=1.5) 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 p-ray s-ray Both 0 50 Angle of incidence, degrees Figure 7.8 Reflectance of p-ray and s-ray from glass. (Reproduced with permission of The McGraw Hill Companies from “Fundamentals of Optics” JENKINS and WHITE, 4th ed. © 1981). At normal incidence, both the p-ray and s-ray are reflected equally. It is seen that at a certain angle (56.3o to be exact), none of the p-ray is reflected and about 15% of the s-ray is reflected. This is called the Brewster angle. Thus, when unpolarized light is incident at the Brewster angle at any interface (glass, water, etc.), the reflected ray is all s-ray and no p-ray, i.e. the reflected light is polarized perpendicular to the plane of incidence. This is one method of creating plane-polarized light starting from unpolarized light. What would then happen to the refracted light inside glass? The refracted light would still be largely unpolarized, however, it would have less of the s-ray component and more 193 of the p-ray component, relatively speaking. Thus, the refracted light is relatively more polarized than the incident ray. This is true of sunlight passing through a glass window pane and entering the house. To understand these phenomena further, let us look at Figure 7.9. Fresnel reflection Air Medium Fresnel reflection Air Figure 7.9 Fresnel reflections from dielectric interfaces. Whatever be the angle of incidence, Fresnel reflection will always take place at both the air-medium interface and the medium-air interface, as shown by the dashed rays. However, for the Brewster condition, the reflected ray and the refracted ray should be perpendicular to each other, as shown in Figure 7.10. Incident light θ1 θ1 Reflected light 90o - θ1 AIR 90o θ2 θ1 Refracted light GLASS Figure 7.10 Illustration for Brewster angle derivation (air to glass). Light is traveling from air (medium 1) and it strikes glass surface (medium 2) at an angle of incidence θ1 . Per the law of reflection, some of the light will be reflected at the same angle θ1 from the normal. Much of the balance will be refracted into glass, obeying Snell’s law, such that it will bend toward the normal making an angle θ2 while in glass. For the reflected light to be completely polarized as an s-ray, the reflected and refracted rays must be perpendicular to each other. Therefore, θ2 = 90o - θ1 194 Following Snell’s law, sin θ1 = n sin θ2 or sin θ1 sin (90o - θ1) = n It is well known from trigonometric principles (see Chapter 5), that sin (90o - θ1) = cos (θ1) Substituting, we obtain, sin θ1 cos θ1 =n tan θ1 or tan B = where B is called the Brewster’s angle. B = tan-1n This equation is valid for light traveling from air into a medium of refractive index n and is known as Brewster’s Law1. Example 7.1 Calculate the Brewster’s angle for water. B B = = tan-1n tan-11.333 53.1o Answer = It can be easily shown that if light is incident upon the air/glass interface at the Brewster angle B, then, it will be incident at the corresponding Brewster angle at the glass/air interface also2. B Fresnel reflection Air o 90 -B B B Fresnel reflection Glass B o 90o-B 90 -B B Air Transmitted ray Figure 7.11 Illustration for Brewster angle derivation. 195 Light is incident at the glass/air interface at an angle of incidence of 90o-B and will undergo Fresnel reflection at an angle of reflection of 90o-B in Figure 7.11. Using the principle of reversibility, the angle of refraction in air will also be B. This will imply that the refracted ray makes 90o-B with the glass-air interface. Thus, the Fresnel reflected ray from the glass/air interface and the refracted ray into air have an angle of 90o between them. Therefore, the Fresnel reflected ray at the bottom interface is also polarized as an s-ray and the transmitted ray has less s-ray component compared to the original incident ray. We have proved that if B is the Brewster angle for light traveling from medium 1 to medium 2, then 90o-B is the Brewster angle for light traveling from medium 2 to medium 1. This is consistent with principle of reversibility discussed in Chapter 5. The general formula for the Brewster angle for light traveling from medium 1 to medium 2 is1 B = tan-1(n2/n1) Needless to say, similar to the refractive index, the Brewster angle also depends on the wavelength. Example 7.2 Calculate the Brewster angle for light traveling from water (n=1.333) to air. B = tan-1(n2/n1) B = tan-1(1/1.333) = 36.9o Answer which is 90o minus the answer from the previous example. Example 7.3 Calculate the Brewster angle for light traveling from water (n=1.333) to glass (n=1.5). B = tan-1(n2/n1) B = tan-1(1.5/1.333) = 48.4o Answer Exercise 7.3A Calculate the Brewster angle for the following cases: (Data2: Refractive index at 589 nm: Diamond 2.4168, Ruby 1.769, Flint glass (dense) 1.669, Ethanol 1.359, Water 1.333.) Light is traveling from medium 1 to medium 2 Problem Medium 1 Medium 2 1 Diamond Flint glass (dense) 2 Ruby Ethanol 3 Water Diamond 4 Ethanol Ruby 5 Ruby Flint glass (dense) 6 Diamond Ruby 7 Flint glass (dense) Ethanol 196 8 9 10 11 12 Problem Flint glass (dense) Ruby Air Ethanol Water Medium 1 Water Water Diamond Flint glass (dense) Ruby Medium 2 When light travels from a medium of higher refractive index to one of lower refractive index, we learned from Chapter 5 that there exists a critical angle below which refraction into the second medium will occur. Since TIR will occur once the critical angle is exceeded, the reflectance will rapidly increase to 1 (100%) at the critical angle. The Brewster angle will be smaller than critical angle. A plot of reflectance of the s- and prays for light traveling from glass to air is shown in Figure 7.122. It is seen that the angles of incidence are confined from 0o to 41.8o, the critical angle from glass to air, unlike from 0o to 90o for the case of air to glass. At 33.7o, the reflected light is polarized with s-ray only and no p-ray is reflected. This is the Brewster angle for light traveling from glass to air. At the Brewster angle, 15% of the s-ray and 0% of the p-ray are reflected. Reflectance Glass (n=1.5) 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 p-ray s-ray Both 0 20 40 60 Angle of incidence, degrees Figure 7.12 Reflectance of p- and s-rays at the glass-air interface. (Reproduced with permission of The McGraw Hill Companies from “Fundamentals of Optics” JENKINS and WHITE, 4th ed. © 1981). Polarization by refraction. As discussed previously, if light is incident at the Brewster angle, the reflected light is polarized as the s-ray only, with no p-ray. For air-glass system, 15% of the s-ray is reflected, meaning 85% is transmitted. When light exits glass into air at the other side, 197 again, due to Brewster conditions as proved previously, 15% of the s-ray is reflected and none of the p-ray is reflected. Thus, the transmitted beam has lost about 28% of the s-ray component [1-0.152 ~ 0.28]and none of the p-ray component, meaning it is preferentially polarized to comprise more of the p-component and less of the s-component, as shown in Figure 7.132. Unpolarized light s-ray B B Air 90o-B B Glass s-ray B 90o-B 90o-B B Air Figure 7.13 Illustration of polarization by refraction. Transmitted ray has less s-ray component and more of the p-ray component. Thus, if one stacks a few glass plates as in a sandwich pattern shown below, the s-ray is reflected off successively at each interface so that the final transmitted ray is essentially p-ray2. Thus, via reflection and transmission, one can separate out the two components that make up unpolarized light, as illustrated in Figure 7.14. AIR Unpolarized light B All reflected rays are s-ray s-ray Essentially all p-ray Transmitted ray AIR Figure 7.14 Illustration of polarization by refraction by a set of glass plates. Now we can look at the principle of Brewster windows. We know that for light traveling from air to glass of refractive index 1.5, the Brewster angle is 56.3o. If a window is positioned in the path of the laser beam at the Brewster angle, 15% of the s-ray is reflected off the window through Fresnel reflection, while none of the p-ray is lost due to 198 reflection. Thus, all of the p-ray is available for amplification via feedback mechanism and multiple reflections between the OC and HR. At every pass, 15% the s-ray is reflected off and out of the optical axis of the tube whereas the p-ray continues to be amplified. It results in polarized light coming out of a laser fitted with Brewster’s window. Figure 7.15 illustrates the concept2,3. s-ray (reflected off and lost every pass) Laser Tube Laser output p-ray Figure 7.15 Brewster window of a laser. Polarization by absorption: Polaroid® type polarizers and dichroic materials such as tourmaline crystal use the principle of selective absorption. To understand the mechanism, assume that you have hundreds of needles (about 2” long) and a grill such as you may find in storm sewers which have long, narrow openings, say 6” long and ½” wide to allow for rain water to go through. This is illustrated in Figure 7.16. Needles held parallel to the grill but oriented in all directions Only needles which were parallel to the grill openings fall through Figure 7.16 A rough illustration to demonstrate selective absorption. A crystal of tourmaline with its axis held vertical absorbs the E-vector in the horizontal direction preferentially over the vertical component. As a result if unpolarized light passes through such a crystal, the output light from the crystal is practically vertically polarized with the horizontal component absorbed as illustrated in Figure 7.171,3. Polaroid® films are fabricated with crystals that are long in one direction and short in another held in place as a sheet using binders and other vehicles. Polaroid® type polarizers are readily available and mounted on a circular slide carrier with angular markings from 0o all the way to 350o (since 360o is the same as 0o), as shown in Figure 7.18. If the polarizer is mounted with the orientation facing up, the light exiting the polarizer will be polarized vertically as in the tourmaline crystal. Polaroid® type polarizers are commercially available and they can be used to polarize unpolarized light or to analyze linearly polarized light. If unpolarized light passes through a polarizer that is aligned to let only vertically polarized light to pass through, 199 Unpolarized light Crystal Vertically polarized light Figure 7.17 Illustration of selective absorption. 0o o 30o 330 o Polaroid® film o 300 60 270o 90o 240o 210o 120o 180o 150o Figure 7.18 Polaroid® film polarizer. then a second polarizer, which we call the analyzer, held at 90o with respect to the first polarizer will practically extinguish the output beam1,2, as shown in Figure 7.19. Polarizer 0o Analyzer 90o Practically no light Unpolarized light Vertically Polarized Figure 7.19 Use of polarizer and analyzer. The law governing the different degrees of extinction is described in the next section. As one rotates the analyzer from 0o to 90o, the light intensity exiting the analyzer goes from bright to dark, and then from 90o to 180o, it goes from dark to bright. In the next quarter revolution, the light intensity exiting the analyzer goes from bright to dark and in the last quarter revolution, it goes from dark to light. Thus, during a full revolution, it goes through two locations of bright, two locations of dark and many other locations of intermediate light intensities. If one suspects that light 200 is linearly polarized, using a Polaroid® type polarizer as an analyzer and by rotating the analyzer, if the brightness of light goes through the cycles described above, then the light was linearly polarized to begin with. If not, one can conclude that either the original light is unpolarized or polarized in a fashion that is not linear. Polarization by scattering: It is our common experience that dust particles, aerosols, fog, etc. scatter light. The degree of scattering depends on the wavelength of light to begin with and the type of scattering depends on the size of particle as well in many cases. More detailed discussion on scattering is provided in the chapter on absorption. Consider a box as shown below filled with air and fine dust particles to simulate atmospheric conditions. If unpolarized light is passed through this box, light that is scattered at 90o with respect to the optical axis of the incident beam will be preferentially vertically polarized, as shown in Figure 7.20 1,2. Incident light Transmitted light Scattered light vertically polarized Figure 7.20 Polarization by scattering. Scattering occurs at all angles by particles and molecules. However, the light that is scattered at 90o is practically vertically polarized whereas the transmitted light will continue to be unpolarized, however with the reduced component of the vertical E-vector. One can realize this phenomenon in nature by looking at sky in the north or south direction during either early in the morning or late in the evening when the sun is either in the east or the west. Since the north or south direction is perpendicular to the east/west direction from where the sunlight is emanating, the light viewed in one of these two directions will be vertically polarized. If you use a linear polarizer as an analyzer, view this light as you rotate the analyzer. You will go through light and dark regions two times each as you rotate the analyzer a full revolution. This proves that the scattered light is linearly polarized1,2. The phenomenon as to why the sky appears blue most of the day and appears red during dawn and dusk is also attributed to scattering and will be explained in the chapter on absorption and scattering. 201 Polarization by reflection off metal surfaces: Experience tells us that highly polished metallic surfaces reflect a higher percentage of light than a surface such as glass at normal incidence, i.e. the reflectance of metals is higher. While glass reflects only about 4% of the incident white light at normal incidence, metals such as silver and aluminum reflect nearly 90% at normal incidence. However, the type of metal, its surface preparation, wavelength of light and angle of incidence play crucial roles in the property of the reflected light2. At angles other than normal incidence, the p-ray and s-ray component of plane-polarized incident light are reflected off metal surfaces with a phase difference leading to elliptically polarized reflected light. The only exceptions are when the E-vector of incident light lies either in the plane of incidence or perpendicular to it, in which case the reflected light continues to be plane polarized, not elliptically polarized. Figure 7.21 illustrates the phenomenon2. Metal reflector Reflected light is elliptically polarized Light linearly polarized in a direction neither parallel nor perpendicular to the plane of incidence θ θr θi Surface normal Figure 7.21 Polarization by reflection off metallic surfaces. (Reproduced with permission of The McGraw Hill Companies from “Fundamentals of Optics” JENKINS and WHITE, 4th ed. © 1981). 7.4 Law of Malus Consider the situation where Polaroid® type polarizers are used as polarizer and analyzer. The polarized is oriented so that the light exiting it is polarized vertically, i.e. at 0o from the vertical axis and the analyzer oriented at θ o from the vertical1. Polarizer 0o P0 Unpolarized light Analyzer P1 θo P2 Vertically Polarized Figure 7.22 Law of Malus. In Figure 7.22, the vertical orientation of the polarizer was chosen for convenience. It can be at any orientation. What is important is to know the difference in angles between 202 the orientations of the polarizer and analyzer is θ o. The power (or irradiance or intensity) of the incident unpolarized light is measured using a power meter as P0, the power of the beam exiting the polarizer is measured as P1, and the power of the beam exiting the analyzer is measured as P2. Clearly, P0 will be the highest among the three measurements and P2 will be the lowest. Even when the angle of the analyzer is set at 0o, P2 will still be lower than P1 because of light absorption by the analyzer. The law of Malus1,2 relates P2 to P1 through the relationship P2= P1cos2θ where θ is the difference in angles between the orientations of the polarizer and the analyzer. Since cosθ is one (1) for 0o and 180o and is zero for 90o and 270o angles of θ, this explains how during a full revolution of the analyzer, two incidences of bright and two incidences of dark light intensity are observed. Example 7.4 Vertically polarized laser light has a measured power of 1.2 mW. An analyzer is oriented at 30o from the vertical. What would be the power of the beam leaving the analyzer? Using P2= P1cos2θ P2= (1.2 mW)cos230o = 0.9 mW or 900 µW Answer Figure 7.23 is a graph of the ratio P2/P1 for various angles of θ is shown in the graph below. It assumes 100% transmission for the analyzer. Law of Malus P2/P1 1 0.8 0.6 0.4 0.2 0 0 100 200 Angle (degrees) Figure 7.23 Illustration of the Law of Malus. 300 203 Example 7.5 A 1 mW horizontally polarized laser beam strikes an analyzer oriented 22o from the horizontal and a second analyzer oriented 52o from the horizontal. What is the power of the beam that exits the second analyzer? Express the answer also as a percentage of the original beam power Analyzer 1 22o Analyzer 2 P2 52o P1 Horizontally polarized light P3 Figure 7.24 Illustration for Example 7.5. Using P2= P1cos2θ P2= (1 mW)cos222o = 860 µW Using P3= P2cos2θ P3= (860 µW)cos230o = 645 µW Answer Note that for the second set of calculations, one needs to use the difference in angles of orientation between the first and second analyzer1. Percentage compared to original beam power = 64.5% Answer Example 7.6 The following data was collected using vertically polarized 2 mW laser light incident upon an analyzer oriented at various angles from the vertical. The transmitted power in each case was measured. Construct a graph using data from Table 7.1 to prove Malus law based on the experimental data. Table 7.1 (Raw Data) Angle(θo) 0 15 30 45 60 75 90 105 120 135 150 165 180 P2 (mW) 1.5 1.35 1.25 0.7 0.4 0.08 0.01 0.085 0.4 0.8 1.1 1.45 1.55 Angle(θo) 195 210 225 240 255 270 285 300 315 330 345 360 P2 (mW) 1.32 1.1 0.8 0.4 0.08 0.02 0.12 0.35 0.7 1.2 1.48 1.53 204 First, corresponding to each row, calculate cos2θ and the ratio P2/P1. For example, corresponding to the entry for 15o, the power transmitted by the analyzer is 1.35. For this case, cos2(θ )= [cos 15o]2 = [0.966]2 = 0.933, and P2/P1 = 1.35/2 = 0.675. Table 7.2 shows the transformations for all the data points. Table 7.2 Angle 0 15 30 45 60 75 90 105 120 135 150 165 180 195 210 225 240 255 270 285 300 315 330 345 360 P2 (mW) 1.5 1.35 1.25 0.7 0.4 0.08 0.01 0.085 0.4 0.8 1.1 1.45 1.55 1.32 1.1 0.8 0.4 0.08 0.02 0.12 0.35 0.7 1.2 1.48 1.53 cos2(θ ) 1 0.933012 0.749999 0.499998 0.249998 0.066986 0 0.066989 0.250004 0.500006 0.750005 0.933016 1 0.933009 0.749993 0.499991 0.249992 0.066982 0 0.066993 0.250011 0.500013 0.750012 0.93302 1 P2/P1 0.75 0.675 0.625 0.35 0.2 0.04 0.005 0.0425 0.2 0.4 0.55 0.725 0.775 0.66 0.55 0.4 0.2 0.04 0.01 0.06 0.175 0.35 0.6 0.74 0.765 Using the data, a graph of cos2(θ ) on the horizontal axis and P2/P1 on the vertical axis is shown in Figure 7.25. The data points fall on a straight line (within the experimental error limits) proving the Law of Malus. The extinction ratio is used to characterize the “goodness” of a linear polarizer. An ideal linear polarizer should transmit the maximum light in a certain direction and no light in a direction perpendicular to it. However, real life polarizers transmit a minimum amount of light in the latter direction. Extinction ratio4 re looks at the two extremes, defined as re = Pmax Pmin where Pmax is the maximum power of light transmitted in any direction and Pmin is the light transmitted when the polarizer is oriented 90o from the original direction. A relatively high re indicates that the polarizer is extremely efficient in blocking the 205 perpendicular component of light. An analyzer is used to determine the minimum and maximum power. Verification of Law of Malus 1 P2/P1 0.8 0.6 0.4 0.2 0 0 0.2 0.4 0.6 0.8 1 2 Cos (Angle) Figure 7.25 Verification of Law of Malus. Exercise 7.4A 1. Vertically polarized laser light has a measured power of 2.2 mW. An analyzer is oriented at 45o from the vertical. What would be the power of the beam leaving the analyzer? 2. Vertically polarized laser light has a measured power of 3.3 mW. An analyzer is oriented at 60o from the vertical. What would be the power of the beam leaving the analyzer? 3. A 2 mW horizontally polarized laser beam strikes an analyzer oriented 42o from the horizontal and a second analyzer oriented 52o from the horizontal. What is the power of the beam that exits the second analyzer? Express the answer also as a percentage of the original beam power. 4. A 800 µW horizontally polarized laser beam strikes an analyzer oriented 15o from the vertical and a second analyzer oriented 30o from the horizontal. What is the power of the beam that exits the second analyzer? Express the answer also as a percentage of the original beam power. 5. At what angle should an analyzer be rotated to from vertical to transmit 40% of the vertically polarized light incident upon it? 6. At what angle should an analyzer be rotated to from vertical to transmit 63% of the vertically polarized light incident upon it? 206 7. At what angle should an analyzer be rotated to from vertical to transmit 85% of the vertically polarized light incident upon it? 8. At what angle should an analyzer be rotated to from horizontal to transmit 68% of the vertically polarized light incident upon it? 7.5 Birefringence Certain materials, such as calcite and quartz, exhibit two indices of refraction. They are called anisotropic, since an optical property such as refractive index varies with the direction. In the case of isotropic materials such as glass, there is only one index of refraction at any given wavelength, no matter which direction. Therefore, when light is incident on glass, some of the light will be reflected and some refracted. Based on geometric optics, there is only one reflected ray and one refracted ray. However, for materials that exhibit birefringence1,2, or double refraction, such as quartz and calcite, since they possess two different indices of refraction, there will be two refracted rays, called ordinary ray (O-ray) and extraordinary ray (E-ray.) Each ray will travel at a different speed. How does one distinguish the O-ray from the E-ray? The answer is simple. The O-ray obeys Snell’s law of refraction, which was discussed in Chapter 5; the E-ray does not and, hence, is named extraordinary. For a calcite crystal, which looks like a parallelogram from a side view with 109o and 71o as interior angles, if the incident light is normal to the crystal face, since the ordinary ray obeys Snell’s Law, it will proceed undeviated. The extraordinary ray will be refracted. The O-ray and the E-ray will emerge from the opposite side of the crystal parallel to each other, however displaced from each other, as shown in Figure 7.26 2. E-ray E-ray O-ray O-ray 71o 109o Figure 7.26 Birefringence. (Reproduced with permission of The McGraw Hill Companies from “Fundamentals of Optics” JENKINS and WHITE, 4th ed. © 1981). Looking from the exit face of the crystal, if one rotates the crystal around the O-ray axis, the E-ray will rotate around the fixed O-ray! How does birefringence relate to polarization? Crystals possess an optical axis. Around this axis the crystal looks symmetrical and the atoms of the crystal are also arranged symmetrically. The normal to any of the faces of a crystal and the optical axis constitute the principal plane of the crystal. Materials such as calcite are capable of bringing about 207 polarization of the O- and E-rays. The O-ray is polarized perpendicular to the principal plane and the E-ray is polarized parallel to the principal plane2,5. Birefringence or double refraction is yet another method of polarizing light. Polarizers based on this concept are commercially available in the form of Nicol, Rochon and Wallaston prisms. These prisms break down the originally unpolarized light into two components, the E-vector of each component vibrating in mutually perpendicular directions. Since calcite material possesses two indices of refraction, if a prism is made of calcite, one would be able to obtain two colored spectra using incident white light. For this to occur, the optical axis of the crystal must be parallel to the first refracting face. This is shown below on the left in Figure 7.27. The figure on the right is for a crystal whose optical axis is parallel to the base, however perpendicular to the first refracting edge2. For the second case, only one dispersion (rainbow) pattern is observed. This is because the O-ray and the E-ray coincide, resulting in dispersed, however, unpolarized light. E-ray Unpolarized O-ray Unpolarized Red Violet Red Violet Red Violet Figure 7.27 Birefringence and dispersion.(Reproduced with permission of The McGraw Hill Companies from “Fundamentals of Optics” JENKINS and WHITE, th4 ed. © 1981). We learned in Chapter 5 the relationship between speed of light in a medium and the refractive index. If v is the speed in the medium and n the refractive index, then v = c n where c is the speed of light in vacuum. With birefingent materials since the indices of refraction are different for the ordinary ray (nO) and extraordinary ray (nE), the velocities of the ordinary and extraordinary rays will be different. For example, at 589 nm, nO for calcite2 is 1.6584 and nE is 1.4864. Consequently, the velocity of the ordinary and extraordinary rays can be calculated to be 1.809 x 108 m/s for the ordinary ray and 2.018 x 108 m/s for the extraordinary ray. Therefore, there is a “fast” axis and a “slow” axis for light to travel. Thus, while unpolarized light entered a birefringent crystal when the E-vector vibrations were in phase, now, all of a sudden, upon exiting the crystal, the E-vector vibrations in mutually perpendicular directions are out of phase. By adjusting the thickness of the medium, one can bring about an out of phase condition of 90o. Thus, if the E-vector was originally oriented at a 45o angle with respect to the crystal axis, light leaving the crystal is polarized in two different directions with a 90o phase difference. This results in circularly polarized light. This is the principle behind quarter-wave retarder plates, one quarter of a wave corresponding to 90o of rotation. By suitable 208 adjustment of plate thickness and the original E-vector angle orientation, one can obtain elliptically polarized light also using retarder plates1,2. 7.6 Laboratory Experiments6 (a) Finding Brewster’s angle for a glass slide If you have a horizontally polarized He:Ne laser, skip step 1 and proceed to step 2. 1. Place a Polaroid® type polarizer in front of the unpolarized He:Ne beam coming from the laser. Align the polarizer so that light leaving the polarizer is polarized in a plane parallel to the plane of incidence. 2. Follow laser safety precautions. Turn on the laser. Place a glass slide vertically on a rotating table that is graduated in degrees as shown in Figure 7.28. Laser Polarizer Glass slide Stand Rotating Table Figure 7.28 Experimental set-up for Brewster angle determination. 3. Position the glass slide so that the reflected laser light shines away from you. For the sake of diagram clarity, just the opposite configuration is shown above. 4. Rotate the table slowly and observe the brightness of the reflected laser beam on the wall. As you rotate, at one point, the reflected beam all but vanishes. Note this location. This corresponds to the Brewster’s angle. Verify using theoretical predictions. (b) Verifying the Law of Malus The following steps are to be followed in sequence if you have an unpolarized He:Ne laser in your laboratory. 1. Place a Polaroid® type polarizer in front of the unpolarized He:Ne beam coming from the laser. Align the polarizer so that light leaving the polarizer is vertically polarized with respect to the plane of incidence. 2. Place a second polarizer so that it is oriented in the same fashion as the original polarizer. Call the second polarizer the analyzer1. 3. Follow laser safety precautions. Turn on the laser. 4. Measure the power P1 using a power meter. 209 5. Measure the power P2 using a power meter. 6. Rotate the analyzer clockwise by 10o. Polarizer 0o Analyzer P1 θo P0 Unpolarized light P2 Vertically Polarized Figure 7.29 Experimental set-up for Law of Malus verification. 7. Repeat steps 5 and 6 until the analyzer has completed a full revolution. 8. Verify the Law of Malus by plotting the cos2θ on the horizontal axis and ratio P2/P1 on the vertical axis. Draw a best-fit straight line through the experimental data points. (Refer to Example 7.6 for guidance). • • • • • • • • Chapter Summary Electromagnetic radiation comprises of two fields, each represented by a vector: the E-vector and the B-vector. The E- and the B-vectors are always perpendicular to each other. Use of the left hand thumb rule will provide the direction of wave travel. In regular unpolarized light the orientation of the E-vector is random and cannot be predicted. Unpolarized light can be linearly (or plane-) polarized via reflection, refraction, selective absorption, scattering and birefringence. When unpolarized light is incident on a dielectric material such as glass, there exists an angle of incidence called the Brewster angle for which the reflected light is polarized only perpendicular to the plane of incidence. The reflected light is all s-ray. The p-ray, which is polarized in a plane parallel to the plane of incidence, is extinguished at the Brewster angle. The Brewster angle is related to the refractive indices of the media that the light encounters. A Brewster window is used to polarize the output of an unpolarized laser. The extinction ratio is an indication of the “goodness” of a linear polarizer and is the ratio of the maximum to the minimum power that the polarizer transmits. The Law of Malus relates the power transmitted by an analyzer oriented at various angles in the path of linearly polarized light. Light may be circularly or elliptically polarized. In either case, the E-vector undergoes a complete 360o rotation when the light wave travels a full cycle. The rotation of the E-vector may be clockwise leading to right polarized wave or counterclockwise leading to left polarized wave. This is true with both circularly or elliptically polarized light. 210 • • • • Birefringent materials possess two different indices of refraction. Thus, when unpolarized light is incident upon a birefringent material, it splits into two rays, called the ordinary ray and the extraordinary ray. The former follows Snell’s law of refraction, while the latter does not. Birefringent materials such as calcite may be used to polarize light. By proper orientation of the crystal principal plane, one can obtain a perpendicularly polarized E-ray and a parallel polarized O-ray, both with respect to the plane of incidence. Dispersion of white light may result in two spectra, one for the E-ray and the second for the O-ray, when birefringent materials are used. A quarter wave retarder plate converts linearly polarized light into either circularly- or elliptically-polarized light. Such materials are again birefingent and possess a “slow axis” and a “fast axis” for light travel resulting in a phase difference between the E-ray and the O-ray. End of Chapter Exercises Calculate the Brewster angle for the following cases: (Data2: Refractive index at 589 nm: Crown glass 1.522, X 1.819, Flint glass (light) 1.588, Ethanol 1.359, and Water 1.333.) Light is traveling from medium 1 to medium 2 Problem 1 2 3 4 5 6 7 8 9 10 11 12 Problem Medium 1 Crown glass Crown glass Water Ethanol Air Crown glass Flint glass (light) Flint glass (light) Crown glass Air X X Medium 1 Medium 2 Flint glass (light) X Crown glass Crown glass Flint glass (light) Crown glass Ethanol Water Water Crown glass Flint glass (light) Crown glass Medium 2 13. When light is incident on a material surface from air, it is seen that the reflected light is perpendicularly polarized with respect to the plane of incidence. If the angle of reflection is 52o, what are the (i) angle of incidence (ii) the Brewster angle, and (iii) refractive index of the medium? 14. When light is incident on a material surface from air, it is seen that the reflected light is perpendicularly polarized with respect to the plane of incidence. If the angle of reflection is 59o, what are the (i) angle of incidence (ii) the Brewster angle, and (iii) refractive index of the medium? 211 15. When light is incident on a material surface from air, it is seen that the reflected light is perpendicularly polarized with respect to the plane of incidence. If the angle of reflection is 62o, what are the (i) angle of incidence (ii) the Brewster angle, and (iii) refractive index of the medium? 16. When light is incident on a material surface from air, it is seen that the reflected light is perpendicularly polarized with respect to the plane of incidence. If the angle of reflection is 49o, what are the (i) angle of incidence (ii) the Brewster angle, and (iii) refractive index of the medium? 17. Vertically polarized laser light has a measured power of 5 mW. An analyzer is oriented at 35o from the vertical. What would be the power of the beam leaving the analyzer? 18. Vertically polarized laser light has a measured power of 2.3 mW. An analyzer is oriented at 20o from the horizontal. What would be the power of the beam leaving the analyzer? 19. A 1.2 mW horizontally polarized laser beam strikes an analyzer oriented 32o from the horizontal and a second analyzer oriented 57o from the horizontal. What is the power of the beam that exits the second analyzer? Express the answer also as a percentage of the original beam power 20. A 500 µW horizontally polarized laser beam strikes an analyzer oriented 18o from the vertical and a second analyzer oriented 36o from the vertical. What is the power of the beam that exits the second analyzer? Express the answer also as a percentage of the original beam power 21. At what angle should an analyzer be rotated to from vertical to transmit 50% of the vertically polarized light incident upon it? 22. At what angle should an analyzer be rotated to from horizontal to transmit 33% of the vertically polarized light incident upon it? 23. At what angle should an analyzer be rotated to from vertical to transmit 75% of the vertically polarized light incident upon it? 24. At what angle should an analyzer be rotated to from vertical to transmit 25% of the vertically polarized light incident upon it? Group Discussion Questions These require the students to form small groups of three or four and discuss each question. Some questions may require you to do a search on the Internet and find the answers. 1. How are polarized sunglasses manufactured? 2. What is the source of the Law of Malus? 212 3. How can polarized light be generated directly (i.e., without reflection, refraction, absorption or scattering)? 4. What happens when a p-ray and an s-ray are combined? 5. What is a quarter wave retarder plate? Using simple materials, how can one construct such a plate? 6. What is the phenomenon behind seeing a secondary rainbow in the sky in addition to a primary rainbow? What is the order of colors for the primary and secondary rainbows? References Sears, F.W, M. W. Zemanksy and H.D. Young. “College Physics.” Reading: AddisonWesley Publishing Company, 1991. 2 Jenkins,F. and H. White. “Fundamentals of Optics.” Singapore: McGraw Hill, 1981. 3 Hecht, J. “Understanding Lasers.” Carmel: Howard H. Sams Company, 1988. 4 Institute for Telecommunication Sciences, http://www.its.bldrdoc.gov/fs-1037/dir015/_2125.htm 5 Welford, W.T. “Optics.” Oxford: Oxford University Press, 1988. 6 Gottlieb, H. “Experiments Using a Helium Neon Laser.” Bellmawr: Metrology Instruments, Inc. (1981) 1 213 Chapter 8 Absorption and Scattering of Light 8.1 Absorption of Light 8.2 Filters 8.3 Scattering of Light 8.4 Optical Windows 8.5 Laboratory Experiments Chapter Summary End of Chapter Exercises Group Discussion Questions Chapter Objectives After completing the chapter, the student should be able to (i) distinguish between absorption and scattering (ii) describe the exponential law of absorption (iii) define optical density of a filter (iv) design a filter system Key words: absorption, optical density, optical filter, scattering, fluorescence, optical density 8.1 Absorption of Light When light is incident (from air) on a medium at the air-medium interface, part of the light may be reflected by Fresnel reflection, as per the following equation for normal incidence1 ρ = (n-1)2 (n+1)2 where ρ is the reflectance. This formula is reasonable approximation for small angles (up to 30o) of incidence. For ordinary glass with refractive index around 1.5, the reflectance works out to be about 4%. As shown in Figure 8.1, much of the remaining light will refract into the medium. In the medium, the light may be absorbed or scattered. If the thickness of the medium is reasonably small, the transmitted light will have appreciable irradiance. Otherwise, if the medium absorbs light strongly, very little light is transmitted. There will be a further reflection loss at the medium-air interface before light refracts back into air. Thus far, we have used the word “light” loosely. One must specify the wavelength of light. White light, as we know, is comprised of many different visible and invisible wavelengths. If a material indiscriminately absorbs all wavelengths of light resulting in an equal reduction of the irradiance of the transmitted light compared to the incident light for all wavelengths, this phenomenon is called general absorption. To the human observer, the transmitted light will appear white, if the incident light is white. Two examples of materials that result in general absorption are thin platinum films and lamp black suspensions. 214 Incident light AIR Reflected light Absorption MEDIUM Scattering Reflected light AIR Transmitted light Figure 8.1 Illustration of reflection, absorption and scattering phenomena. On the other hand, if a material absorbs certain wavelengths strongly compared to other wavelengths, then the phenomenon is called selective absorption. From Chapter 3 on Laser Safety, we learned that laser safety goggles are manufactured from materials that absorb certain wavelengths so that very little light of that wavelength is incident on the human eye. When one uses a red colored He-Ne beam, bluish colored goggles are used. The blue colored pigments in the goggles absorb the red light strongly. In a similar fashion, one can explain why colored materials appear a certain way. For example, a red colored STOP sign looks red because, the pigments or chromophores of the paint absorb the colors of white light except red. The red light is transmitted and some of which is reflected back. The same phenomenon is true with other colored objects. Plants appear green due to chlorophyll, a chemical that enables conversion of solar energy, in the presence of carbon dioxide, to food for the plant. Selective absorption has several applications in chemical analysis, photochemical reactions, filters, etc. The absorbance A of a medium is given by A = εCx where A is the absorbance, ε the molar absorption (extinction) coefficient, C the concentration of species that absorb light and x is the distance into the medium from the surface. Since we are dealing with selective absorption, we are looking at absorbance A at a particular wavelength λ , since it depends strongly on λ. The absorbance itself has no units. While analytical chemists work with absorbance and molar extinction coefficient, in photonics, we deal more closely with transmittance, T, defined as T = I I0 where I is the measured irradiance of transmitted light and I0 the irradiance of incident light. Also, the molar extinction coefficient and the concentration of absorbing species are combined into a single term called α, the linear absorption coefficient, or simply the absorption coefficient. The transmittance is related to the distance x from the surface and the absorption coefficient through the following relationship1-3 referred to as Beer’s Law, Beer-Lambert’s Law or simply exponential law of absorption. 215 T = e-αx This law is strictly valid for low concentration of the absorbing species4. Needless to say the range of T is from 0 to 1 (or 0 to 100%.) A material with a relatively small value of absorption coefficient at a given wavelength lets much of the light through unabsorbed, leading to a relatively high transmittance value. Also, a thin material (i.e. relatively small x) also transmits a larger fraction of light than a thicker film of material. α is measured in inverse length units (cm-1, m-1, etc.) while x is measured in length units (cm, m, etc.) and T has no units. Graphs of T vs. distance for various absorption coefficients and T vs. α for various distances are shown in Figure 8.2-8.3. Transmission for various absorption coefficients Transmission 1 0.8 0.6 0.4 0.2 0 0 1 2 3 4 5 6 Distance (cm) T(alpha=.01) T(alpha=0.1) T(alpha=0.5) T(alpha=1) Figure 8.2 Graph of transmittance versus depth. It is evident from the graphs that for materials that absorb light of a certain wavelength strongly (i.e., those with relatively large values of alpha), less light is available in material layers that are farther away from the surface. This has implications in imaging applications where light uniformity inside thin films is important. An application is photolithography used in the semiconductor industry. It is often difficult to distinguish between absorption of light and scattering of light by measuring the power (or irradiance) of the transmitted beam of light from the medium. In many cases, it is a combination of the two, with the scattering component frequently dominating the absorbing component. The major difference between absorption and scattering phenomena is the permanent loss of a photon by absorption. The photon energy is converted into other forms of energy in the medium, most commonly into heat. On the other hand, during scattering, the direction of the photon is altered, so that it does 216 not exit along with the other photons in the transmitted light. Further discussion of scattering is provided in the next section. Transmission for various distances into the material from the surface Tranmission 1 0.8 0.6 0.4 0.2 0 0 0.5 1 1.5 Alpha (inverse cm) T(dist=1cm) T(dist=2cm) T(dist=4 cm) T(dist=6 cm) Figure 8.3 Graph of transmittance versus absorption coefficient. Example 8.1 (A) Calculate the power of the laser beam at the following depths from the surface for a plastic rod whose absorption coefficient is 0.000429 cm-1.The incident laser beam power is 3.00 mW. (a) 50 cm (b) 100 cm (c) 150 cm (d) 200 cm (e) 500 cm (f) 1000 cm and (g) 5000 cm. (B) Calculated the per cent transmittance in each case (C) Draw a graph of %T vs. depth on a (i) linear graph paper and (ii) semi-logarithmic graph paper (A) In all cases, use the formula P = P0e-αx with P0=3.00 mW and α=0.000429 cm-1 (a) P = (3.00 mW )e-0.000429 x 50 = 2.94 mW Answer (b) P = (3.00 mW )e-0.000429 x 100 = 2.87 mW Answer (c) P = (3.00 mW )e-0.000429 x 150 = 2.81 mW Answer (d) P = (3.00 mW )e-0.000429 x 200 = 2.75 mW Answer (e) P = (3.00 mW )e-0.000429 x 500 = 2.42 mW Answer (f) P = (3.00 mW )e-0.000429 x 1000 = 1.95 mW Answer (g) P = (3.00 mW )e-0.000429 x 5000 217 351 µW Answer = (B) % T = P x 100 P0 (a) % T = 2.94 x 100 = 97.9% Answer 3.00 Similarly, (b) % T = 95.8% (c) % T = 93.8% (d) % T = 91.8% (e) % T = 80.7% (f) % T = 65.1% (g) % T = 11.7% (C) Linear scale graph %T 90 70 50 30 10 0 1000 2000 3000 4000 Depth (cm) 5000 6000 Figure 8.4 Graph of per cent transmittance versus depth. (D) Semi-log scale graph 100 %T 10 0 1000 2000 3000 4000 Depth (cm) 5000 6000 Figure 8.5 Graph of per cent transmittance versus depth (logarithmic scale). 218 The graph is linear on a semi-logarithmic scale and curved on a linear scale5. In reality, a plastic rod would reduce the transmittance more severely due to scattering, which was neglected in the above calculations Example 8.2 Calculate the thickness of material required to reduce the incident beam power to a half. The absorption coefficient of the material is 0.5 cm-1. The starting equation is T = e-αx with T = 0.5 and α=0.5 cm-1. Taking natural logarithm on both sides, we have ln T = ln (e-αx ) Natural logarithm, ln, and the exponent e are inverse functions of each other and would cancel each other out on the right hand side. Thus, ln T = -αx or x = -lnT α = -ln 0.5 0.5 = 1.39 cm Answer Example 8.3 Choose a material so that in 1 mm depth, it would reduce the transmittance to 10%. Using or ln T = -αx α = -lnT x = -ln 0.1 0.1 cm = 23.0 cm-1 Answer Choose a material or a combination of materials sandwiched together with this absorption coefficient. This material absorbs strongly at the given length. Exercise 8.1A In the following problems assume that only absorption takes place and there is no loss of power due to scattering. 219 Complete the following Table 8.1 for a 5 mW incident laser beam power. Problem Number Absorption coefficient Table 8.1 (a) Power at the given depth of 1 2 3 4 5 6 α= 1.1 cm-1 α= 1.1 mm-1 α= 0.85 m-1 α= 2.5 in-1 α= 0.088 cm-1 α= 0.0085 cm-1 2.0 cm 7 mm 150 cm ¼” 5 cm 40 cm (b) % Transmittance at the given depth of 2.0 cm 5 mm 100 cm ½” 5 cm 50 cm (c) % Light lost due to absorption at the given depth of 2.0 cm 4 mm 125 cm ¾” 5 cm 60 cm 7. Calculate the thickness of material required to reduce the incident beam power to a half. The absorption coefficient of the material is 1.5 cm-1. 8. Calculate the thickness of material required to reduce the incident beam power to a quarter. The absorption coefficient of the material is 0.0025 cm-1. 9. Calculate the thickness of material required to reduce the incident beam power to 33%. The absorption coefficient of the material is 0.225 cm-1. 10. Calculate the thickness of material required to reduce the incident beam power to 60%. The absorption coefficient of the material is 0.000455 cm-1. 11. Choose a material so that in 1 inch depth, it would reduce the transmittance to 10%. 12. Choose a material so that in 2 cm depth, it would reduce the transmittance to 50%. 13. Choose a material so that in 0.25 cm depth, it would reduce the transmittance to 80%. 14. Choose a material so that in 2 mm depth, it would reduce the transmittance to 75%. In the above examples and exercises, the calculation of the absorption coefficient was more of academic nature since materials with the exact values may not be available at the given wavelength. However, one can purchase filters suitable of reducing the incident power to a certain fraction at a given wavelength. In fact, those filters may also attenuate light of other (neighboring) wavelengths. The next section provides a detailed description of the mechanism and terminology associated with optical filters. With materials, the absorption coefficient is always positive, and, consequently, light is attenuated in its irradiance (or power) to some extent as it traverses into the material/medium. Can you think of an exception where instead of attenuation, the power 220 of the beam may increase as it travels through the medium? Go back to Chapter 4 to the discussion of light amplification in a laser. The diagram is redrawn in Figure 8.6. Total Reflectance Mirror Lasing Medium Beam undergoes multiple reflections Output Mirror Laser Beam Figure 8.6 Amplification in a laser. Due to stimulated emission, the number of photons is doubled. This happens during each pass of the beam through the active medium. When the number of photons increases, the power associated with the beam also increases. This can happen only if the absorption coefficient of the medium is negative. There will be some reflection losses at the mirrors. However, if the round trip gain is maintained at one (1) then the laser beam will be sustained. If it is less than one, the beam will die out, and if greater than one, the number of photons will continue to increase. For a properly designed and aligned CW laser system, the gain will increase first, and then decrease, to settle down at one and produce a sustained output. Further discussion is provided in the next chapter. In general, what happens when a photon is absorbed by the medium? The photon absorbed will excite the atom or molecule from a lower energy state to a higher energy state. With low pressure gaseous atoms, one observes an emission spectrum, when some of the higher energy state atoms transition to lower energy states emitting a photon. The medium itself will absorb more or less strongly depending upon the wavelength of incident light. If the gas pressure is higher, then the absorbed energy will result in warming the gas1. By looking at either the absorption spectrum or the emission spectrum of low pressure gases, one can understand the chemical identity of the material, a useful technique in analytical chemistry. If the gas is monatomic (i.e. the molecule has one atom only) line spectra are observed, whereas if the gas is polyatomic (i.e., the molecule has two or more atoms), one obtains band spectra due to the overlap of energy levels. If the pressure of the gas is higher, collisions among atoms and molecules are more likely, resulting in energy transfer and less energy available for a photon emission. At low pressures, absorption and emission are competing processes. If the emission wavelength is the same as the absorbing wavelength, this situation is called resonance radiation2. If the wavelength of the emitted photon is higher than that of the absorbed photon this phenomenon is called fluorescence. Iodine vapor exhibits fluorescence when absorbing white light and emitting green light1. Fluorescence can also occur in liquids and solids. The popular antifreeze used in car radiators contains a fluorescent material. UV light from a source known as “black light” can make dye solutions fluoresce. Black light is used in certain applications to inspect objects, for example, to identify rosin flux residues after the electronic assembly has been cleaned. Under black light, areas where the rosin was improperly cleaned from the printed circuit board surface fluoresce, whereas others do not. Rosin (chemical name: abietic acid), as well as many other organic compounds absorb strongly around 365 nm 221 wavelength in the ultra-violet. In the case of solids, for example, on a computer screen, even after the power to the display monitor has been shut off, light is emitted from the surface for a few seconds. This is due to a phenomenon known as phosphorescence1,2. Most materials are transparent to gamma rays and X-rays. Tissue and blood are transparent to X-rays whereas bone and teeth are not. This enables X-ray imaging of the human body. Also, metals such as tin and lead, which make up typical solder alloy used in electronics, are opaque to X-rays so that one can use X-ray imaging to look for solder joint defects such as voids, shorts, opens, etc., in hidden areas. As one looks at larger wavelengths, such as UV around 100 nm wavelength, the components of air absorb the radiation strongly. This region is called vacuum UV (VUV.) Many materials have strong absorption in the ultraviolet and near ultra-violet and sometimes into the visible region. While some materials absorb some of the IR wavelengths, with larger wavelengths such as radio waves, they once again become transparent. This explains how we receive radio waves in our receivers inside our homes. 8.2 Filters We have learned that certain wavelengths are selectively absorbed by materials while certain others are transmitted. This is the principle behind optical filters. A pair of sunglasses is an example where many of the wavelengths of white light are absorbed to some extent transmitting only a portion to the wearer’s eyes. A welder’s glass is another example which absorbs much of the intense radiation allowing barely enough light to be transmitted so that the welder can see the work piece. From working with a He:Ne laser in your laboratory, you would know that the safety glasses are bluish in color. Essentially, the safety glasses are a filter to the red light from the laser. It absorbs in the red portion of the visible spectrum strongly and transmits in the blue region of the spectrum. Similarly, when one works with a purple colored laser, one would use a red colored safety goggles. The absorption characteristics of the material used in the construction of these safety glasses are very different. For example, typical purple (i.e., violet) and red filters may have the characteristics shown in Figure 8.7-8.8 6. % Transmission Purple filter 100 80 60 40 20 0 300 400 500 600 700 Wavelength (nm) Figure 8.7 Typical purple filter. 800 222 % Transmission Red filter 100 80 60 40 20 0 300 400 500 600 700 Wavelength (nm) 800 Figure 8.8 Typical red filter. From these graphs, it is seen that the blue filter transmits significantly in the purple wavelengths, however, absorbs strongly in the red wavelengths. On the other hand, a red filter transmits strongly in the red wavelengths and absorbs significantly in the purple wavelengths. In other words, for the red filter, αred is small and is αpurple large. For the blue filter αpurple is small and αred is large. You may remember from the previous section that α is the absorption coefficient, a measure of how strongly light is absorbed at that wavelength. Clearly α depends on the wavelength, and, filters operate on the principle of selective absorption. Filter manufacturers supply filters along with transmittance data at different wavelengths. An alternate method of denoting absorption or transmittance is through the term defined as the optical density. In the same manner as mass density denotes how dense or packed a material is in a given volume, the optical density at a given wavelength is an indication of how tightly are the absorbing species packed in that medium. Clearly, the optical density, abbreviated as OD, would, in addition to the wavelength, depend upon the thickness of the filter, as well as the concentration of the absorbing species in the filter. The relationship between OD and transmittance T is given by3,6 or T = 10-OD OD = -log T where the logarithm refers to the common logarithm, i.e. base 10. Unlike the absorption coefficient α which has the units of inverse length (i.e. cm-1 or m-1), T has no units. A graph of transmittance versus OD is shown in Figure 8.9. The transmittance calculated from this formula would range from 0 to 1 and should be multiplied by 100 to covert to %T. 223 %T vs OD 100 10 %T 1 0.1 0 1 2 3 4 5 0.01 0.001 OD Figure 8.9 Optical density versus transmittance. Example 8.4 Calculate the % transmittance of optical filters with (a) OD=0.5 and (b) OD=1.5 (a) %T = 100 x 10-OD = 100 x 10-0.5 = 31.6% Answer (b) %T = = = 100 x 10-OD 100 x 10-1.5 3.16% Answer Example 8.5 What should be the OD of filters to transmit (a) 5% and (b) 50% of incident light at a given wavelength. (a) OD = -logT = -log (0.05) = 1.30 Answer (b) OD = = = -logT -log (0.5) 0.301 Answer Example 8.6 You have filters with the following OD at 632.8 nm wavelength: 0.1, 0.5, 1.0, 1.5 and 2.0, one of each OD. How would you combine them to achieve a 2.5% transmittance? OD can be added to obtain a composite value. To obtain T=0.025 OD = -logT 224 = = -log (0.025) 1.60 -> use 1.5 OD filter with 0.1 OD filter Answer Exercise 8.2A 1. To obtain 3% transmittance at a certain wavelength, what should the OD of the filter be at that wavelength? 2. To obtain 10% transmittance at a certain wavelength, what should the OD of the filter be at that wavelength? 3. To obtain 30% transmittance at a certain wavelength, what should the OD of the filter be at that wavelength? 4. To obtain 42% transmittance at a certain wavelength, what should the OD of the filter be at that wavelength? 5. Calculate the per cent transmittance for a filter with OD=0.12. 6. Calculate the per cent transmittance for a filter with OD=0.44. 7. Calculate the per cent transmittance for a filter with OD=1.25. 8. Calculate the per cent transmittance for a filter with OD=2.08. 9. You have filters with OD of 0.3, 0.6 and 1.0 at a given wavelength. How many different transmittance values are obtainable with this set of filters and what are the %T values in each case? 10. You have filters with OD of 0.5, 0.9 and 1.3 at a given wavelength. How many different transmittance values are obtainable with this set of filters and what are the %T values in each case? 11. Given filters with OD of 0.301, 0.477, 0.602 and 0.699 at a given wavelength and one of each kind, how would you obtain a transmittance of 17%? 12. Given filters with OD of 0.301, 0.477, 0.602 and 0.699 at a given wavelength and one of each kind, how would you obtain a transmittance of 8.3%? When we considered the blue and red filters, those are examples of edge filters, i.e., filters that have a wavelength, above or below which transmittance is significant. Edge filters can be short pass, i.e., allowing lower wavelengths light to pass through and stopping higher wavelengths, or, long pass filters, which perform just the opposite way. Filters are also available that would allow a narrow range of wavelengths to be transmitted, absorbing wavelengths outside the band. Commercial filters are available in 100 nm wide bands, for example, a 201-300 nm band pass filter allows light in the 201nm 225 to 300 nm range to pass through, while essentially blocking light outside this band. These are called band pass filters. Filters are also available, which stop light in a range of wavelengths, while transmitting light outside of the band. These are called band stop filters. Filters that have a constant transmittance over a wide range of wavelengths are called neutral density filters3. Example graphs of the transmittance characteristics of typical band pass6 and band stop filters are shown in Figure 8.10-8.11. Band Pass Filter 100 80 % T60 40 20 0 300 400 500 600 Wavelength (nm) 700 800 700 800 Figure 8.10 A typical band pass filter. Band Stop Filter 100 80 %T 60 40 20 0 300 400 500 600 Wavelength (nm) Figure 8.11 A typical band stop filter. By comparing the discussions in sections 8.1 and this section, you would realize that there is a strong connection between the optical density and the absorption coefficient, since both are responsible for attenuating the beam irradiance at a given wavelength. If we equate the transmittance, we obtain 226 or T = e-αx = -αx -OD ln (e ) = ln (10 ) 10-OD giving -αx = -OD ln 10 or α Conversely, OD = 2.303 OD x = αx 2.303 because ln 10 = 2.303 Clearly, the absorption coefficient and OD are linearly related to each other. To increase the OD of a given material, use a greater thickness of the material. 8.3 Scattering of Light This phenomenon is easily observed from every day experiences. For example, a shaft of sunlight illuminating an otherwise darkened room is an example. The light beam (or shaft of light) is clearly visible via the dust or smoke particles that scatter the light. Depending on the level of dust in a movie hall, one can see the path of the light from the projector, again due to dust scattering it. Scattering is also observed with liquid solutions. A glass of water appears clear. If we squeeze juice from a lemon or lime, the water turns turbid. This is another example of light scattering. A method used by the sulfuric acid industry to check the purity of sulfur dioxide glass by passing it through a chamber of light. If no light scattering is observed, the gas is clean enough to be passed on to a reactor to form sulfur trioxide, a precursor to producing sulfuric acid. Tyndall had studied scattering of white light by fine particles. Dust-laden gas could otherwise poison the reactor catalyst. Light scattering is also used by the analytical chemist to determine the length of a molecular chain, and, in turbidimetry used to measure concentrations. Sunlight of normal irradiance of 140 mW/cm2 is considerably reduced due to scattering by dust particles1. The classic example is, during times of volcanic activity, the irradiance of sunlight is reduced drastically in the surrounding areas due to attenuation by dust, smoke and gaseous molecules scattering light. How would one distinguish scattering from absorption of light? After all, both lead to attenuation of the light irradiance. Except for the cases of fluorescence or resonance described earlier, the absorbed light is invariably converted into heat energy by the absorbing molecules, and, the absorbed photons are lost for ever. In the case of scattered light, the photons are redirected and removed from the beam. Absorption and scattering take place simultaneously in most situations, with one of the components dominating the other. Thus, when one measures the irradiance of transmitted light, one is, in reality, measuring the remaining light after the incident light has been absorbed and scattered. Thus, the attenuation equation has to be modified as1,3 227 I = I0e-(αa + αs )x where αa is the absorption coefficient and αs is the scattering coefficient. We must know the contributions from absorption and scattering. In some cases, absorption is more significant than scattering (i.e. αa >> αs); in other cases, absorption may be negligible compared to scattering (i.e. αa << αs ). In the previous section, we assumed the former case to be true in order to solve problems. Example 8.7 A 25 cm long glass tube is filled with smoke particles. An incident beam of light of irradiance 10 mW/cm2 strikes the tube at one end and 6.43 mW/cm2 is measured as the irradiance of the transmitted beam. When all the smoke particles had settled either by condensation or precipitation, the irradiance of the transmitted beam was re-measured. It was found to be 9.05 mW/cm2. What is (a) scattering coefficient, (b) the absorption coefficient, and (c) the total coefficient? The general equation that applies for combined absorption and scattering is = I0e-(αa + αs )x αa + αs = (1) ln[I0 /I] x I or We will solve for part (c) of the problem first, using this equation, as αa + αs = = (1) ln[ 10/6.43] 25 0.0177 cm-1 Answer (b) When the smoke settled, the attenuation is entirely due to absorption. Thus, (a) αa = (1) ln[I0 /I] x αa = = (1) ln[10/9.05] 25 0.00399 cm-1 Answer = Total coefficient - αa = 0.0177 – 0.00399 = 0.0137 cm-1 Answer αs When light beam of high irradiance is used, this could create scattering from small particles, which was not significant at lower irradiances. 228 Exercise 8.3A 1. A 30 cm tube is filled with a low pressure gas. If the tube transmits 75% of the incident light, ignoring scattering effects, what is the absorption coefficient? 2. If the tube in problem 1 is now filled with smoke, and if the scattering coefficient of the smoke is 0.0325 cm-1, what is the per cent transmission now? 3. A medium has an absorption coefficient of 0.00454 cm-1 and a scattering coefficient of 0.0155 cm-1. For a 10 cm long tube calculate the percent transmission and per cent of the light lost due to absorption and scattering. 4. At 20 mW/cm2 incident intensity, the transmitted irradiance is measured to be 18.5 mW/cm2 using a 15 cm tube. At such low irradiances, the attenuation is entirely due to absorption. However, when the light irradiance was increased to 3.2 W/cm2, the transmitted beam was measured to have an irradiance of 2.1 W/cm2. It is known that scattering takes place at such high irradiances. Calculate (a) the total coefficient, (b) the absorption coefficient, (c) the scattering coefficient, (d) the transmittance for a 20 cm tube at 10 mW/cm2 incident irradiance and (e) the per cent transmittance for a 20 cm tube at 3.5 W/cm2 incident irradiance. The chapter on wave optics will elucidate further what happens when light waves impinge small particles to cause diffraction. For now, let us examine the interaction between the wavelength of light and the particle size to determine various phenomena. The first is called Rayleigh scattering. The size of the scattering particle is much smaller (ten times or less) than the wavelength of the incident light. Assuming that absorption is negligible, the scattering coefficient αs is given by1,7 αs = c1 λ4 where c1 depends upon the refractive index of the medium and the concentration of scattering particles. The most significant dependence is on the wavelength in a highly non-linear and inverse fashion as illustrated in Figure 8.12. An arbitrary value of 100 is chosen for the violet wavelength of 400 nm and the scattering coefficients of the other visible wavelengths are calculated accordingly. It is seen that the scattering coefficient of 400 nm light is more than nine times the scattering coefficient of 700 nm light. Thus, during sunrise or sunset, the violet light is scattered so much so that we are unable to see light near the blue end of the spectrum. This explains why the sun is reddish during sunrise or sunset because of the transmitted light. On the other hand, during day, the same scattering particles are still present, causing the bluish end of the spectrum to be scattered towards the viewer on the earth. This explains why the sky is blue during the day1,7,8. The red light passes without much scattering and is outside the viewer’s field of view. The relative irradiance of light of various wavelengths, when sunlight is scattered by small size particles following Rayleigh’s scattering, follows Figure 8.121. When white 229 light is incident on particles such as from tobacco smoke, the smoke appears to have a bluish hue. When the particle size becomes larger, the scattered light no longer appears bluish, however, appears white. The strong dependence of scattering on wavelength drops out for larger size particles such as chalk dust1. Relative scattering coefficient Rayleigh Scattering 120 100 80 60 40 20 0 400 450 500 550 600 650 700 Wavelength (nm) Figure 8.12 Relative Rayleigh scattering coefficient for various wavelengths. When the size of the particles exceeds the Rayleigh criterion, however, is limited to about 25 µm or so in diameter, this leads to a phenomenon called Mie Scattering. The scattering coefficient for Mie scattering depends strongly on the size of the particle and less on the wavelength9. αs = c2R2 where R is the particle radius and c2, the Mie scattering constant, which depends on the concentration of scattering particles, the size of the particle and the wavelength of light9. The graph of scattering coefficient (which is an indication of irradiance of scattered light) versus particle size is shown in Figure 8.13. For ease of computations, the graph disregards any wavelength dependence of the constant c2. The size of the tobacco smoke particles we discussed earlier extends into the Mie scattering regime. When the particle size becomes much larger, the wavelength dependence drops out entirely. The spreading of white light that one notices, when the particle size is much higher than the wavelength of light, is due to diffraction, and, is discussed in a later chapter. Thus, when particle sizes are small, as in Rayleigh scattering regime, the wavelength of light is very important. When the particle size is much larger than the wavelength of light, as in rain drops illuminated by a car headlight, the scattered light is no longer wavelength dependent. 230 Relative scattering coeffcient Mie Scattering 3000 2500 2000 1500 1000 500 0 0 5 10 15 20 25 Particle diameter (micrometers) Figure 8.13 Relative Mie scattering coefficient for diameter particles. Exercise 8.3B 1. White light passes through air which is filled with dust after a volcanic eruption. Assuming that light attenuation is solely through scattering, determine the transmittance of a 10 m path with an average scattering coefficient of 0.2 m-1. 2. As one travels from near the earth surface to, say 500 miles into space, how would your predict the scattering coefficient to change from close to the earth surface to far away? 3. For pure air, devoid of any particles or aerosols, if violet light (λ=400 nm) has a scattering coefficient of approximately 0.048 km-1, predict the scattering coefficient of the following colors of light: (a) yellow (λ=590 nm), (b) green (λ=515 nm) and red (λ=633 nm). Hint: Use the Rayleigh scattering coefficient graph. 4. For pure air, devoid of any particles or aerosols, if violet light (λ=400 nm) has a scattering coefficient of approximately 0.048 km-1, predict the scattering coefficient of the following colors of light: (a) bluish-green (λ=477 nm), (b) lime-green (λ=540 nm) and deep red (λ=680 nm). Hint: Use the Rayleigh scattering coefficient graph. 5. A tube filled with 10 µm diameter size particles exhibits a Mie scattering coefficient of 5 m-1 at a certain wavelength of light. Estimate the scattering coefficient for the following size diameter particles using the Mie scattering theory: (a) 5 µm (b) 15 µm and (c) 25 µm at the same wavelength of light. 6. A tube filled with 5 µm diameter size particles exhibits a Mie scattering coefficient of 0.5 m-1 at a certain wavelength of light. Estimate the scattering coefficient for the following size diameter particles using the Mie scattering theory: (a) 10 µm (b) 15 µm and (c) 20 µm at the same wavelength of light. 231 It is our common experience that dust particles, aerosols, fog, etc., scatter light. The degree of scattering depends strongly on the wavelength of light, as we have seen with Rayleigh scattering, and, the type of scattering depends on the size of particle as well, as in the case of Mie scattering. As discussed in Chapter 7, when unpolarized light passes through a scattering volume, the scattered light is linearly polarized. 8.4 Optical Windows While filters attenuate certain wavelengths by selective absorption or interference (a topic to be discussed in a later chapter), one needs materials that transmit a high percentage of light incident on them in certain applications. The simplest example is the Brewster window or any window that lets light out of s source. Also, in applications where certain materials need to be irradiated by light of definite wavelength, they need to be placed in a cell or chamber, and, the incident light would enter the cell through a window. In all these cases, the window material needs to possess a high transmittance or low absorption at the given wavelength. Thus, windows and filters perform opposite functions. While normal window glass of the borosilicate type is suitable for many applications in the visible range of the spectrum, for high transmittance applications, as well as for wavelengths in the ultraviolet and infrared, one needs specialized materials such as fused silica (SiO2), quartz (SiO2), calcium fluoride (CaF2), magnesium fluoride (MgF2), sodium chloride (NaCl), potassium chloride (KCl), cesium iodide (CsI), potassium bromide (KBr), lithium fluoride (LiF), barium fluoride (BaF2), etc. Needless to say, the price of these specialized window materials is much higher than glass. Precision windows are available to various degrees of flatness, from one-quarter wavelength flat (λ/4) to λ/10 flatness10. For example, a λ/10 window material suitable for 300 nm UV wavelength application can be expected to have a flatness of 30 nm! Additionally, precision windows guarantee a thickness tolerance of 0.1 mm or better for nominal 3 mm thick windows. The window is sold as is or coated. One purpose of coating is to make the window anti-reflective. Such coatings are known as ARC, antireflective coatings. They function on the principle of interference of light waves, a topic discussed under wave optics in a later chapter. Calcium and magnesium fluoride materials have become popular on account of their high transmittance over a wide range of wavelengths10, from deep UV to mid-IR range. Fused silica and quartz, on the other hand, while less expensive than calcium or magnesium fluoride, transmit well only in the 200 nm to 3 µm range. While moisture susceptibility is one concern with calcium fluoride, it has withstood harsh environments successfully10. The principle of transparency of materials at select wavelengths has been used in laser interferometry. Using this technique, the details of which are discussed in the Chapter on wave optics, the thickness of coating materials can be accurately determined. In the case of deposition or etching of a thin layer of material in the semiconductor industry, the method serves as an end-point determination technique, i.e., one would know when to stop the process based on the interferometer chart data known as the interferogram. 232 8.5 Laboratory Experiments (a) Absorption (Prepare different concentrations of dye solutions. For example, a drop of blue food color in 500 ml of water may be one concentration. Prepare ten different concentrations so that the range of transmittance in the 5-95% range is covered for a 2 cm optical length of the solution. (This may require some trial and error. See details below). 1. Following safety precautions, align a He: Ne laser on an optical bench. 2. Measure the incident power of the beam. Record in Table 8.2. 3. Place a square test tube about 2 cm by 2 cm and 5 cm high in the path of the laser beam on a mount ensuring safety precautions. 4. Measure the power of the beam transmitted by the test tube. Record in Table 8.2. 5. Fill the tube to about 4 cm height with distilled water. Replace it on the mount. 6. Measure the power of the beam transmitted by the test tube. Record in Table 8.2. 7. Fill the tube to about 4 cm height with dye solution of a certain concentration. Replace it on the mount. 8. Measure the power of the beam transmitted by the test tube. Record in Table 8.2. 9. Repeat steps 7 and 8 for a different concentration of the dye solution. 10. Continue until all concentrations are exhausted and transmittance obtained in step 8 covers the 5-95% range. Laser on stand Rail Test tube Detector Power Meter Readout Figure 8.14 Experimental set-up for absorption/scattering. 11. Plot a graph of T versus number of drops on (i) a linear graph paper, and (ii)semi-log graph paper with T on the logarithmic scale. 12. What are your conclusions from these graphs? 233 Table 8.2 Measured Power Original Beam from laser Test tube only Test tube with water Test tube with dye conc. 1 Test tube with dye conc. 2 Test tube with dye conc. 3 Test tube with dye conc. 4 Test tube with dye conc. 5 Test tube with dye conc. 6 Test tube with dye conc. 7 Test tube with dye conc. 8 Test tube with dye conc. 9 Test tube with dye conc. 10 P0= P1= P2= P3= P4= P5= P6= P7= P8= P9= P10= Transmittance ------------------------------100% T1=(P1/P0 )x 100 = T2=(P2/P0 )x 100 T3=(P3/P0 )x 100 T4=(P4/P0 )x 100 T5=(P5/P0 )x 100 T6=(P6/P0 )x 100 T7=(P7/P0 )x 100 T8=(P8/P0 )x 100 T9=(P9/P0 )x 100 T10=(P10/P0 )x 100 (b) Scattering 1. Following safety precautions, align a He: Ne laser on an optical bench. 2. Measure the incident power of the beam. Record in Table 8.3. 3. Place a square test tube about 2 cm by 2 cm and 5 cm high in the path of the laser beam on a mount ensuring safety precautions. 4. Measure the power of the beam transmitted by the test tube. Record in Table 8.3. 5. Fill the tube to about 4 cm height with distilled water. Replace it on the mount. 6. Measure the power of the beam transmitted by the test tube. Record in Table 8.3 7. Obtain 1 gram of a finely powdered non-toxic substance such as fine sand or finely ground chalk powder. Add the powder to the water. Keep shaking the test tube as you return the tube to the mount. 8. Note the power reading immediately and observe the power meter. The power meter reading should continue increasing as the powder settles slowly. Note the approximate time it takes for the power to reach 75% of the value from step 6. 9. Remove the test tube from the mount. Connect the leads from the power supply to a strip chart recorder (in place of the digital multimeter DMM). Adjust the scale so that the plotter pen stays within the range of the paper. Calibrate the graph so that the vertical axis measured beam power in milliwatts. Adjust the speed of the paper so that the time observed in step 8 is achievable on the strip chart recorder paper in about 30 cm distance. 234 10. Shake the tube well as your return it to the mount. Turn on the strip chart recorder. The pen will track the transmitted power. Continue recording until the strip chart recorder shows a constant transmitted power at which point discontinue the experiment. 11. Submit the chart recorder graph along with your comments and observations to your instructor. 12. What is the initial transmittance of the turbid solution? What is the final transmittance after much of the solids settled? Table 8.3 Measured Power Original Beam from laser Test tube only Test tube with water Test tube, water with powder, time zero Test tube, water with powder, time final P0= P1= Transmittance ------------------------------100% T1=(P1/P0 )x 100 = Pf= Tf=(Pf/P0 )x 100 Depending upon the size distribution of particles, the measured Pf value may never totally reach the P0 value. This is attributed to Brownian motion of very fine particles, and, in some cases, due to colloidal formation in other cases. • • • • Chapter Summary Absorption and scattering of light in a medium leads to its attenuation. Quite frequently, these processes take place simultaneously, and, it is often difficult to assign the relative contributions of each. Selective absorption results in loss of photons. The energy of the photon lost is converted most commonly into heat by the absorbing medium. In other cases, such as low pressure gases or liquids or certain solids, the absorbed photon is reemitted, either at the same wavelength (resonance), or at a longer wavelength (fluorescence and phosphorescence.) An analysis of absorption and emission spectra of these materials enables the identification of the chemical species, as well as the concentration. The principle of selective absorption of certain wavelength photons is used in the principle of optical filters. These filters may act as cut-off filters allowing light either above or below a cut off wavelength to be transmitted. Band pass filters allow a narrow range of wavelengths to be transmitted, whereas band stop filters transmit light outside a narrow band of wavelengths. Neutral density filters allow a constant and high transmittance over a wide range of wavelengths. In any case, the transmittance characteristics of a filter at different wavelengths must be provided by the manufacturer. Absorption of light follows the exponential law T=e-αx. Transmittance, T, is defined as the ratio of the power (or irradiance) of the transmitted light to the incident light. 235 • • • • • • • Scattering of light follows a similar exponential law T=e-αsx where αs is the scattering coefficient. The major difference between scattering and absorption is, for the former, the photon is redirected, rather than absorbed. The combined exponential law, which includes both absorption and scattering, is I= I0e-(αa + αs )x Both the absorption and the scattering coefficients are dependent to different degrees on the wavelength of light. For particles which are (ten times or less) smaller than the wavelength of light, the scattering phenomenon is described by Rayleigh’s theory in which the scattering coefficient depends on the inverse fourth power of the wavelength. This makes the scattering coefficient of blue light times about nine times higher than red light, explaining the observation of blue sky during the day and red sky at dawn or dusk. For larger size particles up to 25-30 µm in diameter, the scattering phenomenon explained by Mie’s theory, describes the process as less dependent on the wavelength and more on the size of the particle. In this case, the scattering coefficient varies as the square of the particle diameter at a given wavelength of light. When the size of the particles become much larger, for example, tenths of millimeters in diameter or larger, the wavelength dependence totally drops out and a phenomenon called diffraction, described in a later chapter, dominates the scattering process. Light scattering has applications in chemistry, for example, in determining chain lengths of polymer molecules and concentrations of species and in radar and sonar. End of Chapter Exercises In the following problems assume that only absorption takes place and there is no loss of power due to scattering. Complete the following Table for a 3 mW incident laser beam power Problem Absorption (a) Power at the (b) % Number coefficient given depth of Transmittance at the given depth of 1 α= 2.1 cm-1 0.5 cm 1.0 cm -1 2 α= 0.0075 cm 50 cm 60 cm (c) % Light lost due to absorption at the given depth: 1.5 cm 70 cm 3. Calculate the thickness of material required to reduce the incident beam power to a half. The absorption coefficient of the material is 0.5 cm-1. 4. Calculate the thickness of material required to reduce the incident beam power to a quarter. The absorption coefficient of the material is 0.0035 cm-1. 236 5. Choose a material so that in 2 inch (50 mm) depth, it would reduce the transmittance to 20%. 6. Choose a material so that in 2 mm depth, it would reduce the transmittance to 75%. 7. To obtain 13% transmittance at a certain wavelength, what should the OD of the filter be at that wavelength? 8. To obtain 52% transmittance at a certain wavelength, what should the OD of the filter be at that wavelength? 9. Calculate the per cent transmittance for a filter with OD=0.18. 10. Calculate the per cent transmittance for a filter with OD=1.78. 11. You have filters with OD of 0.3, 0.5 and 0.7 at a given wavelength. How many different transmittance values are obtainable with this set of filters and what are the %T values in each case? 12. Given filters with OD of 0.301, 0.477, 0.602 and 0.699 at a given wavelength and one of each kind, how would you obtain a transmittance of 3.3%? 13. A 40 cm tube is filled with a low pressure gas. If the tube transmits 85% of the incident light, ignoring scattering effects, what is the absorption coefficient? 14. If the tube in problem 13 is now filled with smoke, and if the scattering coefficient of the smoke is 0.0345 cm-1, what is the per cent transmission now? 15. White light passes through air which is filled with fine dust. Assuming that light attenuation is solely through scattering, determine the transmittance of a 1 km path with an average scattering coefficient of 0.00150 m-1. 16. For pure air devoid of any particles or aerosols, if red light (λ=700 nm) has a scattering coefficient of approximately 0.0052 km-1, predict the scattering coefficient of the following colors of light: (a) yellow (λ=590 nm), (b) green (λ=515 nm) and red (λ=633 nm). Hint: Use the Rayleigh scattering coefficient graph. Group Discussion Questions These require the students to form small groups of three or four and discuss each question. Some questions may require you to do a search on the Internet and find the answers. 1. How would you explain to a two year old why the sky is blue? 2. Compare the terminology (i) cut-off filter, (ii) band pass filter, and (iii) band stop filter as they apply to optics vs. electronics. 3. Via measurements how can you separate out the contributions from absorption and scattering? 237 4. What are some medical applications of light absorption? 5. What are some medical applications of light scattering? 6. What is Raman scattering? What are its applications? References Jenkins,F. and H. White. “Fundamentals of Optics.” Singapore: McGraw Hill, 1981. 2 Glasstone, S. and D. Lewis. “Elements of Physical Chemistry.” London: MacMillan and Company Limited, 1970. 3 Hecht, J. “Understanding Lasers.” Carmel: Howard H. Sams Company, 1988. 4 Hyperscience web site http://elchem.kaist.ac.kr/vt/chem-ed/spec/beerslaw.htm 5 Vasan, Srini. “Technical Mathematics (with Applications in Electronics and Photonics.” Victoria: Trafford Publishing, 2003. 6 Oriel Corporation Technical Literature on Filters, http://www.oriel.com/tech/curves.htm 7 Wikipedia Encyclopedia, http://www.wikipedia.org/wiki/Rayleigh_scattering 8 Sears, F.W, M. W. Zemanksy and H.D. Young. “College Physics.” Reading: AddisonWesley Publishing Company, 1991. 9 Oregon Medical Laser Center, http://omlc.ogi.edu/calc/mie_calc.html 10 DDC Technologies, Inc. Product Catalog, http://www.ddctech.com 1 238 Chapter 9 Pulsed Laser Characterization 9.1 Laser Gain 9.2 Pulsed Lasers 9.3 Pulsing Techniques 9.4 Laboratory Experiments Chapter Summary End of Chapter Exercises Group Discussion Questions Chapter Objectives After completing the chapter, the student should be able to (i) describe how gain occurs in a laser (ii) characterize laser pulses through pulse repetition frequency, pulse width, duty cycle, rise time etc. (iii) describe pulsing techniques of Q-switching, modulation and mode (phase) locking (iv) describe briefly functioning mechanisms of a few common pulsed lasers Key words: gain, round trip gain, gain coefficient, cavity loss, resonator, Doppler broadening, pulse width, pulse repetition frequency (PRF), rise time, fall time, time period, pulse power, average power, duty cycle, mode locking, phase locking, Qswitching, acoustic modulation, electro-optic modulation 9.1 Laser Gain In Chapter 4, we discussed basics of laser operation including spontaneous emission, population inversion, stimulated emission, feedback mechanism and amplification. In this section, we will look more closely at parameters responsible for sustaining laser output for continuous wave (CW) and pulsed lasers. From Chapter 8 on absorption, we know that light passing through most materials undergoes attenuation, given by Beer-Lambert’s Law as T = e-αx where T is the transmittance, α may represent the absorption coefficient, the scattering coefficient or a combination of the two and x is the distance into the medium. Referring back to Chapter 4, we learned that, normally, the energy level distribution in an atom is such that the ground state atoms have the highest population and the higher energy states have a lower population. In fact, higher the energy state, the lower the population. For the laser to work, population inversion has to take place, upsetting the equilibrium. If the laser transition is, say, between E3 and E2 levels of a four level laser as shown in 239 Figure 9.1 1,2, 4, then, the relative population of the E3 level must be higher than that of the E2 level. E4 Radiationless decay E3 Lasing E2 Radiationless decay Ground State E1 Figure 9.1 Four level laser energy diagram. We also learned that absorption and emission are competing processes. If amplification has to occur, the rate of stimulated emission must be faster than the rate of photon absorption. For this to happen, the absorption coefficient α in the above equation for transmittance must be negative. Replacing -α with β and substituting G (gain) in place of T, we rewrite Beer-Lambert’s Law as G = e βx where β is now interpreted as a net gain coefficient, rather than an absorption coefficient. G is the amplifier gain or “small signal” gain2 for light passing through the medium once, a measure of light amplification. Using the feedback mechanism of the lasing medium and the mirrors as shown below, the round trip gain of a laser, GR, is related to the “small signal” gain G through the following approximate equation3,4 GR = G2ρHρO(1-η) = e 2β LρHρO(1-η) where the ρ ‘s represent the reflectance of the two laser mirrors (see Figure 9.2), L represents the optical length of the laser and η represents the fraction of light lost due to collisions and other mechanisms during a round trip. GR here represents the ratio of beam irradiances (or power) at the completion of a round trip compared to the beginning of a round trip. Mirror (ρH) Lasing Medium Beam undergoes multiple reflections Output ρO Laser Beam Figure 9.2 Amplification of laser light. The gain coefficient itself is a complicated function of several variables, and, is given by the following proportionality pertinent to the four level laser outlined above3. 240 β ~ (N3-N2)c2 ν2τ where c is the speed of light, ν the frequency of light, τ the average time that the atom spends in the excited state prior to making a spontaneous transition, and, N3 and N2 are the respective populations of the energy states corresponding to E3 and E2, the lasing levels, respectively. While these arguments are presented here for a four level laser system, they are valid for lasing transitions for all lasers. From this equation, it is clear that to achieve a large positive value of the gain coefficient, one needs (i) population inversion, i.e. N3>>N2 (ii) smaller frequencies of light (or larger wavelengths) and (iii) smaller spontaneous life times3. Observation (i) is obvious in terms of population inversion. If the populations are equal, or, if the lower level is populated more so than the higher level, there will be no gain for the laser (i.e., gain less than one, implying loss.) Observation (ii) shows that high gains are more easily achievable with visible and IR lasers and relatively more difficult with shorter wavelength lasers using conventional lasing principles. Observation (iii) is problematic because while a higher life time of the excited state is more desirable for the sake of population inversion, at the same time, it should be kept short to improve the gain. Therefore, one has to strike a balance between these two competing parameters3. When a sustained laser output power is feasible (i.e. GR >1), the gain coefficient corresponding to the round trip with amplification, called threshold (or saturation) gain coefficient, will be lower than the small signal gain coefficient4. This is because during the round trip there will some loss of population inversion, compared to the “small signal” gain which corresponds to a single pass through the lasing medium. Example 9.1 A crystal rod, about 90 mm in length has a gain coefficient of 0.025 mm-1. The total reflector has a reflectance of 99.8% and the output mirror has a reflectance of 60%. Find (i) the gain and (ii) the round trip gain. Assume 10% losses. (i) Use G = = = e βx e 0.025 x 90 9.49 Answer (b) Use GR = = = G2ρHRρOC(1-ε) (9.49)2 x 0.998 x 0.60 x (1-0.10) 48.5 Answer As mentioned previously, if the round trip gain (GR)is less than one, the laser beam cannot be sustained and the power will die out. If GR is greater than one, the power will continue to increase. This is not a desirable situation either, unless it ultimately drops and then stays at one, where the beam maintains a steady output power. 241 Let us examine how the round trip gain GR and the laser power build up for a continuous wave (CW) laser from the moment the laser power is turned on3,4. Clearly, as we have learned before, several processes will have to occur sequentially such as excitation (or pumping), spontaneous emission, stimulated mission and amplification. Table 9.1 demonstrates this phenomenon using exemplar values for a continuous wave 2 mW power laser. The time values are based on arbitrary units. Time GR 0 1 2 3 4 5 6 7 8 9 0 0.4 0.8 1.1 1.3 1.5 1.6 1.7 1.8 1.9 Table 9.1 Power (mW) Time GR 0 10 0 11 0 12 0 13 0 14 0 15 0.3 16 0.6 17 0.9 18 1.2 19 Power (mW) 2 1.8 1.6 1.4 1.3 1.2 1.1 1 1 1 1.4 1.5 1.6 1.7 1.8 1.9 1.95 1.99 2 2 It is seen from Table 9.1 that, despite the excitation mechanism that was present from zero time units, lasing begins only at t=6 time units, as evidenced by the laser power. Clearly population inversion is achieved to such an extent that any losses are overcome (note GR=1.6 now). With increasing time, even though the round trip gain decreases, laser output power steadily increases until GR=1 again, and the laser output power becomes steady. This clearly explains why it takes a few seconds for a CW laser to produce an output beam from the time the power is turned on. The laser is just not “warming up”; rather, all the processes are set in place, namely, excitation, population inversion, spontaneous emission, stimulated emission, amplification and finally maintenance of the round trip gain at one. We see that the steady output of the laser of 2 mW power begins at 18 time units. A more rigorous mathematical treatment is available3 and is beyond the scope of this book. It is to be noted that the gain coefficient and the gain are predominant for a distinct wavelength1-4. The gain decreases below and above this nominal wavelength. This can be explained by various mechanisms. First, due to the imperfection of the spontaneous emission process, atoms in the higher excited states have slightly differing lifetimes leading to slightly differing frequencies, which in turn exhibit as line broadening for the laser output. Secondly, a phenomenon called Doppler broadening contributes additionally. The Doppler Effect can be easily explained using sound waves. For example, if you are standing next to a rail road track and you see a train coming toward you blowing its whistle. The frequency of the whistle sounds high pitched as the train approaches you 242 fast, and, its pitch drops off significantly as the engine passes you by and heads away from you. The reason for the increased frequency in the case of the approaching engine is due to the speed of the waves that impinges on your ear drum increasing from a combination of the speed of sound in air (330 m/s) and the speed of the train. Thus, the sound waves are “compressed” as they head toward you. “Compressed” implies shorter wave length for the sound, or higher frequency (i.e. higher pitch, meaning shriller.) On the other hand, when the engine is heading away from you, the sound waves are elongated, i.e. larger wavelength and lower frequency or pitch. In much the same fashion, if all exited atoms were stationary when they emit photons, they would emit the same wavelength. In reality, atoms exhibit random motion causing Doppler broadening of the wavelength of photons emitted as they undergo spontaneous emission. The third phenomenon called collision broadening is more pronounced in high pressure lasers. A laser cavity or optical resonator provides the feedback mechanism by for lasing. Some common configurations of stable laser resonators are shown in Figure 9.3 1-4: plane parallel confocal (i.e. common focus point for both mirrors.) hemispherical Figure 9.3 Some examples of cavity configurations. One mirror serves as the total reflectance mirror, the other as the output mirror. The resonator quality, denoted by the letter Q, is an indication of the loss level in the cavity. A resonator with high Q has low loss. Q depends on the wavelength. A large value of Q at a wavelength results in the laser emitting output of that wavelength. The confocal resonator is commonly used in CW gas lasers. Needless to say, precise alignment and a high degree of cleanliness of mirrors are needed for lasing to occur. For a confocal cavity as shown in Figure 9.3, the simplest mode is one that travels along the optical axis and undergoes multiple reflections at the mirrors. This is called the 00 mode, or more precisely as TEM00 mode, TEM representing Transverse Electromagnetic 243 Mode (as opposed to an axial mode.) The intensity distribution along the beam diameter follows a Gaussian distribution, as shown in Figure 9.4. Gaussian distribution Intensity -4 1 0.8 0.6 0.4 0.2 0 -2 0 2 Distance from beam center, r 4 Figure 9.4 Intensity distribution from a Gaussian mode (TEM00). The beam is clearly not uniform in its intensity due to diffraction effects, a topic discussed in Chapter 10. This being the case, the radius of the beam is defined as the location where the intensity falls to 1/e2 of the peak value, the peak occurring at the beam center. The value of e is 2.71828… and therefore, 1/e2 is 0.135335… Thus, for a laser beam operating in the TEM00 mode, the intensity distribution is Gaussian, with the highest intensity occurring at the center, and, the intensity falling off to about 13.5% of the highest at a radius w, called the 1/e2 radius. The corresponding diameter is called the 1/e2 beam diameter. The TEM00 mode has the lowest diffraction loss. Higher modes exist and are shown in Figure 9.5 1-4. TEM00 mode TEM01 mode TEM11 mode TEM10 mode Figure 9.5 Some examples of TEM modes. The definition of the TEM mode number is based on the number of minimum points of the beam intensity as one traverses the beam in the horizontal and vertical directions, respectively. For example, for the TEM00, one encounters a maximum intensity at the center as one traverses the beam in either the horizontal or vertical direction. Thus “00” denotes zero minima in the intensity in either direction. For the TEM01 mode, no minimum intensity is encountered in the horizontal direction and one minimum is seen in the vertical direction, and hence the nomenclature “01.” For a laser exhibiting a 244 combination (i.e. superposition) of two or more transverse modes, the beam has several minima in both directions. Known as multimode lasers, these have the primary 00 mode and several superimposed transverse modes. The approximate beam divergence of a laser with a Gaussian intensity distribution profile can be calculated knowing the wavelength and the diameter of the laser cavity aperture using the equation2-4. θ ~ 4λ πd where θ is the beam divergence in radians, λ the wavelength of light and d the aperture diameter. Radian is a different unit for measuring angles, with one radian approximately equaling 57.3o. The approximate beam divergence equation is based on Rayleigh’s diffraction equation discussed in the Chapter on wave optics. The reason why this is approximate is because the actual multiplier is 1.22, whereas 4 divided by π is about 1.27. In another approximation, 1.22 or 1.27 is approximated to 1.0 for ease of calculations. Example 9.2 Calculate the beam divergence of a He:Ne laser outputting 632.8 nm light with a 2 mm diameter for the aperture. Using λ=632.8 nm, d=1mm, we obtain θ ~ 403 µradians or 0.0230o Answer While we mentioned in Chapter 4 that a laser beam is highly directional, there is some divergence. While the divergence is constant, the spreading of the beam becomes appreciable over very long distances. Exercise 9.1A 1. A crystal rod 4 inches (102 mm) in length has a gain coefficient of 0.025 mm-1. The mirror reflectances are 99.9% and 45%. Find (i) the gain and (ii) the round trip gain. Assume 12% losses. 2. A crystal rod 125 mm in length has a gain coefficient of 0.035 mm-1. The reflectances are 99.9% and 50%. Find (i) the gain and (ii) the round trip gain. Assume 12% losses. 3. Calculate the beam divergence of a laser outputting 488 nm light with a 3 mm aperture diameter. 4. Calculate the beam divergence of a laser outputting 514.5 nm light with a 3 mm aperture diameter. 245 9.2 Pulsed Lasers At this point, the student should have a better understanding of the overall laser operation including gain and mode. A laser whose output appears in the form of pulses is known as a pulsed laser. This is in sharp contrast to continuous wave (CW) lasers whose output is continuous. For certain applications, one needs high power to be delivered to the target in a short period of time. This is feasible only with pulsed lasers. While the individual energy of each pulse may not appear to be appreciable, the duration of the pulse can be so short that the power of the pulse delivered can be very high, bringing about novel physical and chemical processes in the irradiated areas. Examples of pulsed lasers include Nd: YAG, carbon dioxide, nitrogen, excimer, dye laser, etc. With pulsed lasers, sufficient energy has to build up between the firing of pulses that a very high population inversion has to take place2. Consequently, the round trip gain also builds up to much higher values compared to a CW laser discussed previously. Some lasers such as the excimer can operate only in the pulsed mode due to difficulties with the excitation process and life time of the excited species. Others can produce only pulses because in CW mode, excessive heat builds up. In other cases, population inversion can be accomplished only intermittently because of the longer pumping times required to elevate the lower lasing level atoms. The typical gain and laser pulse power are shown in Table 9.2 using exemplar values, for a laser with relatively low pulse repetition rate. As before, the power to the laser is turned on at time zero. The laser peak pulse power was chosen as 2.3 kW. Table 9.2 Time Time Time Time (µs) GR P(kW) (µs) GR P(kW) (µs) GR P(kW) (µs) GR P(kW) 0 0 0 10 4.7 2.3 20 1 0 30 4.7 2.3 1 1 0 11 4.9 2 21 1 0 31 4.9 2 2 1 0 12 5 1.7 22 1 0 32 5 1.7 3 2 0.3 13 4 1.4 23 2 0.3 33 4 1.4 4 2 0.6 14 3 1.2 24 2 0.6 34 3 1.2 5 3 0.9 15 2 1 25 3 0.9 35 2 1 6 3 1.2 16 1 0.8 26 3 1.2 36 1 0.8 7 4 1.4 17 0.8 0.6 27 4 1.4 37 0.8 0.6 8 4 1.7 18 0.7 0.3 28 4 1.7 38 0.7 0.3 9 4 2 19 0.7 0 29 4 2 39 0.7 0 The pulse is “on” from 2+ units on the time scale corresponding to the round trip gain barely exceeding one. Around 19 units of time, when the round trip gain is approaching the minimum, the pulse shuts off because of low population inversion. It would come “on” around 22+ time units when the round trip gain barely exceeds one again, and, the population inversion is more favorable. The second pulse would cut out soon after 38+ time units. The process would repeat itself cyclically. The next section describes in detail the pulse shape, the pulse repetition frequency and related parameters. 246 A pulse is a burst of energy. It lasts for certain duration, known as the pulse width tw. The duration between appearances of successive pulses is known as the time period T. Conversely, the number of pulses that appear each second, known as the pulse repetition frequency (PRF) is related to the time period T in an inverse fashion, as2,5 PRF = 1 T T = 1 . PRF or The time period T used in this context should not be confused with the time period associated with the photon. An ideal pulse would appear as follows. A train of three pulses appearing in time sequence is illustrated in Figure 9.6. Pulse on tw Pulse on Pulse on Pulse off T Leading edge T Trailing edge Figure 9.6 Illustration of definitions associated with pulses. We called these “ideal” pulses because the pulse is turned on instantaneously and switched off instantaneously. In other words, it takes zero time for the pulse power to reach its high value when it is turned on, and, it takes zero time for the power to go down to zero when the pulse is turned off. Clearly, the electrical switching circuits controlling the laser cannot respond that quickly and hence these pulses are termed “ideal.” The time period can be measured from the leading edge of a pulse to the leading edge of the next, or, from the trailing edge of a pulse to the trailing edge of the next. The pulse is “on” for a duration tw, the pulse width, and it is off for a duration corresponding to T-tw. An ideal square pulse is one for which the time period is twice the pulse width, i.e. the pulse is on and off for the same duration of time. Let us look at real pulses. A real pulse takes time to turn on and reach its peak energy value. Similarly, it takes time for the real pulse to change from the maximum energy to zero. The maximum energy is termed the amplitude of the pulse. The time taken for the pulse energy to increase from 10% of the maximum to 90% of the maximum is one method of characterizing the rise time5 tr. The time it takes for the pulse energy to decrease from 90% of the maximum to 10% of the maximum is one method of characterizing the fall time tf. The time duration, when the pulse energy remains at 50% 247 or higher of the maximum, is termed the pulse width5, tw, also known as FWHM, full width at half maximum. Figure 9.7 illustrates these definitions for a real pulse. 90% of maximum 50% of maximum Maximum energy (Amplitude) 10% of maximum tw Rise time Fall time Figure 9.7 Illustration of rise time, pulse width and fall time. The ratio of the pulse width to the time period is known as the duty cycle2,5. Duty cycle = tw T Duty cycle can be expressed as a percentage. It is an indication of how often the pulse is on. Pulse energy Example 9.3 Calculate the rise time, fall time, pulse width, time period, pulse repetition frequency and the duty cycle of the following wave form. 100 90 80 70 60 50 40 30 20 10 0 0 5 10 15 Time (ns) Figure 9.8 Illustration of Example 9.3. Verify if you obtain the following answers: • Rise time = 2.6 ns • Fall time = 3.7 ns • Pulse width = 2 ns • Time period = 23 ns • PRF = 1/23 ns = 43.5 MHz • Duty cycle = 2/23 = 0.087 or 8.7% 20 25 30 248 Example 9.3 illustrated the temporal (i.e. along the time axis) characterization. What about pulse energy, pulse peak power and average laser power? We made similar calculations in Chapter 3 for pulsed lasers. Let us revisit in light of the new terminology. Laser pulse (peak) power is defined as the power delivered by each pulse while it is on. It is also called the peak or maximum power. The pulse itself is considered “on” during the pulse width tw, and, it is “off” during T-tw, where T is the duration between appearances of pulses. The reciprocal of T is the pulse repetition frequency PRF. The average laser power is lower than the pulse (peak) power since the power is averaged over the on and off cycles of the pulse. The duty cycle defines the fraction of the time that the pulse is on compared to the total (on plus off) cycle time. The following formulas are useful2,5. Laser pulse power (Ppeak) = Energy per pulse (Ep) Pulse width/duration (tw) Average laser power (Pavg) = Total energy of pulses in 1 s = (PRF)Ep 1 second Laser pulse power (Ppeak) = Average laser power (Pavg) Duty Cycle Energy per pulse (Ep) = Ppeak x tw Energy per pulse (Ep) = Pavg PRF Duty cycle = Pavg Ppeak PRF = Pavg Ep The derivations of these formulas follow a logical pattern based on fundamental definitions of energy, time, power and frequency and is left as an exercise for the student. Example 9.4 An excimer laser puts out pulses of 14 ns pulse width, each pulse having energy of 250 mJ. If the laser is operating at a PRF of 10 Hz, calculate (i) the pulse power, (ii) the average power, (iii) the duty cycle. (i) Laser pulse power (Ppeak) = Energy per pulse (Ep) = 250 mJ = 17.9 MW Pulse width/duration (tw) 14 ns Answer (ii) Average laser power (Pavg) = (PRF)Ep = 10 x 250 mJ = 2.5 W Answer 249 (iii) Duty cycle = tw T = PRF x tw = 10Hz x 14 ns = 1.4 x 10-7 Answer Example 9.5 A carbon dioxide laser puts out an average power of 200 W. If it operates at a duty cycle of 20% and if takes 1 ms between pulses, calculate (a) peak power, (b) PRF, (c) laser pulse width and (d) energy per pulse. The facts are Pavg = 200 W; Duty cycle = 0.2; T= 1 ms PRF = 1/1ms = 1 kHz Average laser power (Pavg) = (a) Ppeak = Pavg Duty cycle Laser pulse power (Ppeak) x Duty Cycle = 200 W = 1 kW Answer 0.2 (b) PRF = 1/T = 1/1 ms = 1 kHz Answer (c) tw Duty cycle = tw T = T x Duty cycle = 1 ms x 0.2 = 200 µs Answer (d) Energy per pulse (Ep)= Ppeak x tw = 1 kW x 200 µs = 200 mJ Answer Nd:YAG and carbon dioxide lasers are capable of relatively high PRF discussed in this example. The methods of pulsing the laser, as well as the lifetimes associated with spontaneous emission, largely dictate the pulse width and the PRF attainable. Exercise 9.2A 1. An excimer laser outputs 248 nm wavelength photons each lasting 20 ns. If the pulse energy is 150 mJ, (a) what should be the PRF to obtain an average power of 1.05 W? (b) How far apart would the pulses be spaced in time? (c) What is the duty cycle? (d) What is the laser pulse power? 2. An excimer laser outputs 193 nm wavelength photons each lasting 15 ns. If the pulse energy is 200 mJ, (a) what is the average power at a PRF of 5 Hz? (b) How far apart would the pulses be spaced in time? (c) What is the duty cycle? (d) What is the laser pulse power? 3. An IR laser puts out an average power of 100 W. If it operates at a duty cycle of 20% and if takes 4 ms between pulses, calculate (a) peak power, (b) PRF, (c) laser pulse width and (d) energy per pulse. 4. (a) What is the pulse energy of a crystal based laser that operates at 2 kHz PRF with an average 10 W power? (b) What is the duration between successive pulses? (c) Can you calculate the laser pulse power? 250 5. Calculate the rise time, fall time, pulse width, time period, pulse repetition frequency and the duty cycle of the following wave form. 100 Energy 80 60 40 20 0 0 5 10 15 20 25 30 35 40 45 50 55 60 Time (microseconds) Figure 9.9 Illustration for Exercise 9.2A, problem 5. 6. Calculate the rise time, fall time, pulse width, time period, pulse repetition frequency and the duty cycle of the following wave form. 100 Energy 80 60 40 20 0 0 5 10 15 20 25 30 Time (ns) Figure 9.10 Illustration for Exercise 9.2A, problem 6. 7. A pulsed laser puts out 8 W of average power and 1 kW of laser pulse power. What is its duty cycle? 8. On the average a pulsed laser puts out 20 W of power. If the PRF is 20 Hz, what is the energy per pulse? 251 9. The pulse width of a laser pulse is 100 µs. If the duty cycle is 16.7%, calculate (a) the PRF and (b) the duration between pulses. 10. (a) If successive laser pulses appear every 500 µs, what is the PRF? (b) If each pulse lasts 10 µs, what is the per cent duty cycle? (c) If the laser pulse power is 2 kW, what is the energy per pulse? (d) What is the average power of the laser? 11. On the average a pulsed laser puts out 10 W of power. If the PRF is 100 Hz, what is the energy per pulse? 12. The pulse width of a laser pulse is 2.5 ms. If the duty cycle is 16.7%, calculate (a) the PRF and (b) the duration between pulses. 9.3 Pulsing Techniques In Section 9.1, we briefly discussed laser modes. Axial modes are formed when the spacing between the laser mirrors is an integral multiple of one half wavelengths. Transverse modes are also formed which cause variations in the laser beam intensity across its cross section. A few TEM modes were discussed diagrammatically in Section 9.1. If modes are not controlled or selected, the output beam is multimodal. Multimodal output exhibits spectral width. In order to suppress modes and allow only a single mode, and, thereby nearly monochromatic light to be output, one uses a device known as an etalon1,2,4,6. An example of an etalon is a Fabry-Perot etalon which is basically a block with two reflectors inserted into the lasing medium. By adjusting the spacing or the angle of the reflectors, laser oscillation at a desired wavelength can be achieved, while suppressing other wavelengths. The underlying principle is interference, discussed in the Chapter on wave optics. While single mode concept applies equally well to CW and pulsed lasers, it is more common with CW lasers. While a laser is more monochromatic than conventional light sources, the light inside the cavity oscillates at different frequencies or wavelengths known as spectral modes. For multimodal waves there is no phase relationship among them. Therefore, they vary in a random fashion. If the different random modes can be combined in such a fashion that they are in phase, this is known as mode locking1-4. For a mode locked laser with n modes, the laser output would consist of pulses, each with approximately n times the power of a laser which is not mode locked. To produce very short duration (i.e., femto- to nanosecond), high power pulses, mode locking1-4 (also known as phase locking) is commonly employed. If the different axial modes in the laser can somehow be added together, now all the modes will be in phase, and a continuous stream of pulses will be emitted. Figure 9.11 illustrates this principle. In the first figure, a random axial wave (marked as Wave # 1) is shown. In the second figure, a second axial wave is shown. The second wave (Wave #2) is in phase with wave #1, however, of slightly different wavelength. If these two waves are added together, because of slightly differing wavelengths, they either reinforce or cancel out each other in different places. The resultant wave is shown in the third figure (Wave #1 and Wave #2 252 Wave #1 15 10 5 0 0 100 200 300 400 500 600 700 800 500 600 700 800 700 800 -5 - 10 - 15 Wave #2 15 10 5 0 0 100 200 300 400 -5 - 10 - 15 Figure 9.11 Illustration of mode locking principle. Wave #1 + Wave #2 added 25 20 15 10 5 0 0 100 200 300 400 500 600 -5 - 10 - 15 - 20 - 25 Pulse on Pulse off Pulse on 253 added.). One can see the beginnings of a well defined area where the resultant wave is absent1-4. When more and more waves of the same phase, however differing wavelengths are added together, it truly results in a train of pulses as illustrated below the third figure. By suitable phase locking via modulation, pulsed lasers with pulse widths of a few hundred femtoseconds to a few picoseconds have been constructed. Another method of obtaining laser pulses is through Q-switching1-4 or Q-spoiling. As discussed earlier, the resonator quality, denoted by the letter Q, is an indication of the loss level in the cavity. A resonator with high Q has low loss. Q depends on the wavelength. A large value of Q at a wavelength results in the laser emitting output of that wavelength. By interrupting the Q of a laser for a short period of time and then reinstating it, a laser pulse can be generated. When the Q is spoiled, population inversion continues while no lasing occurs. This allows continuous energy build up to occur. When the Q is restored, all the built up energy is released as a high energy pulse. Turning the Q on and off can be achieved, for example, using a disk with slots mounted to a motor, or by electro-optics means. As the motor rotates, the slots interrupt the beam. Q switches are commonly used with solid state lasers such as ruby and Nd:YAG. Modulation1,2,4 is another technique for Q-switching. Modulation can be accomplished either using sound waves that are inaudible (hundreds of MHz) or by electro-optic means. For the former case, a transducer converts an electric signal into a mechanical wave. A mechanical wave traveling through a crystal changes its refractive index. The crystal then behaves like a diffraction grating. (See the Chapter on wave optics for a discussion of the diffraction phenomenon.) The crystal can deflect light beams. A modulated lower frequency signal can successfully turn the acoustic wave on and off, thus, creating among other phenomena, Q-switching, mode locking, etc. Using bleachable dyes, for example, a technique known as cavity dumping has also been used to create laser pulses. In recent years, a technique known as chirped pulse amplification has gained prominence7,8. For example, consider a laser pulse with 2 mJ energy and pulse duration of 10 fs. The peak pulse power can easily be calculated to be 200GW. Let us, for the sake of calculations, assume that a pulse of laser light can be focused to a spot 50 µm in diameter. The peak irradiance or power density at the spot can be calculated to be over 10 PW/cm2, which far exceeds the irradiance threshold for damage to materials in contact by several orders of magnitude. The solution has been to elongate the pulse, amplify it and to recompress it so that its original pulse width is restored. The system often uses gratings, lenses and mirrors and titanium:sapphire amplifier as part of the set up. One of the novel developments in the past twenty years is the evolution of the excimer laser. As discussed in Chapter 4, an excimer stands for an excited dimer, a two-atom molecule. The atoms are derived from a noble gas such as argon, krypton or xenon, and a halogen such as chlorine, fluorine or bromine. Population inversion is achieved with a high degree of efficiency, since the dimer exists only in the excited state, and not in the ground state. The ground state consists only of pure noble gas and pure halogen, and not a dimer molecule. Besides the noble gas and halogen, the gas mixture consists of inert 254 gases such as helium and/or neon at higher than atmospheric pressures. The higher pressure of the gas mixture used in excimer lasers sharply contrasts with many other gas lasers such as the He:Ne. While sub-atmospheric pressures are desirable in order to minimize collisions and line broadening of the output, however, with higher pressures (at or above atmospheric) more molecules are available in the active medium so that higher laser output powers can be achieved. In order to obtain a stable, high energy pulsed laser output, the electric field must be in a direction perpendicular to the optical axis. Such lasers are known as Transverse Excitation at Atmospheric (TEA) pressure lasers1,2. Due to the high pressure, the light is nearly not as monochromatic as low pressure lasers. Molecular collisions cause line broadening and a larger spectral width, leading to the ability to operate the laser in a mode locked format, with consequent high energy excimer laser pulses. Pulsed carbon dioxide lasers also operate in the TEA mode. An excimer laser can be pumped using an electron beam pulse1. The electron beam itself must be generated in vacuum and delivered into the high pressure gas chamber. The mirrors used in excimer lasers need to be protected from the corrosive halogen gases. While one mirror is highly reflective, the output mirror of the excimer laser reflects only about a few per cent back into the laser cavity. Due to the ease of population inversion and associated high gain, this level of low reflectivity is adequate. Another way of pumping the excimer laser is using a discharge mechanism similar to a carbon dioxide laser1. By pulsing the electrical power using a thyratron circuit one obtains laser pulses as output. The thyratron circuit consists of large capacitors connected to a DC source. The charging and discharging of these capacitors produce laser pulses. The pulse width of the excimer laser ranges from a few to few tens of nanoseconds, with pulse energies ranging from a few hundred millijoules to around a joule. The corresponding peak pulse powers are extremely large. Such powers bring about novel photochemical processes on materials. They find applications in laser surgery of the eye, angioplasty, semiconductor manufacturing operations such as photolithography, etching and annealing, analytical chemistry and general surface treatment. One of the drawbacks of the excimer laser is its poor collimation characteristics. The beam divergence of up to 0.2o is much higher than that of a He:Ne or argon ion laser. A pulsed carbon dioxide laser operates similar to an excimer in terms of the TEA mode of operation. With high power pulses, they find applications in high temperature metals processing such as drilling, cutting, welding, etc. They are also used in skin surface treatment and surgery. They also find use in military applications as in distance finding of targets, and in analytical chemistry. Crystal based lasers such as the Nd:YAG can produce one hundred or more joules of energy per pulse, with peak pulse powers of the order of a few GW. Its applications are similar to that of the carbon dioxide laser, due to its ability to heat up material quickly. The excitation mechanism involves a pulsed flash lamp resulting in pulsed laser output. The power supplies used with flash lamps are also pulsed. This method of pumping produces millisecond duration pulses with several tens to hundreds of kilowatts of pulse power. In order to obtain nanosecond pulse widths, Q-switching is used, and, mode locking can result in tenths of nanosecond duration pulses. One normally works with the 255 1.064 µm wavelength output. However by frequency doubling or tripling via the use of crystals, one can convert the output to 532 nm or 355 nm, respectively. Diode laser pumping has become popular in view of minimum system cooling requirements. A pulsed dye laser is pumped by a frequency doubled Nd:YAG laser, an excimer laser, a nitrogen laser or flash lamps. As discussed in Chapter 4, the main advantage of the dye laser is its tuneability. By changing the dye solution, one can obtain a variety of output wavelengths which find applications in analytical chemistry. Diode lasers using gallium arsenide semiconductor can operate in the pulsed mode outputting up to a kilowatt of pulse power. The spectral width of the output can be as high as a few nm. Pulsing the current that flows into the semiconductor junction results in pulsed laser output. As discussed in Chapters 4 and 6, the diode lasers find applications in fiber optics communication. 9.4 Suggested Laboratory Experiments Measurement of pulse width and pulse energy 1. Ensure that you have an energy meter capable of reading individual pulse energy with a fast response rate at the wavelength of your pulsed laser. Use correction if needed if the wavelength of your laser is different from that for which the meter was calibrated. Make sure that the meter is rated to handle the instantaneous (i.e. maximum) pulse power. 2. Hook up the energy meter to the read out. Measure the energy of an individual pulse as Ep. Record it in Table 9.3. 3. Disconnect the readout from the energy meter. Hook the energy meter to an oscilloscope. Adjust the vertical and horizontal scales such that you obtain a pulse profile that fits in the screen, as shown below. The laser should be adjusted to output pulses with a very low PRF. Establish a conversion between a vertical division on the oscilloscope and pulse energy measured in step 2. Base line Figure 9.12 Typical strip chart recorder output of a pulse. 256 4. Using the time base scale of the oscilloscope, determine the time scale associated with each horizontal division of the chart. 5. Determine the locations of the half power points. Count the number of time base divisions between the two half power points. Convert the time base divisions into actual time. This provides the pulse width, tw. Record it in Table 9.3. 6. Determine the time associated with the pulse rising from 10% of the maximum power to 90% of the maximum power. This is the rise time tr. Record it in Table 9.3. 7. Determine the time associated with the pulse falling from 90% of the maximum power to 10% of the maximum power. This is the fall time tf. Record it in Table 9.3. 8. Calculate the peak power using Ppeak = Ep/tw . Record it in Table 9.3. 9. Adjust the PRF of the laser at 10 Hz. Determine if the oscilloscope peak pulse height is different from that measured in step 1. If different, apply the conversion from step 3. 10. Calculate the average power using Average laser power (Pavg) =(PRF)Ep Record it in Table 9.3. 11. Calculate the duty cycle using Duty cycle = Pavg Ppeak Record it in Table 9.3. 12. Repeat steps 9, 10 and 11 for a slightly higher PRF. Record answers in Table 9.3. PRF (Hz) Ep Table 9.3 Individual single pulse energy = Pulse width= Rise time= Fall time= Individual pulse peak power= Ppeak Pavg Duty cycle Does the peak pulse power decrease with increasing PRF for your laser? Explain. Chapter Summary • A laser will produce sustained output if the round trip gain is maintained at one. If less than one, the beam will die out. The round trip gain is related to the gain coefficient, reflectivity of the laser mirrors and cavity losses. 257 • • • • • • For a pulsed laser, the gain continues to build when the pulse is not on. When the pulse is turned on, the gain will fall and the pulse will continue until the gain falls below a certain threshold corresponding to favorable population inversion. An optical resonator has two mirrors providing gain. The mirrors may be arranged in plane parallel, confocal or hemispherical forms. A pulse is characterized by its leading edge, trailing edge, pulse width, amplitude, rise time, fall time, pulse energy and laser pulse (peak) power. A train of pulses is characterized by the time period, pulse repetition frequency, average power and duty cycle. Pulsing techniques include Q-switching, mode (phase) locking, modulation and/or pulsing the power supply. Due to the high power deliverable from pulsed lasers, pulsed lasers find many industrial applications in medicine, cutting, welding, soldering, drilling, and military, semiconductor processing and analytical chemistry. End of Chapter Exercises 1. A crystal rod 125 mm in length has a gain coefficient of 0.020 mm-1. One mirror has a reflectance of 99.2% and the other has a reflectance of 55%. Find (i) the gain and (ii) the round trip gain. Assume 10% losses. 2. A crystal rod 100 mm in length has a gain coefficient of 0.035 mm-1. The mirrors have reflectance of 99.4% and 52%, respectively. Find (i) the gain and (ii) the round trip gain. Assume 9% losses. 3. Calculate the beam divergence of a laser outputting 633 nm light with a 1.8 mm aperture diameter. 4. Calculate the beam divergence of a laser outputting 532 nm light with a 2 mm aperture diameter. 5. An IR laser puts out an average power of 150 W. If it operates at a duty cycle of 33% and if takes 4 ms between pulses, calculate (a) peak power, (b) PRF, (c) laser pulse width and (d) energy per pulse. 6. (a) What is the pulse energy of a crystal based laser that operates at 1 kHz PRF with an average 15 W power? (b) What is the duration between successive pulses? (c) Can you calculate the laser pulse power? 7. A pulsed laser puts out 10 W of average power and 1 kW of laser pulse power. What is its duty cycle? 8. On the average a pulsed laser puts out 50 W of power. If the PRF is 20 Hz, what is the energy per pulse? 258 9. The pulse width of a laser pulse is 50 µs. If the duty cycle is 16.7%, calculate (a) the PRF and (b) the duration between pulses. 10. (a) If successive laser pulses appear every 400 µs, what is the PRF? (b) If each pulse lasts 12.5 µs, what is the per cent duty cycle? (c) If the laser pulse power is 2 kW, what is the energy per pulse? (d) What is the average power of the laser? 11. On the average a pulsed laser puts out 10 W of power. If the PRF is 10 Hz, what is the energy per pulse? 12. The pulse width of a laser pulse is 1 ms. If the duty cycle is 16.7%, calculate (a) the PRF and (b) the duration between pulses. Group Discussion Questions These require the students to form small groups of three or four and discuss each question. Some questions may require you to do a search on the Internet and find the answers. 1. Why do some lasers operate only in the pulsed mode? 2. Describe the principles behind frequency doubling crystals. 3. What is chirped pulse amplification method? 4. Describe mode locking (phase locking). 5. What is a TEA laser? 6. Research on the highest power laser that you can find in the open literature. References Laurence, C. “Laser Book: A New Technology of Light.” New York: Prentice Hall (1986) 2 Hecht, J. “Understanding Lasers.” Carmel: Howard H. Sams Company, 1988. 3 Luxon, J.T. and D.E. Parker. “Industrial Lasers and their Applications.” Englewood Cliffs:Prentice Hall, 1985. 4 O’Shea, D.C., W.R. Callen and W.T. Rhodes. “Introduction to Lasers and their Applications.” Reading: Addison-Wesley Publishing Company, 1978. 5 Floyd, T.L. “Electronics Circuits Fundamentals.” Englewood Cliffs: Prentice Hall, 2002. 6 Young, M., “Optics and Lasers.” Berlin: Springer-Verlag, 2000 7 The Femtosecond Research Center, http://dutch.phys.strath.ac.uk/FRC/stuff/dictionary/cpa/cpa.html 8 Stanford University Department of Physics, http://wwwproject.slac.stanford.edu/lc/local/ systems/Lasers/LaserPhysics/cpa.htm 1 259 Chapter 10 Geometric Optics: Part 1 10.1 Lens Maker’s Equation 10.2 Thin Lens Equation 10.3 Imaging with Single Lens 10.4 Imaging with Multiple Lenses 10.5 Suggested Laboratory Experiments Chapter Summary End of Chapter Exercises Group Discussion Questions Chapter Objectives After completing the chapter, the student should be able to (i) calculate focal length and focal power using lens maker’s equation (ii) use thin lens equation to predict image location and magnification. (iii) perform ray tracing in single and multiple lens systems Key words: thick lens, thin lens, focal points, radius of curvature, object, lens, real image, virtual image, paraxial rays, ray tracing, lateral magnification, erect image, inverted image 10.1 Lens Maker’s Equation1-4 A brief discussion of lenses was presented in Chapter 5. Refer back to the pertinent section to understand the difference between positive and negative lenses. In either case, a lens comprises of two surfaces, each with a different radius of curvature. Each surface has a center of curvature. The distance from any point on the surface to the center of curvature C is given by the radius of curvature R, as shown in Figure 10.1. R R C C R Concave surface Optical Axis R Convex surface Figure 10.1 Examples of curved surfaces, center of curvature and radius of curvature The curved surface is part of a sphere, although we present it on a two-dimensional piece of paper. The figure on the left side represents a concave surface, while the figure on the right represents a convex surface. In any case, the radius is always perpendicular to the surface at the point of contact. 260 The reciprocal of the radius of curvature R is known as the curvature of the surface. The curvature,ξ is given by ξ= 1 R A small radius of curvature results in a large curvature. This is obvious by examining two drops of, say, water. A small drop of water (i.e., small radius of curvature) is highly curved, whereas a larger drop, say, a soap bubble, is relatively less curved. In the limit, a flat surface has the center of curvature at infinity, resulting in an infinite radius of curvature and zero curvature (i.e., perfectly flat). We need to know these concepts because, as discussed in Chapter 5, lens can be plano-convex, plano-concave, bi-convex, bi-concave, etc. The “plano” refers to that surface that is flat with R-> ∞ and ξ-> 0. Throughout the next two chapters we are going to assume that light travels from left to right. If there is an exception to this rule, it will be clearly outlined. The second assumption that we will make is that light from the object travels very close to the optical axis. These are known as paraxial rays. There is an implicit assumption that the object height is “small” compared to the diameter of the lens. Accordingly, we can assign signs to the radii of curvature for convex and concave surfaces. The sign convention is to adopt the radius of curvature, R, of a convex surface as positive and negative for a concave surface. In general, the two surfaces that make up a lens can have positive, negative or infinite radius of curvature. A thick lens is one where the two surfaces are separated by a certain thickness which is appreciable compared to the diameter or the radii of curvature. The thin lens, on the other hand, is such that the two curved surfaces are in contact, and the thickest portion of the lens is very small in comparison to the lens diameter or the radii of curvature of the surfaces (see Figure 10.2). Figure 10.2 Examples thick lenses (left two) and thin lenses (right two). The remainder of this and the following chapter will deal exclusively with thin lenses since the mathematical treatment is much simpler. As noted in Chapter 5, one can use a single curved surface and use it as a concave or convex mirror. We also determined that the convex mirror can produce only a virtual image of the object, i.e., the image cannot be caught on a screen; rather, it is visible inside the mirror. The concave mirror, on the other hand, can produce a real or a virtual image, depending on the location of the object from the mirror and its radius of curvature. A real image can be caught on a screen. Applications of the two types of mirrors were discussed in Chapter 5. Clearly the mirror 261 surfaces are coated with silver, gold or aluminum to result in a high reflectance, upwards of 99%. In the case of lenses, the material is glass (sometimes plastic, as in contact lenses), and the reflection from the surfaces are limited to Fresnel reflections only, with reflectance around 4% for paraxial rays. The lens maker’s equation relates the curvatures of the surfaces to a composite focal length through the refractive index of the lens material. Clearly, refraction through curved surfaces, a topic we did not discuss in Chapter 5, is the phenomenon that takes place in lenses. Assume that the lens is in air. For thin lenses and for paraxial rays, we define a focal length f as the distance from the center of the lens to its focal point. Further discussion of primary and secondary focal points for a lens is presented in the next section. It is sufficient to know at this point the focal point is the locus of either a set of originally parallel rays, or a set of parallel rays generated by the lens. The focal length is related to the individual radii of curvature that make up the lens R1 and R2 and the index of refraction of the lens through the following relationship: 1 f = (n-1) 1 - 1 R1 R2 . This is the famous lens maker’s equation. The implicit assumptions are (i) the lens is thin, (ii) the rays are close to the optical axis, (iii) the thinness of the lens lends us to the approximation that light refracts just once at a plane located in the center of the two curved surfaces, and, (iv) the lens is surrounded by air. The proof associated with deriving the lens maker’s equation is presented in the next chapter. Example 10.1 Calculate the focal lengths for the following lenses, with the given radii. The refractive index of the material is 1.5 at the wavelength of interest. (a) (b) R=7 cm C R=15 cm C R=10 cm Figure 10.3 Illustration for Example 10.1. (a) Note that R1=15 cm and R2=-7 cm (concave) 1 f or f = (n-1) 1 - 1 R1 R2 = = = = (1.5-1.0) [(1/15) – (-1/7)] 0.5 [ (1/15) + (1/7)] 0.105 1/0.105 = 9.55 cm Answer . 262 (a) Note that R1=-10 cm (concave) and R2=∞ 1 f or f = (n-1) 1 - 1 R1 R2 = = = (1.5-1.0)[(-1/10) – 1/(∞)] 0.5[-1/10] -0.05 = -20 cm Answer . As stated in Chapter 5, positive lenses have the thickest portion in the middle and thinnest at the edges; negative lenses have just the opposite features. The lens in example (a) above has a positive focal length, and is therefore called a positive lens. The lens in (b) has a negative focal length and hence is called a negative lens. The focal power of a lens is defined as the inverse of the focal length. When the focal length is expressed in meters, then the focal power expressed in inverse meters is called the Diopter power of the lens. If the focal length is expressed in cm or mm, it must be converted to meters first to calculate the focal power. Power of a lens in Diopters (D) = 1 (f must be expressed in meters) f Example 10.2 Calculate the power for the lenses in Example 10.1. (a) Power = 1/(0.0955 m) (b) Power = 1/(-0.2 m) = 10.5 D Answer = -5 D Answer Example 10.2 illustrates that positive lenses have positive powers, while negative lenses have negative powers. Eye glass powers follow the same terminology. A large magnitude of the power indicates that the lens is more powerful, whereas a small value indicates a small corrective action on the part of the lens. Reading glasses always have positive powers since the person needs to magnify the print in order to be able to read. Most glasses needed to correct vision problems associated with far sight have negative powers. Thus far we have focused on lenses whose curved surfaces are part of a sphere. It is conceivable that lens can be made whose surface is part of a cylinder, i.e., curved in one direction and flat in the other. Known as cylindrical lenses, these are used to correct special vision problems such as astigmatism. Whereas a spherical lens is curved to the same degree in all directions, a cylindrical lens is curved only in one direction. The radius of curvature in the perpendicular direction is essentially infinity. Thus, a cylindrical lens can focus parallel beam of light only in one direction, i.e., in the direction in which it is curved. A spherical lens, on the other hand, will focus a parallel beam of light essentially to a spot. Also, bifocal, trifocal and multi-focal lenses are now available, which aid in correcting vision at various distances. The radii of curvature with multi- 263 focal lenses vary continually to accomplish multiple focal lengths, and therefore can be expected to be expensive to fabricate compared to a simple, single focal length lens. Does it matter if one switches the surfaces? If the lens is flipped 180o, it does not matter. If one exchanges the two surfaces, it does matter. The following examples illustrate the points. Example 10.3 Rework Example 10.1(a) such that for the biconvex lens, the 7 cm radius surface is now the first surface facing the light and the 15 cm radius surface is the second surface. Note that R1=7 cm and R2=-15 cm (concave) 1 f or f = (n-1) 1 - 1 R1 R2 = = = = (1.5-1.0) [(1/7) – (-1/15)] 0.5 [ (1/7) + (1/15)] 0.105 1/0.105 = 9.55 cm Answer . Clearly, it does not matter, if the lens is turned around. Example 10.4 Rework Example 10.1(a) such that the curved surfaces are exchanged as shown below. R=7 cm R=15 cm Figure 10.4 Illustration for Example 10.4. Clearly, this is now a biconcave lens. Note that R1=-7 cm and R2=15 cm (concave) 1 f or f = (n-1) 1 - 1 R1 R2 = = = = (1.5-1.0) [(-1/7) – (1/15)] 0.5 [ (-1/7) + (-1/15)] -0.105 -1/0.105 = -9.55 cm Answer . 264 While the magnitude of the focal length is the same as in Example 10.1(a), the sign is just the opposite. This lens has a negative focal power as well. Thus, it is important for the optics technician to use the lens maker’s equation correctly, following the sign conventions for the radius of curvature. Exercise 10.1A Complete the following table. Determine in each case, if the lens is positive or negative. Use n=1.5 for the lens material. Problem Number 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Radius of curvature of first surface 30 mm, convex 5”, concave 12 cm, concave 12 cm, convex 16 cm, convex 10 cm, concave 8 cm, convex 9 cm, concave 5 cm, convex Flat 7 cm, concave Flat Flat 2.5 cm, convex Radius of curvature of second surface 20 mm, concave 4”, concave 8 cm, concave 8 cm, concave 20 cm, concave 10 cm, convex 8 cm, convex 11 cm, convex Flat 145 mm, convex Flat 2.5”, concave Flat Flat f, focal length (include sign and units.) Focal power (include sign and units.) 10.2 Thin Lens Equation1-4 After we discussed the lens maker’s equation to determine the focal length, the next question is its meaning with reference to the two curved surfaces. Is it measured to the left side of the lens or to the right side of the lens? For a thin lens, it does not much matter. There are two points on either side of the center of the lens and at a distance f from the center of the lens. These two points are called focal points. For a positive lens (biconvex, plano-convex, etc.), the point on the left is called the primary focal point F. For a negative lens (biconcave, plano-concave, etc.), the point on the right is called the primary focal point F. For a positive lens, the point on the right is called the secondary focal point F’, whereas the point on the left is called F’ for a negative lens, as shown in Figure 10.5. We discussed briefly in Chapter 5 the properties of these points. It is worthwhile summarizing the three major rules that would be useful in ray tracing later on in this chapter. Remember, light always travels from left to right, from a calculation standpoint. Furthermore, the lens being thin, will hereinafter be shown by a line with a marking L to denote that it is a lens. The light is assumed to undergo refraction at the center plane of 265 the lens, once, rather than twice, one at each curved surface. This assumption simplifies the mathematics and the geometry considerably. o F o F’ f o F’ f o F f f Figure 10.5 Location of primary (F) and secondary (F’) focal points. Rule 1: Parallel rays of light incident on a thin lens, will after refraction through the lens, pass through F’ for a positive lens, or appear to emerge from F’ for a negative lens, as shown in Figure 10.6. F F’ F’ L F L Figure 10.6 Illustration of Rule 1 for a positive lens (left) and a negative lens (right). Rule 2: If light rays pass through F for a positive lens, or appear to be headed toward it for a negative lens, then, after being incident on the lens, the rays, after refraction through the lens, will become parallel, as shown in Figure 10.7. F F’ L F’ F L Figure 10.7 Illustration of Rule 2 for a positive lens (left) and a negative lens (right). Rule 3: If rays of light pass through the center of the lens, they will proceed undeviated, as shown below. This is true for both positive and negative lenses (see Figure 10.8). 266 F F’ L F’ F L Figure 10.8 Illustration of Rule 3 for a positive lens (left) and a negative lens (right). Based on thin lens assumptions, these rules enable one to perform ray tracing, i.e., follow the path of two or more distinct rays emanating from the object, follow them through the lens, and determine where the rays intersect. This would enable prediction of image location, as well as image height and orientation, and, if the image is real or virtual. If the image is obtained on the same side as the object, then it is virtual. To “see” this virtual image, look back into the lens towards the object side. This image can never be captured on a screen, or projected on a wall such as in a slide show or in movie projection. If the image is located on the side opposite to the object, then the image is considered real, i.e., it can be captured on a screen (example: movie screen.) The distance from the object to the lens is termed the object distance and is denoted by s. The distance from the image to the lens is termed the image distance and is denoted by s’. In either case, the distance is measured from the lens center. For a thin lens, this is not a problem since the lens thickness is small enough to be ignored. For a positive lens, the following guidelines will be useful. In Table 10.1, 2F denotes a point located at a distance 2f to the left of the lens center, and 2F’ denotes a point located at a distance 2f to the right of the lens center. Table 10.1 Image location for various object locations for a single, positive lens. Object distance Image distance Type of image Magnification/Orientation from lens, s from lens, s’ of image Essentially ∞ At F’ Real Highly diminished/ inverted > 2F Between F’ and Real Diminished/inverted 2F’ 2F 2F’ Real Same size as object/ inverted Between 2F and F >2F’ Real Enlarged/ inverted At F Infinity Real Enlarged Between F and Varies Virtual Enlarged/ erect lens For a positive lens, the image may be either real or virtual, or enlarged or diminished, or inverted (meaning upside down compared to the object) or erect (meaning same orientation as object). With a negative lens, no real image can be obtained; the image is always virtual and diminished. It helps to remember these guidelines so that when 267 calculations are performed using the thin lens equation (see below) or ray tracing (see next section), you would know if your answer is at least in the ball park. The thin lens equation enables prediction of an unknown, given two of the following quantities: s, s’ or f. In its original form, it states 1 f = 1 s + 1 s’ where f denotes the focal length of the thin lens s denotes the distance of the object from the lens s’ denotes the distance of the image from the lens. The equation may also be rewritten as f = ss’ s + s’ Remember to use a positive value for f, if you are using a positive lens and negative value, if you are using a negative lens. Additionally, if the object is real, s is positive, negative otherwise. If the image is real, s’ is positive; s’ is negative if the image is virtual. One would benefit from recasting the thin lens equation so that it is useful for calculating the image distance, when the other variables are known, or for calculating the object distance, when the other variables are known. Via formula transformation, we obtain the following two equations, as s’ = s = sf . s–f s’f . s’ – f Example 10.5 An object is located 12 cm to the left of a positive lens of focal length 8 cm. Where would the image be located? Is the image real? Since s=12 cm and f=8 cm, we use s’ to obtain s’. = sf . s–f 268 s’ = 12 x 8 (12-8) = 24 cm Answer Since s’ turned out to be positive, the image is real and is located 24 cm to the right of the lens, assuming, as usual, that light is coming from left to right, from the object toward the lens. Example 10.6 An object is located 12 cm to the left of a negative lens of focal length 8 cm. Where would the image be located? Is the image real? Since s=12 cm and f=-8 cm, to obtain s’. s’ = 12 x (-8) [12-(-8)] = -4.8 cm Answer Since s’ turned out to be negative, the image is virtual and is located 4.8 cm to the left of the lens, assuming, as usual, that light is coming from left to right, from the object toward the lens. When you repeat this last example with a negative lens for various object distances, you will always find that the image is virtual. Thus, with a single negative lens, one will always obtain a virtual image which can never be captured on a screen. However, if one looks from the right side of the lens, one can see the image with the naked eye. This is what makes the image virtual. Example 10.7 An object is located 6 cm to the left of a positive lens of focal length 8 cm. Where would the image be located? Is the image real? Since s=6 cm and f=8 cm, to obtain s’. s’ = 6x8 (6-8) = -24 cm Answer Since s’ turned out to be negative, the image is virtual and is located 24 cm to the left of the lens, assuming, as usual, that light is coming from left to right, from the object toward the lens. 269 Comparing Example 10.7 with Example 10.5, it is seen that with a positive lens, one can obtain images that are virtual or real, depending on the object location and the focal length. The observations are consistent with the summary presented in the table earlier. The thin lens equation has enabled us calculate the image location and image type for both types of lenses. How does it help predict the image size or orientation? By orientation, we mean if the image is erect (oriented the same way as the object) or inverted (oriented the opposite way as the object). With a single lens, a virtual image is always erect, as we will prove in the next section via ray tracing. Similarly, with a single, positive lens, the real image is always inverted. Let us use this case, apply the principles of similar triangle and calculate the image size (or magnification of the object). B 2F A P F Q F’ L 2F’ A’ B’ Rule 1 Rule 3 Figure 10.9 Application of similar triangles principle to calculate lateral magnification. In Figure 10.9, L is a positive lens. An object AB stands on the optical axis, A, being the foot of the object and B its head. Ray BP obeys Rule 1 since it is parallel to the optical axis to begin with. After striking the lens L at point P, it will bend so that it goes through F’. A second ray, BQ, follows Rule 3, goes through the center of the lens Q undeviated. The rays following the two rules will intersect at a point B’ to the right of the lens. B’ is the location of the image of the point B. If one draws a perpendicular from B’ to the optical axis, it will intersect at point A’, which determines the location of the image of the foot of the object. Refer the pertinent Appendix to refresh the concepts of geometry. Look at triangles BAQ and B’QA’. They are considered similar each angle of one triangle is equal to a corresponding angle of the second. If triangles are similar, then the ratios of corresponding sides are equal, specifically AB = A’B’ AQ QA’ Note that AQ=s (object distance from lens) and QA’=s’ (image distance from the lens). Rearranging after substitution, we obtain A’B’ = s’ AB s Since the object and image are of opposite orientation, A’B’ = -B’A’ 270 The convention here is to measure a height from down to up as positive and from up to down as negative. Image lateral magnification, M = B’A’/AB (heights measured in the same fashion). Thus, M = -s’ s and Image Height = M x Object Height. While we derived these formulas using a real image, the results hold for virtual images as well. Clearly, if you are projecting a movie on to a screen, the lateral magnification M will have to be very high. Note that the lateral magnification M is independent of the object size. Example 10.8 Calculate the lateral magnification for Example 10.5. Calculate the image height, if the object height is 5 mm. M = -24 = -2 Answer (Minus sign just indicates that image is inverted) 12 Image Height = (-2)x(5 mm) = -10 mm Answer Again, the negative height of the image just indicates that the image is inverted. Note that the image is still real as indicated by s’ value of +24 cm. Example 10.9 Calculate the lateral magnification for Example 10.6. Calculate the image height, if the object height is 5 mm. M = -(-4.8) = 0.4 Answer (Plus sign just indicates that image is erect) 12 Image Height = (0.4)x(5 mm) = 2 mm Answer Example 10.10 Calculate the lateral magnification for Example 10.7. Calculate the image height, if the object height is 5 mm. = -(-24) = 2 Answer (Plus sign just indicates that image is erect) 12 Image Height = (2)x(5 mm) = 10 mm Answer M Example 10.10 is an illustration of the principle behind a simple magnifying glass (or a simple microscope). If you hold an object so that it is within the focal length of a positive lens, an enlarged, virtual, erect image is observed. In this example, the image appears bigger, enabling small objects to be magnified, such as fine print. 271 Example 10.11 If I need to obtain a real image that is four times taller than the object, using a 10 cm focal length positive lens, where should I place the object? Where would the image be obtained? It is given that M=-4 (note that a real image with a single lens is always inverted) Thus, -s’ = -4 s or s’ = 4s Given that f=10 cm, use f = ss’ s + s’ Substituting known values, we obtain 10 cm = s(4s) s+4s Simplifying and solving for s, we obtain s=12.5 cm. Answer Since s’=4s, s’ = 4 x 12.5 = 50 cm Answer The object should be placed 12.5 cm to the left of the lens; the image will be obtained on a screen 50 cm to the right of the lens. The image will be four times taller than the object and will be inverted. Exercise 10.2A Fill in the blanks Number 1 s 12 mm 2 3 4 5 6 7 s’ 10 cm 20 cm 20 cm 10 cm 12 mm 10 cm -10 cm -10 cm -2 cm f Lens 10 mm 10 mm Positive Image Object Image (erect or height Height inverted) 1 mm Positive 2 mm Positive 1 cm ? 1 cm Negative 2 cm Negative 15 mm 2 mm 5 cm Positive Image (real or virtual) M -8 mm 272 8 9 10 11 5 cm 12 -7 cm -13 mm 13 14 15 12 cm 10 cm 16 17 18 19 20 mm 15 cm 10 cm 15 cm s Positive -3 5 mm Negative 0.25 2 cm Negative 0.333 20 mm -15 cm 10 cm -15 cm 20 Number 75 mm -5 cm -58 mm -10 cm 14 cm -26 mm -10 cm 5 cm s’ 35 mm f 1.4 mm Negative 7 mm Positive 1.1 cm Negative 1 mm Negative 1.5 cm Positive 5 mm Positive 10 km Negative Positive 25 mm 1 cm Negative 1.2 cm Positive 1.5 mm Object height Lens Image (real or virtual) M Image (erect or inverted) -7.5 mm Image Height Remember the implicit assumptions with the thin lens equation. It is valid, first of all for thin lenses and for paraxial rays, i.e., rays coming off the object that are parallel to the optical axis and very close to it. If these assumptions are violated, the predictions from the thin lens equation will only be approximate. 10.3 Imaging with Single Lens The procedure described in this section is known as ray tracing. We will use the three rules discussed in the previous section. We will arbitrarily call that graphical method as regular ray tracing method. The word regular is used to denote the fact that the three rules are based on three distinct rays with distinct properties, i.e., either parallel to the optical axis originally, or, passing through (or headed toward) F, or, passing through the lens center. We will also discuss a second technique called oblique ray tracing using an oblique ray that does not necessarily follow the first three rules; rather it will follow Rule 4 that will be discussed. Using Rule 4 and any of the previous three rules will enable ray tracing using the oblique method. 273 Ray tracing fundamentals using regular ray tracing method. While we will be able to obtain reasonable results using a piece of graph paper and applying any two of the three rules mentioned in the previous section 10.2, it is advisable to apply all three rules, at least in the beginning. This will ensure that all three rays will pass through the image point. If one of them does not, then, clearly, there was an error in the application of one of the three rules. Example 10.12 Use regular ray tracing method to solve Example 10.5 (i.e., an object is located 12 cm to the left of a positive lens of focal length 8 cm. Where would the image be located? Is the image real?) Assume that the object is 2 cm tall. Rule 1 12 10 8 6 4 2 0 Rule 3 L F F’ Rule 0 2 10 20 30 40 50 Figure 10.10 Regular ray tracing illustration for Example 10.12. The three rays following the three rules are clearly marked in Figure 10.10. The lens is shown by a thin line at x=22 units and marked L. The optical axis is at y=6 units (arbitrarily chosen). The object is shown is using an up arrow, and is 2 units tall and located at x=10 units. The image is seen at x=46 units (i.e., 24 cm from lens center), inverted, and 4 cm tall. Answer. Example 10.13 Use regular ray tracing method to solve Example 10.5 (i.e., An object is located 12 cm to the left of a negative lens of focal length 8 cm. Where would the image be located? Is the image real?) Assume that the object is 2 cm tall. The lens is located at x=22 units; the object is at x=10 units. The image is formed about 5 cm to the left of the lens (analytical solution using thin lens equation gives 4.8 cm). The image is about four-tenths of the size of the object, agreeing with the solution from Example 10.9. 274 12 10 8 6 4 2 0 Rule 1 Rule 2 Object F’ F Rule 3 0 10 20 30 40 50 Figure 10.. 11Regular ray tracing illustration for Example 10.13. In either of the examples, the graphical (ray tracing) method agrees with the thin lens equation solution. In ray tracing problems, it is always a good idea to perform the calculations first using the thin lens equation so that one would have more confidence on the answers obtained using the ray tracing method. Ray tracing using the oblique method. The word oblique in simple terms means slanted. In other words, an oblique ray is not parallel to the optical axis. It starts from the object and is inclined at a certain angle with respect to the optical axis. In order to proceed with ray tracing of an oblique ray, one must understand a new rule, called Rule 4, first. It is not really new; rather it is modified from Rule 1 as follows. Rule 4. A parallel set of rays (not necessarily parallel to the optical axis) will, after going through a lens, pass through a point on the plane corresponding to F’ for a positive lens, or appear to emerge from a point on the plane corresponding to F’ for a negative lens. In Figure 10.12, the set of parallel rays either pass through the point P (positive lens shown on the left), or appear to emerge from point P (negative lens, right side). The point P in either case lies on a plane that is drawn perpendicular to the optical axis and passing through the focal point F’. Example 10.13 Using the oblique ray method find the image location for a 2 cm tall object, located 12 cm to the left of a positive lens of focal length 8 cm. 275 Note that we have already solved this problem using (i) the thin lens analytical method, and (ii) regular ray tracing method. Thus, we should have a good idea of where the answer lies. Parallel rays Parallel rays F F’ F’ F P P Positive Lens Negative Lens Figure 10.12 Illustration of Rule 4 that facilitates oblique ray tracing. F’ Plane 12 C 10 8 L A D 6 4 2 B E Rule 4 0 0 10 20 30 40 Rule 3 50 Figure 10.13 Illustration for Example 10.13. The lens is shown by L in Figure 10.13. The focal plane through F’ is denoted by dashed line and labeled as such although the point F’ itself lies on the optical axis and 8 cm to the right side of the lens. Procedure: Draw a ray from the top of the object (A) through the lens center D, obeying Rule 3 from before. Choose any oblique ray (i.e., ray at an angle other than zero or ninety degrees with respect to the optical axis) at point A, connect it to an arbitrary point on the lens B. AB is thus the oblique ray striking the lens. Clearly, at this point, we have no idea which direction it emerges from to the right side of the lens. In order to determine 276 this, draw a ray of light CD, such that CD is parallel to AB and passes through the lens center. Since CD passes though the lens center D, evoke Rule 3 from before and demand that it goes through undeviated, i.e., extend CD and call the intersection point of CD with the plane through F’ as E. The point E lies on the focal plane F’, and CD and AB were parallel to each other before striking the lens. Following Rule 4, the parallel set of rays AB and CD mist come together at point E on the focal plane. Connect the point B with the point E, and extend it forward. The ray BE is the new direction of the oblique ray AB after it passed the lens. The intersection of BE and the ray AD (extended) gives the location of the image. The answer matches the previous answers derived from the thin lens equation (analytical) and the regular ray tracing. Example 10.14 Using the oblique ray method, find the image location for a 2 cm tall object, located 8 cm to the left of a negative lens of focal length 8 cm. Plane through F’ Rule 4 L 12 10 8 6 4 2 0 A B D E 0 10 Rule 3 20 30 40 50 ` Figure 10.14 Illustration for Example 10.14. As before, the lens is shown as a line at x=22 units, the object is at x=10 units and the plane through F’ passes through x=14 units in Figure 10.14. Ray AD from the object follows Rule 3. Ray AB is an oblique ray that strikes the lens at the point B. Draw a line parallel to AB and passing through the lens center D. Let it intersect the plane through F’ at the point E. Connect E to B and extend it to the right of the lens which shows the path of the oblique ray to the right side of the lens. The intersection of BE and AD determines the image location, which matches with the previous two methods. The choice of ray tracing method, regular versus oblique, depends on the exact problem. With just a single lens, use either method. As we will see in the next chapter, some problems require that you use oblique method only. Exercise 10.3A Use (a) regular ray tracing and (b) oblique ray tracing methods to determine the image location for the following cases: 277 1. s=10 cm, f=4 cm, object height = 1 cm. 2. s=10 cm, f=-4 cm, object height = 1 cm. 3. s=10 cm, f=-14 cm, object height = 2 cm. 4. s=10 cm, f=14 cm, object height = 2 cm. 5. s=25 cm, f=20 cm, object height = 2 cm. 6. s=10 cm, f=10 cm, object height = 2 cm. 10.4 Imaging with Multiple Lenses (a) Analytical Method: Using the thin lens equation, one calculates the image distance for the first lens, ignoring the presence of all other lenses. Then, one determines the distance of the first image thus obtained from the second lens. This distance is now the object distance for the second lens. Ignoring all but the second lens, one uses the thin lens equation again to determine the image location from the second lens, and so on. Realize that it is conceivable that object distance could itself turn out to be negative in certain instances due to the existence of virtual, intermediate objects. The final image itself (after imaging through all the lenses sequentially) can be real or virtual, magnified or diminished, erect or inverted. It helps to draw a rough diagram to keep track of distances and their signs. Example 10.15 An object is located 10 cm to the left of a positive lens of focal length 4 cm. A second positive lens of focal length 5 cm is located 13 cm from the first lens. Using thin lens equation, determine for the final image (a) its location, (b) its characteristics (real vs. virtual, inverted vs. erect) and (c) its height compared to the object. L1 L2 13 cm 10 cm f1=4 cm f2=5 cm Figure 10.15 Illustration for Example 10.15. Consider L1 only; ignore L2 in Figure 10.15. For the first lens, s1=10 cm, f1=4 cm. The thin lens equation for s1 gives s1’ = s1 f 1 . s1 – f1 = 10 x 4 / (10-4) = 6.67 cm to the right of the first lens Lateral magnification from the first lens = -s1’/s1 = M1 = -6.67/10 = -0.67 The image from the first lens is inverted, diminished and 6.67 cm to its right. 278 For imaging through the second lens, ignore the presence of the first lens. Also observe that the image from the first lens is now the object for the second lens. s2 f2 s2’ = = = 13 – 6.67 = 6.33 cm 5 cm s2 f 2 . s2 – f2 s2 ’ = 6.33 x 5 / (6.33-5) = 23.8 cm Lateral magnification due to second lens M2= -s2’/s2 = -23.8/6.33 = -3.76 The negative sign indicates that the second lens inverts the orientation of the first image. Overall lateral magnification M = M1M2 = (-0.67) x (-3.76) = 2.52 The positive sign of M indicates that the final image is the same orientation as the original object (i.e., erect) and is 2.52X taller. The final image is located 23.8 cm to the right of the second lens or 46.8 cm from the object. Answer. Example 10.16 An object is located 10 cm to the left of a positive lens of focal length 4 cm. A second, negative lens of focal length 5 cm is located 5 cm from the first lens. Using thin lens equation, determine for the final image (a) its location, (b) its characteristics (real vs. virtual, inverted vs. erect) and (c) its height compared to the object. L1 L2 2.51 cm 10 cm 5 cm 6.67 cm Figure 10.16 Illustration for Example 10.16. Consider L1 only; ignore L2 in Figure 10.16. For the first lens, s1=10 cm, f1=4 cm. The thin lens equation for s1 gives s1’ = s1 f 1 . s1 – f1 = 10 x 4 / (10-4) = 6.67 cm to the right of the first lens Lateral magnification from the first lens = -s1’/s1 = M1 = -6.67/10 = -0.67 The image from the first lens is inverted, diminished and 6.67 cm to its right. 279 For imaging through the second lens, ignore the presence of the first lens. The object for the second lens is to the right side of it. Therefore, s2 = 5 – 6.67 = -1.67 cm, a virtual object! f2 = -5 cm s2’ = s2 f 2 . s2 – f2 s2 ’ = (-1.67) x (-5) / [-1.67-(-5)] = 2.51 cm The final image is 2.51 cm to the right of the second lens. M2 = -2.51/(-1.67) = 1.50 Overall M = M1M2 = (-0.67) x (1.50) = -1.00 The final image is real, inverted, the same size as the original object and is located 2.5 cm to the right of the second lens. Answer. Exercise 10.4A 1. An object 8 mm tall is located 90 mm to the left of a positive lens of focal length 45 mm. A second positive lens of focal length 20 mm is located 120 mm to the right of the first lens. Using thin lens equation, determine for the final image (a) its location, (b) its characteristics (real vs. virtual, inverted vs. erect) and (c) its height compared to the object. 2. An object 8 mm tall is located 90 mm to the left of a positive lens of focal length 45 mm. A second, negative lens of focal length 20 mm is located 120 mm to the right of the first lens. Using thin lens equation, determine for the final image (a) its location, (b) its characteristics (real vs. virtual, inverted vs. erect) and (c) its height compared to the object. 3. An object 8 mm tall is located 90 mm to the left of a negative lens of focal length 45 mm. A second positive lens of focal length 40 mm is located 120 mm to the right of the first lens. Using thin lens equation, determine for the final image (a) its location, (b) its characteristics (real vs. virtual, inverted vs. erect) and (c) its height compared to the object. 4. An object 8 mm tall is located 6 cm to the left of a positive lens of focal length 8 cm. A second positive lens of focal length 3 cm is located 10 cm to the right of the first lens. Using thin lens equation, determine for the final image (a) its location, (b) its characteristics (real vs. virtual, inverted vs. erect) and (c) its height compared to the object. 5. An object 1 cm tall is placed 10 cm to the left of a positive lens of focal length 3.5 cm. A second positive lens is placed 17 cm to the right of the first lens. Its focal length is 4 280 cm. A third positive lens of focal length 5 cm is placed 12 cm to the right of the second lens. Using thin lens equation, determine for the final image (a) its location, (b) its characteristics (real vs. virtual, inverted vs. erect) and (c) its height compared to the object. 6. An object 1 cm tall is placed 10 cm to the left of a negative lens of focal length 3.5 cm. A second positive lens is placed 17 cm to the right of the first lens. Its focal length is 4 cm. A third positive lens of focal length 5 cm is placed 12 cm to the right of the second lens. Using thin lens equation, determine for the final image (a) its location, (b) its characteristics (real vs. virtual, inverted vs. erect) and (c) its height compared to the object. (b) Ray tracing with multiple lenses Using either the regular ray tracing or the oblique method one can apply the rules and graphically trace the rays through each lens in succession. Again, as we did with the thin lens equation, we can ignore the presence of all other lenses when tracing the rays through a particular lens. An example each of the regular ray tracing and oblique method is provided for a two lens system. The procedure is extendable to three or more lenses as well. Example 10.17 An object is located 10 cm to the left of a positive lens of focal length 4 cm. A second positive lens of focal length 5 cm is located 13 cm from the first lens. Using regular ray tracing method, determine for the final image location. Assume that the object is 2 cm tall. Rule 1, 2L2 12 10 8 6 Rule 3, L1 4 2 0 0 L1 Rule 3, 2L2 L2 I1 Rule 1, 1L1 10 20 30 40 50 60 Figure 10.17 Illustration for Example 10.17. For the sake of clarity, only two of the three rules of regular ray tracing method were used for either lens. For the first lens, Rules 1 and 3 were used to obtain an intermediate 281 image, which is inverted and real, and located slightly less than 7 cm to the right of the first lens. This is shown by the label I1 in Figure 10.17. Using Rules 1 and 3 for the second lens, the intermediate image was further imaged through the second lens. The final image is about 5 units tall, real and erect and located about 24 cm from the second lens and to its right. Example 10.18 An object is located 10 cm to the left of a positive lens of focal length 4 cm. A second positive lens of focal length 5 cm is located 13 cm from the first lens. Using oblique ray tracing method for the second lens, determine for the final image location. Assume that the object is 2 cm tall. Oblique ray from L2 12 L1 10 Rule 3, L1 C 8 6 4 Rule2, L1 2 0 0 10 L2 F2 plane Rule1, L2 D E 20 G 30 40 50 60 Figure 10.18 Illustration for Example 10.18. In Figure 10.18, the imaging for first lens is done using Rules 2 and 3. For example, the ray following Rule 2 through the first lens becomes parallel to the right of L1, and when it crosses L2, it follows Rule 1 for L2. Unlike the previous example, we do not attempt to find the location of the intermediate image. Instead, the ray that follows Rule 3 from L1 is used as an oblique ray to be imaged through L2. This ray intersects L2 at the point E. To facilitate this, a ray CD is drawn parallel to it, however, passing through the center of the second lens. This line intersects the F2 plane of L2 at point G. Connect E and G and extend it. The final image location is shown in the graph, and it agrees with the regular ray tracing method answer of the previous example. Exercise 10.4B Perform ray tracing using first (a) regular method for all lenses, and (b) regular method for the first lens, followed by oblique method for lenses other than the very first, for the following cases. Determine the final image location and characteristics (size, orientation, real vs. virtual) for each problem. 282 Problem 1 2 3 4 5 6 s1 12 cm 12 cm 12 cm 12 cm 5 cm 5 cm f1 5 cm -5 cm 5 cm -5 cm 10 cm -10 cm L1L2 20 cm 20 cm 20 cm 20 cm 20 cm 20 cm f2 4 cm 4 cm -4 cm -4 cm 4 cm 4 cm L2L3 N/A N/A N/A N/A 10 cm 10 cm f3 N/A N/A N/A N/A 7 cm -7 cm In this chapter, we considered cases where the thin lens approximations hold. We used objects that were much smaller in height. We had rays emerging from the head of the object that were essentially paraxial. We represented the lens itself by a line with no thickness. The diameter of the lens was much larger in comparison to the object height, so that during ray tracing rays following Rules 1 and 2, the ray will be incident inside the rim of the lens. It should be stressed that while we discussed about inverted images, the word “inverted” essentially means cases where there is a lateral inversion: top of the object is the bottom of the image, and vice versa, and left side of the object is the right side of the image, and vice versa, and so on. We did not consider aspects of how bright or clear the image would be. For the cases discussed thus far, the rim of the lens controls the amount of light that travels from the object to the lens. In the next chapter we will look at cases where apertures are used between the object and a lens or between lenses to accomplish this. Also, we assumed that the image will be formed at a unique location and would appear uniform in clarity and intensity. This arises from small angle approximation used in the governing equations. We will also see in the next chapter that even with monochromatic light the image undergoes aberrations through a lens. Aberrations were ignored in this chapter. 10.5 Suggested Laboratory Experiments Finding focal length of a positive lens: • Follow safety precautions for laser operation. • Set up an optical axis. • Expand the laser beam using a negative lens. • Place a suitable object in the path of the expanded beam. This object must have unique characteristics so that lateral inversion in terms of left and right and top and bottom can be realized in the image formed by a positive lens. (An example of an object may be frosted glass with a letter such as F or R written on it with a fine black marker). • Take a positive lens. Find its approximate focal length by focusing the ceiling light on to the floor. Measure the distance from the lens to the image of the ceiling light on the floor. This is the approximate focal length of the lens. • Place the lens at various distances from the object starting from 3X the focal length, 2.5X the focal length, 2X the focal length, 1.8X the focal length and 1.5X the focal length. Obtain the image on a screen in each case. Measure the image distance in each case as well as the image height. Measure the object height once. 283 • Calculate the lateral magnification in each case and compare with predictions. Calculate the focal length using the thin lens equation. Find the average of the five values. For an object distance of 1.67X the average focal length calculated, use the thin lens equation to predict the image distance. Verify with experimentation if the answer is correct. Imaging with multiple lenses: • Use the same object as above. • Set up an optical axis. • Obtain two positive lenses; determine their approximate focal lengths using the ceiling light imaging technique described above. • Image the object using each lens separately; obtain at least three sets of s and s’ values; calculate the focal length of each lens using the thin lens equation and average the focal length for each lens. • Using the average focal length of each positive lens thus calculated, determine two set ups of s1, and d (the distance between the two lenses) values for which a real image can be obtained. Use the thin lens equation. Of these one should provide an enlarged image of the object and the other, a diminished image. • Verify via experimentation if the images for the two cases obtained are in the same locations as predicted and that the magnifications are the same as predicted. • • • • • • • • Chapter Summary Flat surfaces such as ones made of glass bring about displacement of light through refraction as it is transmitted and reflect a small portion through Fresnel reflections. A curved surface permanently changes the direction of light as light passes through it due to refraction. Two curved surfaces brought together result in a lens. The lens maker’s equation relates the radii of curvature of the two surfaces to a net focal length of the composite. The inverse of the focal length in meters is called the focal power and measured in units of Diopter. A thin lens is one whose thickness is much smaller than its diameter or the radii of curvature that make up the two surfaces of the lens. Lenses thicker in the middle have positive focal lengths, and hence called positive lenses. Lenses thinner in the middle have negative focal lengths, and hence called negative lenses. An image is real if it can be captured on a screen, virtual otherwise. A single negative lens can produce only virtual images. A single positive lens can produce real or virtual images. A single negative lens always produces an erect, virtual image. A single positive lens can produce an erect, virtual image or a real, inverted image, depending on the focal length and the object distance from the lens. 284 • • • • • The thin lens equation is a relationship among the object distance from the lens (s), the image distance from the lens (s’) and the focal length of the lens (f). We always consider light to be traveling from left to right, unless otherwise specified. The thin lens equation can be applied sequentially for each lens in a train of lenses. This enables prediction of the location and characteristics of intermediate images as well as the final image. Ray tracing provides a geometrical/graphical procedure for determining the image location and characteristics. Applying the rules of ray tracing, one can perform a regular ray tracing or oblique ray tracing. In either case, at least two rays starting from the head of the object is traced through a lens or a system of lenses. The intersection point of those rays gives the location of the image. The equations and geometrical procedures provided in this chapter hold for thin lenses and for paraxial rays. They do not provide information on the intensity of the image or the uniformity of image clarity. Those aspects will be discussed in the next chapter. End of Chapter Exercises Complete the following table. Determine in each case, if the lens is positive or negative Problem Radius of Radius of f, focal length Focal power Number curvature of curvature of (include sign (include sign first surface second surface and units.) and units.) 1 40 mm, 20 mm, concave concave 2 5”, convex 4”, concave 3 10 cm, concave 8 cm, convex 4 12 cm, convex 11 cm, convex Fill in the blanks Number s s’ f Lens Image M Image Object Image (real or (erect or height Height virtual) inverted) 5 20 10 Positive 1 mm mm mm 6 12 10 Positive 2 mm mm mm 7 20 10 Negative 1 cm cm cm 8 20 40 Positive 1 cm cm cm Use (a) regular ray tracing and (b) oblique ray tracing methods to determine the image location for the following cases: 9. s=10 cm, f=-7 cm, object height = 2 cm. 10. s=5 cm, f=14 cm, object height = 2 cm. 11. s=25 cm, f=-10 cm, object height = 2 cm. 12. s=30 cm, f=10 cm, object height = 2 cm. 285 13. An object 5 mm tall is located 80 mm to the left of a negative lens of focal length 55 mm. A second positive lens of focal length 30 mm is located 150 mm to the right of the first lens. Using thin lens equation, determine for the final image (a) its location, (b) its characteristics (real vs. virtual, inverted vs. erect) and (c) its height compared to the object. 14. An object 10 mm tall is located 8 cm to the left of a positive lens of focal length 5 cm. A second positive lens of focal length 3 cm is located 5 cm to the right of the first lens. Using thin lens equation, determine for the final image (a) its location, (b) its characteristics (real vs. virtual, inverted vs. erect) and (c) its height compared to the object. Perform ray tracing using the regular method for the following cases. Determine the final image location and characteristics (size, orientation, real vs. virtual) for each problem. Problem s1 f1 L1L2 f2 L2L3 f3 15 8 cm 5 cm 20 cm 4 cm N/A N/A 16 12 cm -10 cm 20 cm 4 cm 10 cm -3 cm 17 12 cm -4 cm 30 cm 5 cm N/A N/A 18 10 cm -5 cm 25 cm -7 cm 25 cm 4 cm Group Discussion Questions These require the students to form small groups of three or four and discuss each question. Some questions may require you to do a search on the Internet and find the answers. 1. Show by ray tracing what happens if the object is located at the primary focal point of a positive lens. 2. Show by ray tracing what happens if the object is located at the secondary focal point of a negative lens. 3. What is an application when one uses a single positive lens with the object located within its primary focal point? 4. What would result if a positive lens and a negative lens were cemented together with transparent glue that does not affect the optical transmission? Assume that the surfaces that are glued together match perfectly in curvature and diameter. 5. If, let’s say that you are trying to obtain a real image of an object. Other than fine tuning the location of the lens and the screen, what else can you do to produce the sharpest image possible on the screen? References Sears, F.W, M. W. Zemanksy and H.D. Young. “College Physics.” Reading: AddisonWesley Publishing Company, 1991. 2 Jenkins,F. and H. White. “Fundamentals of Optics.” Singapore: McGraw Hill, 1981. 3 Welford, T. “Optics.” Oxford: Oxford University Press, 1988. 4 Bureau of Naval Personnel. “Basic Optics and Optical Instruments.” New York: Dover Publications, Inc., 1969. 1 286 Chapter 11 Geometric Optics: Part 2 11.1 Imaging using Apertures 11.2 Optical Instruments 11.3 Aberrations 11.4 Suggested Laboratory Experiments Chapter Summary End of Chapter Exercises Group Discussion Questions Chapter Objectives After completing the chapter, the student should be able to (i) determine the limiting aperture, entrance and exit pupils in a system (ii) ray trace using chief and marginal rays. (iii) describe how optical instruments work (iv) explain types of aberrations caused by a lens Key words: front stop, rear stop, middle stop, limiting aperture, stop, entrance pupil, exit pupil, chief ray, marginal ray, angular magnification, aberration, chromatic aberration, spherical aberration, coma, distortion. 11.1 Imaging using Apertures1-3 In the previous chapter, we had rays close to the optical axis. We assumed that rays from an object will strike the lens surface and did not pay attention to the diameter of the lens. Both the analytical method and ray tracing gave us accurate description of the image location and characteristics. In reality the rim of the lens(es) dictates the amount of light successfully passing from the object through the lens system to the image. Also, the active diameter of a lens is approximately 90% of the actual lens diameter since the edges thin out. In many optical systems, one wants to control this amount of light by placing a physical aperture or stop in the system so that one is not dependent on the lens rim for controlling the light intensity at the image plane. A stop or aperture is a mechanical device that restricts the amount of light that passes through it. An aperture wheel has a set of different diameter openings so that one could select discrete aperture diameters such as 16 mm, 12mm, etc. A mechanical iris serves the same purpose except that one could obtain different aperture diameters in a continuous manner from a minimum up to a maximum value. An aperture or iris is usually painted black so that the amount of light that falls outside the aperture area is absorbed rather than reflected. Let us work with a single lens and a single stop to begin with. Refer to Figure 11.1. The stop, when placed in front of the lens is known as a front stop. If it is placed behind the lens, then it is called a rear stop. If we have two lenses and the aperture is placed 287 between the lenses, then it is called the middle stop. If the stop is placed close to the image plane, then it is called the field stop. In any case, the goal of using a stop is to restrict the amount of light. Stop Lens Lens Stop Lens Stop Lens Figure 11.1 Illustration of front stop (left), rear stop (middle) and middle stop (right). Stops can be used with positive or negative lenses. When a stop is in place, while the thin lens equation will still provide us details of the image location and characteristics, one can not use the ray tracing rules that we learned from the previous chapter arbitrarily. This is because one or more of the rays following the rules may physically collide with the stop. When that happens, the ray cannot go through! Therefore, we develop as different set of procedures for ray tracing in this chapter, valid when a stop is used. We need to realize that certain modifications should be made to the established ray tracing procedures from the previous chapter. Light travels from left to right as always. o We will not use Rule 1 or Rule 2. Rule 3 still holds, i.e., light passing through the center of the lens will proceed undeviated. o In the problems in this chapter, there will be a stop. o One needs to image all elements of the system through the optics to the left of it (i.e., preceding optics). That element, the image of which subtends the smallest angle at the axial object point (i.e., the foot of the object which stands on the optical axis), is indeed the limiting aperture of the system. o Once the aperture has been correctly identified, either using the thin lens equation and trigonometry, or using ray tracing methods from the previous chapter, one needs to determine the entrance pupil and the exit pupil of the system. The entrance pupil is the image of the limiting aperture through the optics to the left of it. The exit pupil is the image of the limiting aperture through the optics to the right of it (i.e., succeeding optics). o Knowing the location and size of the entrance pupil, the limiting aperture and the exit pupil, the new ray tracing procedure involves starting from the top of the object, and systematically proceeding through these three elements. A variety of examples are provided in the following pages. Let us consider a front stop located in front of a single positive lens, as shown in Figure 11.2. The stop is the first element as light travels from left to right, and the lens is the second. One needs to image each of these elements through optics that are to the left of it. This will let us know which of these elements indeed controls the amount of light through the system. 288 Example 11.1 Determine for both cases below, which element controls the amount of light from the axial object point “x.” Case A Case B Stop Stop 20 mm x 40 mm 40 mm Positive Lens f=25 mm 40 mm x 40 mm 75 mm 30 mm Optical Axis Positive Lens f=45 mm 75 mm Figure 11.2 Illustration of front stop and lens with different diameters and locations. In Figure 11.2, x denotes an axial object point, i.e., a point object which is located on the optical axis of a lens. Two situations are shown. While the stop and the lens are located at the same distance in both cases (figures on the left and right), the opening and lens diameter are different. The question to answer in either case is: which of the two elements, the stop or the rim of the lens, controls the amount of light originating from the object and proceeding to the right? In order to understand this, construct triangles based on the dimensions shown above. Since the stop does not have a lens to the left of it, it will image on to itself. Since the lens does not have another lens to the left of it, it will image on to itself. Case A x θs 10 mm x 40 mm θL Case B 20 mm Optical axis x θs 75 mm 20 mm x θL 40 mm 15 75 mm Figure 11.3A Illustration of front stop and lens with different diameters and locations . The radii of the lens and stop as well as the distances from the objects are shown in Figure 11.3A. The angle that the radius of the stop subtends at x is denoted by θs and the angle the radius of the lens subtends at x is denoted by θL. For Case A, θs = tan-1(10/40) = 14.0o 289 θL= tan-1(20/75) = 14.9o For Case B, θs = tan-1(20/40) = 26.6o θL= tan-1(15/75) =11.3o In either case, one should look at the smaller of the two angles, θs and θL. The lower of the two angles will determine the element that controls the amount of light from the object point into the system. For Case A, the stop restricts the amount of light. The stop is called the limiting aperture now. For Case B, the lens restricts the amount of light. The lens is called the limiting aperture now. The phrase “limiting aperture,” as defined earlier, is used to denote that physical element which restricts or controls the amount of light from an axial object point through an optical system. While the word aperture means an opening, we will use the word “stop” to denote an opening and the phrase “limiting aperture” to denote the element that restricts the amount of light. Depending on the diameters of the elements and their distances from the object, a lens or a stop can be the “limiting aperture” as aptly demonstrated above. Also, when the distances are changed, the limiting aperture may also change. The student is encouraged to rework Case A where the stop is located at 30 mm (instead of 40 mm) from the object, while the lens is still at 75 mm from the object. How about a negative lens? So long as a front stop is used, the calculations shown in the example above still hold and the answers will be the same. For either type of lens, for Case A, the stop is the limiting aperture. The image of the stop obtained using a lens in front of it (i.e., to its left) is the stop itself, since there is no lens in front of it. The stop is now also called the entrance pupil for this particular case. In other words, if you were to place your eye to the left of the stop and view the system from the object axial point x, you will see the stop opening, not the lens. This is the view “as the axial object looks at the system following it.” Hence the stop here serves as the entrance pupil. For case B, the lens is the limiting aperture. The image of the lens obtained using a lens in front of it (i.e., to its left) is the original lens itself since there no lens in front of the original lens. For case B, the lens is called the entrance pupil. In other words, if you were to place your eye to the left of the stop and view the system from the object axial point x, you will see the lens in its entirety. So far we have looked at two new terminology, limiting aperture and entrance pupil. To reiterate, the limiting aperture is that physical element that controls the amount of light in an optical system and the entrance pupil is the image of the aperture obtained through optics to the left of it. What happens to the light after it passes successfully through he entrance pupil and the aperture? It has to leave the system. The light exits the system through the exit pupil which is obtained by imaging the limiting aperture using optics following it. In general, the entrance and exit pupils need not be a physical element. 290 Example 11.2 Determine the exit pupils for Example 11.1, case A only. Case A Case B Stop 20 mm x 40 mm Stop 40 mm Positive Lens f=25 mm 40 mm x 40 mm 75 mm Figure 11.3B Exit pupil determination 30 mm Optical Axis Positive Lens f=45 mm 75 mm -Example 11.2 For case A, we know that the stop is the limiting aperture and for case B the lens is the limiting aperture. Exit pupil determination involves imaging the limiting aperture through the optics that follows it. For case A, we have to image the stop through the lens. For Case B, the lens is the last element in the system and there is no additional lens to the right of the original lens to image through. For case B, the lens is the exit pupil To image the stop through the lens for Case A, use either the thin lens equation or ray tracing. The stop is symmetrical about the optical axis. Therefore, only one half (example, the top half) of the stop opening of radius 10 mm needs to be considered. Method 1. Using thin lens equation: For this, s=35 mm, f=25 mm. Thus, s’ = 35 x 25 (35-25) = 87.5 mm and M = -87.5/35 = -2.5 Diameter of the image = 2.5 x 20 mm = 50 mm. Thus, the exit pupil is a circular opening of diameter 50 mm, located 87.5 mm to the right of the lens. The image is inverted, however, since the stop is a circular opening one 291 cannot tell the difference of the orientation of the image. However, it is important to keep track of this in future. Method 2: Ray tracing (regular method). Refer to Figure 11.4. Stop 60 50 40 30 20 10 0 Lens Exit Pupil B’ A B 0 A’ 50 100 Figure 11.4 Exit pupil determinationfor Example 11.2. Now that we know how to determine the limiting aperture, entrance pupil and exit pupil of a front stop, what do we do with this knowledge? In an optical system with a stop, a light ray that passes through the center of the entrance pupil, the center of the limiting aperture and the center of the exit pupil has a special name. It is called the chief ray. On a two-dimensional plane, the rays that pass through the extremities of the entrance pupil, the limiting aperture and the exit pupil are called marginal rays. While only one marginal ray is adequate to predict image location via ray tracing, two marginal rays are often used, one starting from the object and first passing through the top of the entrance pupil, and the other starting from the top of the object and passing through the bottom of the entrance pupil. The following example illustrates ray tracing using the chief and marginal rays for a system with a limiting aperture, entrance and exit pupils. For a system with no stop, the lens serves as the limiting aperture and takes on the role of the entrance pupil, the limiting aperture and exit pupil. For this case, the ray tracing rules of the previous chapters will suffice. However, once there is a stop one has to rigorously follow the steps outlined in the example and others to come. Example 11.3 For Example 11.2, assume that an object 5 mm tall is located standing on the point x. For the situation in Case A, determine the image location via ray tracing. Use the information from Example 11.2 for the exit pupil location. 292 We know that the stop is the entrance pupil since it is the first element facing the object. The stop is the limiting aperture as previously determined. We will use the exit pupil information from Example 11.2. Refer to Figure 11.5. Object Limiting Aperture & Ent. Pupil Marginal AA’ Lens 60 50 40 30 20 10 0 A Final Image Exit Pupil Marginal BB’ B’ C D G B A’ E 0 50 100 150 Figure 11.5 Ray Tracing for Example 11.3. Once the limiting aperture, entrance pupil, exit pupil, the object and the lens are plotted on the graph paper, the procedure is as follows. • • • Start from the top of the object. Draw the chief ray which heads towards the center, D of the entrance pupil and limiting aperture. Extend it until it meets the lens at E. At the lens, it is refracted. Therefore, stop the ray at the lens. The ray will then head toward the center of the exit pupil, G. Start from the top of the object. Draw the marginal ray toward A, extend it to meet the lens at C. At the lens, it will be refracted to head toward A’. Start from the top of the object. Draw the marginal ray toward B, extend it to meet the lens at E. At the lens, it will be refracted to head toward B’. The chief ray and the two marginal rays will meet at a common point where the image is drawn. Verify the location using thin lens equation. Let us now look at a single negative lens with a front stop. The entrance pupil is the front stop. The exit pupil is the image of the stop through the negative lens, as shown in the example below. Example 11.4 Locate graphically the location of the exit pupil. The axial object point is denoted by x. 293 Stop 20 mm 60 mm x Optical Axis 40 mm Negative Lens f=-25 mm 35 mm Figure 11.6 Illustration for Example 11.4. From Example 11.1, we know that the stop is the limiting aperture for the axial object point. 60 50 40 30 20 10 0 Limiting Aperture and Ent. Pupil Exit Pupil A F’ A’ B 0 10 Negative Lens B’ 20 30 40 50 Figure 11.7 Determination of exit pupil forExample 11.4. Since the limiting aperture is the first element of the system, it is the entrance pupil as well. The exit pupil is obtained by imaging the points A and B of the aperture through the negative lens. Thus, we obtain points A’ and B’, respectively. A’B’ thus represents the exit pupil. Example 11.5 For Example 11.4, assume that an object 5 mm tall is stands on the point x. Via ray tracing using chief and marginal rays, locate the image. Here the location of the exit pupil is redrawn from the previous example since the limiting aperture location, opening and lens focal length are the same as before. The chief ray is shown using the solid line. It starts from the top of the object and proceeds toward the center of the entrance pupil and limiting aperture. The ray is extended to strike the lens. At point D, where it strikes the lens, it is joined (backward) with point C, the center of the exit pupil. 294 Object Limiting Aperture & Ent. Pupil 60 50 40 30 20 10 0 A Exit Pupil Neg. Lens E A’ C B’ D B 0 20 G 40 60 80 Figure 11.8 Ray tracing forExample 11.5. The marginal rays are shown by dashed lines for clarity. The marginal ray through A strikes the top of the entrance pupil. Extend it until it strikes the lens at point E. Do not be concerned that it is hitting the forbidden area of the exit pupil above the point A’. At point E, connect it (backward) to point A’ and extend it backward further. The marginal ray through B’ is similarly extended beyond B to point G on the lens. Again do not be concerned that the ray strikes a point below B’. Join G (backward) to point B’ and extend it backward further. The common intersection of the three lines, will determine image location. Thus far, we have looked at a front stop. Next, we consider a rear stop. In this case the lens is the first element. Let us consider two cases, as before. Remember, if we have to image the rear stop through the lens, we have to temporarily assume that light travels from right to left. This is the only exception to the rule that light travels from left to right. Example 11.6 Determine which element in each case is the limiting aperture. Lens diameter=50 mm; stop diameter = 20 mm for either case (Figure 11.9). Case A: The two elements are the lens and the stop. Both need to be imaged through the preceding optics. For the lens, there is no preceding optic. Therefore, the triangular construction consisting of the axial object x, the center of the lens and the top of the lens is simple, and is shown in Figure 11.10. θL = tan-1(25/30) = 39.8o 295 Case A Lens, f=20 mm Case B Lens, f=20 mm Stop x Axial Object 30 mm Stop x Axial Object 30 mm 40 mm 100 mm Figure 11.9 Illustration for Example 11.6. θL 25 mm 30 mm Figure 11.10 Illustration for Example 11.6, case A. The image of the rear stop through the lens is obtained as follows using the thin lens equation (remember light is traveling now from right to left): and s = 10 mm, f = 20 mm. s’ = 10 x 20 / [10-20] = - 20 mm M = -(-20)/10 = 2 The image of the stop through the lens is virtual, 2X larger and located 20 mm to the right of the lens, as shown below: Lens Stop Image of stop x 40 mm 30 mm 10 mm 20 mm Figure 11.11 Illustration for Example 11.6, case A. The triangle consisting of the axial object point x, the top and the center of the stop’s image will look as shown in Figure 11.12. θs = tan-1 (20/50) = 21.8o 296 Between the two angles, this angle is smaller. Therefore, the stop is the limiting aperture. It also serves as the exit pupil. Its image through he preceding optics, located at 50 mm from the object and diameter 40 mm, is the exit pupil. We will use this information in Example 11.7. θs 20 mm x 50 mm Figure 11.12 Illustration for Example 11.6, case A Case B: The two elements are the lens and the stop. Both need to be imaged through the preceding optics. For the lens, there is no preceding optic. Therefore, the triangular construction consisting of the axial object x, the center of the lens and the top of the lens is simple, and is shown below. θL 25 mm 30 mm Figure 11.13 Illustration for Example 11.6, case B θL = tan-1(25/30) = 39.8o, as before The image of the rear stop through the lens is obtained as follows using the thin lens equation (remember light is traveling now from right to left): s = 70 mm, f = 20 mm. s’ = 70 x 20 / [70-20] = 28 mm Lateral magnification, M = -28/70 = -0.4 Thus the image of the stop through the lens is located 28 mm to the left of the lens and is only 8 mm in diameter (20 x 0.4). This places the image only 2 mm to the right of x. The triangle consisting of the axial object point x, the top and the center of the stop’s image will look as shown in Figure 11.14. θs = tan-1 (4/2) = 63.4o Clearly, the first angle is smaller. Therefore, the lens is the limiting aperture. 297 θs 4 mm x 2 mm Figure 11.14 Illustration for Example 11.6, case B Example 11.7. For Example 11.6, case A, assume that an object 5 mm in height stands on the point denoted by x. Use ray tracing via chief and marginal rays to determine the image location. Let us use the information that we have from the previous example: (i) the rear stop serves as the limiting aperture and exit pupil, and (ii) the entrance pupil is located at 50 mm from the object and has a diameter of 40 mm. Object Limiting Aperture & Lens Exit Pupil Ent. Pupil Final Image 60 A 50 C A’ 40 D K 30 J G 20 B’ B 10 0 0 20 40 60 80 100 Figure 11.15 Illustration for Example 11.7 The object, lens, entrance pupil and exit pupil locations are drawn to scale in Figure 11.15. 298 Chief ray: Start from the top of the object. Draw a ray heading toward the center of the entrance pupil K. Draw a solid line to the point D where the ray meets the lens and a dashed line to the right of it. The dashed line merely denotes that light actually does not follow the path since the lens will refract it at point D. Join the point D with the point J, the center of the exit pupil and continue the ray forward. Marginal ray A: Start from the top of the object. Draw the marginal ray heading toward the point A, the top of the entrance pupil. However, the lens will refract it at point C. Join C with A’, the top of the exit pupil and extend the marginal ray forward. Marginal ray B: Start from the top of the object. Draw the marginal ray heading toward the point B, the bottom of the entrance pupil. However, the lens will refract it at point G. Join G with B’, the top of the exit pupil and extend the marginal ray forward. The three rays will meet at a common point that gives the image location. Let us consider a rear stop with a negative lens. As usual one should determine which element, the lens or the stop is indeed the limiting aperture. Example 11.8 Determine which element in each case is the limiting aperture. Lens diameter=50 mm; stop diameter = 20 mm for either case. Case A Lens, f=-20 mm Case B Lens, f=-20 mm Stop x Axial Object 30 mm Stop x Axial Object 40 mm 30 mm 100 mm Figure 11.16 Illustration for Example 11.8. Since we have considered three other examples, we will skip some of the intermediate steps. The student is urged to work out to verify if the following answers are obtained. Case A: θL = tan-1(25/30) = 39.8o Image of the rear stop through the negative lens will be at 6.7 mm to the right of the lens with an image diameter of 13.3 mm. θs = tan-1(6.67/36.7) = 10.3o 299 This angle being smaller, the stop is the limiting aperture and its image, as determined above, is the exit pupil. Case B: θL = tan-1(25/30) = 39.8o Image of the rear stop through the negative lens will be at 15.6 mm to the right of the lens with an image diameter of 4.44 mm. θs = tan-1(2.22/45.6) = 2.8o This angle being smaller, the stop is the limiting aperture. Example 11.9 For Example 11.8, case A, assume that a 6 mm object stands on the point marked x. Using chief ray and the marginal rays, perform ray tracing. The elements to be drawn are (i) the object 6 mm tall, (ii) the negative lens 50 mm in diameter and 30 mm to the right of the object, (iii) the limiting aperture (and exit pupil) 20 mm in diameter and 40 mm from the object, and (iv) the entrance pupil, 36.7 mm from the object and 13.3 mm in diameter, as shown in Figure 11.17: Object Image Neg. Lens Ent. Pupil Exit Pupil 60 50 C 40 A’ A D 30 20 G H B B’ 10 0 0 10 20 30 40 50 . Figure 11.17 Illustration for Example 11.9. 300 Chief ray: From the top of the object, proceed toward G, the center of the entrance pupil. However, it is intercepted by point D on the lens. The ray will be refracted such that it will head toward H, the center of the exit pupil. Since the lens is negative, extend HD backward. This is shown by dotted line for clarity. For the sake of clarity, only the top marginal ray is shown. Start from the top of the object. Proceed toward A, the top of the entrance pupil. The lens will intercept this marginal ray at point C. Connect C to A’, the top of the exit pupil. (Don’t be concerned that the ray is falling outside the entrance pupil opening since it is not a physical stop.) Extend A’C backward. This is shown by a dotted line. The intersection of the dotted lines from the chief ray and the marginal ray established the location of the virtual image. A similar procedure can be used to track the marginal ray through B. Exercise 11.1A In each of the cases below, locate the limiting aperture either using trigonometry or ray tracing. Then, using ray tracing via the chief ray and the two marginal rays, locate the image. Problem Object Height Object to lens distance Lens focal length Lens diameter 1 5 mm 7 cm 5 cm 5 cm 2 3 5 mm 5 mm 7 cm 7 cm 5 cm -5 cm 5 cm 5 cm 4 5 5 mm 5 mm 7 cm 12 cm -5 cm 5 cm 5 cm 5 cm 6 6 mm 12 cm -5 cm 5 cm 7 8 6 mm 6 mm 7 cm 7 cm -5 cm 10 cm 5 cm 5 cm 9 10 6 mm 6 mm 7 cm 7 cm -10 cm 10 cm 6 cm 6 cm Stop distance from lens/Type of stop 1 cm/Front 1 cm/Rear 2 cm/Front 1 cm/Rear 7 cm/Front 7 cm/Front 7 cm/Rear 5 cm/Front 5 cm/Rear 5 cm/Rear Stop diameter 1 cm 1 cm 1 cm 1 cm 1 cm 2 cm 2 cm 2 cm 2 cm 2 cm We devoted many pages to a single lens, single stop system, encompassing front stops, rear stops, positive lenses and negative lenses, and following the ray tracing principles developed for an entrance pupil, limiting aperture, and exit pupil system. Let us examine a two-lens system with a middle stop. As you can imagine the complexity of ray tracing process increases tremendously. We will work out three examples here. In each case, the 301 first step is the correct identification of the limiting aperture, followed by evaluation of the entrance and exit pupil locations and sizes. The last step involves systematic ray tracing of the chief ray and the two marginal rays, through the entire system. Example 11.10 For the two-lens, middle stop system shown in Figure 11.18, perform ray tracing using the chief ray and the two marginal rays, through the following system: (The figure is not to scale) f1=10 mm L1, diameter=20 mm f2=25 mm L2, diameter=20 mm STOP Object 3 mm high dia.=10mm 35 mm 6 mm 14 mm Figure 11.18 Illustration for Example 11.10. Step 1: Identification of the system limiting aperture. Image each element through the optics to the left of it. Since this involves ray tracing to image the middle stop and L2 through L1, the graph below to determine the aperture looks busy. Refer to Figure 11.19. The image of L1 is itself since there is no optics to the left. When the foot of the object is joined to the bottom of L1 (shown by a dotted line), an angle marked θL1 is obtained. The image of the stop AB is virtual, and by regular ray tracing we determined the image as A’B’. Connecting the foot of the object to the point B’, we obtain an angle which we mark as θs, “s” representing the image of the stop through L1. Finally, the lens L2 (CD) is imaged through L1 to obtain its real, inverted image D’C’. When the foot of the object is connected to C’, we obtain an angle, which we will call as θL2. Which of three angles is the smallest: θs, θL1, or θL2? By examination we see that θs is the smallest, making the middle stop the limiting aperture for the entire system. 302 60 Image of L2 Image of stop 50 D’ L1 Stop A’ 40 L2 C A θL1 B θs 30 20 C’ D θL2 10 B’ 0 0 10 20 30 40 50 60 Figure 11.19 Determination of aperture for Example 11.10. In the two other examples to follow, we will use the thin lens equation in combination with trigonometric operations to determine the smallest angle, and thus the aperture. Knowing that the stop is the limiting aperture, imaging the stop through L1 would give us the entrance pupil, whereas imaging the stop through L2 would give us the exit pupil. We have already done the latter, giving us A’B’ as the exit pupil. Imaging on a separate piece of graph paper or by using the thin lens equation, we will obtain the entrance pupil A”B”. The object, the lenses, the stop, the entrance and exit pupils are marked on the graph below to enable ray tracing of the chief and marginal rays, as follows. Step2. Ray Tracing: Refer to Figure 11.20. Chief ray: Proceed from the top of the object toward N, the center of the entrance pupil. However, the lens L1 intercepts the chief ray at point E and refracts it such that it will head toward P, the center of the limiting aperture. Extend EP toward the lens L2 to intersect at point K. The ray will get refracted at K such that the points H (the center of the exit pupil) and K are connected and extended forward. The ray from HJK is the chief ray and will proceed to the right of L2. Marginal ray through A: Connect the top of the object to A’. L1 will intercept it at C. Connect C with A, the top of the limiting aperture. Extend to meet L2 at L. Connect L with A”, the top of the exit pupil. The ray A”DL is the direction of the marginal ray through A and it will proceed to the right of L2. 303 Marginal ray through B: Connect the top of the object to B’. L1 will intercept it at M. Connect M with B, the bottom of the limiting aperture. Extend to meet L2 at G. L2 will refract the ray such that the point G should be joined with B”, the bottom of the exit pupil. The ray B”KG is the marginal ray through B. The chief ray HJK, and the marginal rays A”DL and B”KG will go through a common point defining the image location. Verify using the thin lens equation that the image location and height agree with ray tracing. Exit Pupil Limiting Aperture Ent. Pupil L1 L2 40 35 A A” 30 C A’ D 25 A E G H 20 P J K 15 N K L B M 10 B” B’ 5 0 0 10 20 30 40 50 60 Figure 11.20 Ray tracing for Example 11.10. Example 11.11 For the two-lens, middle stop system shown in Figure 11.21, perform ray tracing using the chief ray and the two marginal rays, through the following system: Step 1. Identification of the aperture Instead of ray tracing to locate the limiting aperture, we will use thin lens equation and trigonometry to identify which of the three angles is the smallest. 304 Element 1, L1 will image on to itself since it is the first element of the system. f1=5 mm L1, diameter=10 mm f2=15 mm L2, diameter=20 mm STOP Object x 3 mm high dia.=10mm 20 mm 20 mm 20 mm Figure 11.21 Illustration for Example 11.11. (Figure 11.21 is not to scale) θL1 5 mm x 20 mm Figure 11.22 Aperture determination Example 11.11. θL1 = tan-1(5/20) = 14.0o Element 2 is the stop. It is imaged through L1, with f=5 mm and s=20mm to give s’ = 6.67 mm, M = -0.333 and image diameter of 3.33 mm, giving us the triangle θs x 1.67 mm 13.3 mm Figure 11.23 Aperture determination Example 11.11. θs = tan-1(1.67/13.3) = 7.2o Element 3 is L2. It is imaged through L1, with s=40 mm and f=5 mm, giving s’=5.71 mm, M=-0.143 and image diameter of 2.86 mm. The corresponding triangle is shown in Figure 11.24. θL2 = tan-1(1.43/14.3) = 5.7o 305 Among the three angles, this is the lowest. Thus L2 is the limiting aperture, as well as the exit pupil since L2 is the last element. The entrance pupil is the image of L2 through L1, which we just determined. The entrance pupil is located 14.3 mm to the right of the object with diameter of 2.86 mm and is the inverted image of L2. θL2 1.43 mm 14.3 mm Figure 11.24 Aperture determination Example 11.11. Step 2: Ray tracing. Draw the object, the lenses, stop and the entrance pupil to scale as shown in Figure 11.25. Object Ent. Pupil L1 Stop L2(limiting Aperture, exit pupil) Image 30 A 25 F2 Plane M 20 Y L B’ C 15 D A’ K X G H N 10 Z P 5 B 0 0 20 40 60 80 100 Figure 11.25 Ray tracing for Example 11.11. Chief ray: Start from the top of the object. Proceed toward D, the center of the entrance pupil. Extend to meet the lens at G. L1 will refract the ray and direct it toward K, the center of the limiting aperture and exit pupil (L2). The ray will proceed toward X. 306 Marginal ray through A: Start from the top of object. Orient the ray toward A’, the bottom of the entrance pupil. Extend it to meet the lens at H. Direct the ray toward the top of the exit pupil A. In order to determine the location of this marginal ray to the right of L2, use the oblique method. Draw the F2 plane as shown by the dotted line (MN). Draw a construction line PK (dashed line) passing through L2 center and parallel to HA. This construction line will intersect the F2 plane at point M. Connect A with M and extend it toward Z. This is one marginal ray. Marginal ray through B. Start from the top of object. Orient the ray toward B’, the top of the entrance pupil. Extend it to meet the lens at C. Direct the ray toward the bottom of the exit pupil B. In order to determine the location of this marginal ray to the right of L2, use the oblique method. Use the F2 plane as shown by the dotted line (MN). Draw a construction line LK (dashed line) passing through L2 center and parallel to CB. This construction line will intersect the F2 plane at point N. Connect B with N and extend it toward Y. This is the second marginal ray. The chief ray marked X and the marginal rays marked Y and Z go through a common point, giving us the image location. Verify using analytical method, the location and size of the image and compare with the graphical procedure above. Example 11.12 For the two-lens, middle stop system shown below, perform ray tracing using the chief ray and the two marginal rays, through the following system: (The figure is not to scale) f1=15 mm L1, diameter=20 mm f2= -10 mm L2, diameter=20 mm STOP Object x 3 mm high dia.=15mm 30 mm 10 mm 25 mm Figure 11.26 Illustration for Example 11.12. Step 1 Aperture determination: Element 1 is L1 (see Figure 11.27). θL1 = tan-1(10/30) = 18.4o Element 2 is the stop. Image the stop through L1; s=10, f=15. Therefore s’= -30 mm and M= 3, giving an image diameter of 45 mm. 307 θL 10 mm 30 mm Figure 11.27 Aperture determination for Example 11.12. θs 22.5 mm 60 mm Figure 11.28 Aperture determination for Example 11.12. θs = tan-1(22.5/60) = 20.6o Element 3 is L2. Image L2 through L1; s=35, f=15. Therefore s’= 26.3 mm and M=-0.75, giving an image diameter of 15 mm. θL2 7.5 mm 3.75 mm Figure 11.29 Aperture determination for Example 11.12. θs = tan-1(7.5/3.75) = 63.4o Thus, the smallest angle corresponds to L1. L1 is the limiting aperture and thus also the entrance pupil. The image of L1 in L2 is the exit pupil. Again, by thin lens equation, using s=35, f= -10mm, we obtain s’=-7.78 mm (to the left of L2), M=0.222 and exit pupil diameter= 4.44 mm. Step 2: Ray tracing. Refer to Figure 11.30. Draw the elements, L1, L2, object and exit pupil to scale as follows: Chief ray: Start from top of object. Proceed toward the center W of L1 (the entrance pupil and aperture). The ray travels undeviated until it hits the lens L2 at Q. At Q, connect Q with C, the center of the exit pupil shown by dashed line. The ray proceeds toward X to the right of L2. This is the chief ray. Marginal ray through A: Start from the top of the object. Proceed toward A, the top of the entrance pupil. We need the oblique method to trace this marginal ray to the right of L1. Draw the F1 plane, shown by the dotted line MN. Draw a construction ray (dashed line) JM passing through the center W of L1 and parallel to the marginal ray through A. 308 JM intersects F1 plane at point M. Connect A with M and extend to L2 to meet at R. Align the point R with A’, the top of the exit pupil. Show this by a dashed line. Extend the marginal ray forward beyond L2. Mark it as Z. Marginal ray through B: Start from the top of the object. Proceed toward B, the bottom of the entrance pupil. We need the oblique method to trace this marginal ray to the right of L1. Use the F1 plane, shown by the dotted line MN. Draw a construction ray (dashed line) KN passing through the center W of L1 and parallel to the marginal ray through B. KN intersects F1 plane at point N. Connect B with N and extend to L2 to meet at P. Align the point P with B’, the bottom of the exit pupil. Show this by a dashed line. Extend the marginal ray forward beyond L2. Mark it as Y. The intersection of the dashed lines from which the chief ray X, and the marginal rays Y and Z appear to emerge from, give us the location of the final image. Verify using the analytical method if the location and size of the image agrees with the graphical solution above. Object L1(Limiting Aperture and Ent. Pupil) Exit Pupil Stop F1 plane Image L2 30 A 25 K 20 M A’ Y W 15 C P Q 10 J B’ N R X 5 B Z 0 0 20 40 Figure 11.30 Ray tracing for Example 11.12. 60 80 309 Exercise 11.1B 1. Identify the limiting aperture for the following system (figure not to scale): f1= -10 mm L1, diameter=20 mm STOP Object 3 mm high dia.=10mm 35 mm 6 mm f2=25 mm L2, diameter=20 mm 14 mm Figure 11.31 Illustration for Exercise 11.1B, problem 1. 2. Perform ray tracing using the chief ray and the two marginal rays for problem 1. 3. Identify the limiting aperture for the following system (figure not to scale): f1= -15 mm f2= 10 mm L1, diameter=20 mm L2, diameter=20 mm STOP Object x 10 mm high dia.=15mm 30 mm 10 mm 25 mm Figure 11.32 Illustration for Exercise 11.1B, problem 3. 4. Perform ray tracing using the chief ray and the two marginal rays for problem 3. 5. Identify the limiting aperture for the following system (figure not to scale): f1=15 mm f2=5 mm L1, diameter=10 mm L2, diameter=20 mm STOP Object x 3 mm high dia.=10mm 20 mm 20 mm 20 mm Figure 11.33 Illustration for Exercise 11.1B, problem 5. 6. Perform ray tracing using the chief ray and the two marginal rays for problem 5. 310 11.2 Optical Instruments1-4 The most common, yet complicated optical instrument is the human eye. We briefly discussed the parts of the human eye with regards to laser safety. The retina is the image plane where the object is imaged on to by the lens, which is filled with a transparent jelly. The lens center is hard and the outer portions are soft so that the radius of curvature is adjustable so as to focus different distance objects. The bulk of the eye ball is filled with vitreous humor, a dilute water-based jelly whose refractive index is close to that of water. The refractive index of the lens is somewhat between that of water and typical glass, in the neighborhood of 1.42 to 1.44. The aperture of this optical system is the pupil of the eye or the iris, which is continually adjustable from practically closed to about 8 mm in diameter. The range of distance over which an object can be seen by the unaided eye (or with corrective lenses for those requiring it) spans from the near point to the far point. The near point is the closest distance that one can bring the object to the eye so that the object can still be seen clearly. If the object is brought any closer, then it becomes blurry, the eye is strained, sometimes causing headache. The near point itself will increase with age, from a few centimeters for a pre-teenager to tens of centimeters in seniors (50+ years of age.) For the rest of the chapter, we will assume somewhat arbitrarily, that the near point is at 25 cm from the eye. The far point of the eye while mathematically at infinity is practically several hundred meters away from the eye. Vision problems include myopia (short sight, or near sight), hypermetropia (long sight or far sight) and astigmatism. A myopic person can see closer objects easily and does not need reading glasses, whereas, when the object is far away the image is formed in front of the retina, and not on it. Thus, a negative power lens is required to correct this problem to be able to see distant objects, and when driving. A hypermetropic person can see far away objects really well, whereas, when the object gets closer, the image becomes blurry, since it is formed behind the retina, and not on it. Thus, this person would need reading glasses, which use positive power lenses. Bifocal lenses incorporate both a positive and a negative lens to correct for both types of problems. Astigmatism, on the other hand is a situation when vertical lines on a grid are in focus while the horizontal lines are not, and vice versa. Cylindrical lenses, which possess only one radius of curvature, (a piece of broken glass from a bottle is an example of a cylindrical lens) enable correcting astigmatism. Earlier, we discussed lateral magnification, M, which is a ratio of the image height to the object height. One literally measures these heights with a ruler to calculate the lateral magnification, or, knowing the object height and M calculated from the thin lens equation methodology, one can predict the size of the image that will be obtained on a screen. This approach is useful to predict image size when using a movie projector, a slide projector or an overhead (transparency) projector. However, the way the human eye views an image and interprets how large it is when magnified is very different from interpretation of the lateral magnification, M. The human eye makes a comparison of viewing the object unaided when the object is at the near point of 25 cm versus viewing 311 the same object through an optical instrument when the virtual image now falls close to the near point of 25 cm. This comparison is feasible through the use of the angular magnification Mang, which is defined below. The simple microscope: This is the simplest of the optical instruments involving a single, positive lens. It is also called a magnifying glass or a magnifier, and is used in a variety of situations, such as map reading, examining forensic evidence such as fingerprints, palm reading, etc. As we learned in the previous chapter, an object located at the F point of a positive lens will result in a virtual image, infinitely enlarged, and erect, and located at infinity. An image located at infinity will often be blurry and not very useful. Thus, one moves the magnifier so that the object is slightly within the F point (i.e., between F and the lens, however, very close to F). This would appear as a clear, virtual, and highly magnified image to the human eye. For the sake of analysis and calculation of the angular magnification, let us, for all practical purposes assume that the object is at the F’ point, and therefore, the image is at infinity, as shown in Figure 11.34. Note that in this case f is measured in centimeters only. Image at infinity Object Human Eye h F θ‘ f Figure 11.34 Illustration of a simple magnifier. F’ f Compare this to a situation when the same object is viewed by the unaided eye and the object is located at the near point of the eye of 25 cm from the eye, as shown in Figure 11.35. In both cases, the object height h, is measured in centimeters only. If the human eye is placed close to the right of the lens in Figure 11.34, the image subtends an angleθ’ with the eye. In Figure 11.35, the unaided eye looks at the object, and the object subtends an angle θ with the eye. Since we deal with small size objects that require the use of a magnifier, the angles involved are small enough so that we evoke the small angle approximations: 312 tan θ’ ~ θ ‘= h (from Figure 11.34) f tan θ ~ θ = h (from Figure 11.35) 25 θ Human Eye Object h 25 cm Figure 11.35 Illustration of an object at the near point. The ratio of the angles subtended at the aided eye (i.e., with the optical instrument) to the unaided eye is called angular magnification Mang, given by Mang = θ ‘= h x 25 θ f h = 25 f with f measured in centimeters only since 25 is the near point location in centimeters. The angular magnification has no units (similar to the lateral magnification). However, the angular magnification is a true measure of how large an object appears to the aided eye, compared to the unaided eye. Example 11.13 Calculate the angular magnification of a 100 mm focal length positive lens. f = 100 mm = 10 cm Mang = 25 = 2.5 Answer 10 One simply denotes the angular magnification in this case as 2.5X. With a single lens, a practical angular magnification of this magnitude is feasible. Larger angular magnifications require the use of a compound microscope discussed next. This is because lens aberrations (see section 11.3) restrict one from obtaining a clear image with a large angular magnification, if only one lens is used. The compound microscope: This is typically used by the scientist or technician examining biological specimens, or in the industry to examine minute objects such as electronic circuitry, small solder connections, laminate cross sections, etc. A high angular magnification such as 100X, 250X or 400X is feasible when one employs two positive lenses. The lens closest to the 313 object is called the objective lens and the second lens is called the eye piece, which is closer to the eye. Refer to Figure 11.36. The principle behind the compound microscope is as follows: The object to be magnified is located just outside the focal point of the objective lens. This results in a real, magnified and inverted image which is located beyond twice the focal length of the objective. The eye piece is placed such that the real image formed from the objective is within the eye piece’s focal length. The final image will be highly enlarged, inverted and virtual, and is seen through the eye piece. For the compound microscope, the objective lens has a focal length fo. The object has to be placed inverted just outside its focal point, i.e., just beyond Fo point. A real inverted image, marked I1, is formed. The lateral magnification of the objective Mo is given by Mo = -so’ so Fo | Fo’ | so 2Fo’ | Fe | I2 I1 F’ e | so’ Figure 11.36 Compound microscope This can be simplified to Mo = -fo . so-fo The real image I1 is now the object for the eyepiece whose focal length is fe. The angular magnification of the eye piece is given by Mang,e = 25 fe The eye piece images I1 as a virtual image at location I2, maintaining the orientation and enlarging it even further. The overall magnification of the compound microscope is given by 314 Moverall = MoMang,e = -25fo fe(so-fo) the product of the lateral magnification of the objective lens and the angular magnification of the eyepiece. Knowing the object location from the objective, the focal lengths of the objective and the eyepiece, the overall magnification can be calculated easily. Note that all required dimensions must be expressed in centimeters only. The eye piece is also known as the ocular. Example 11.14 Calculate the overall magnification of a home made compound microscope with an objective of focal length 1 cm and eye piece focal length of 0.5 cm for an object placed 2 mm beyond the focal point of the objective. Calculate the distance between the two lenses Moverall = MoMang,e = -25fo fe(so-fo) with fo=1 cm, fe = 0.5 cm and so-fo = 0.2 cm. Substituting gives us Moverall = = - 25 x 1 0.5 x 0.2 -250X Answer The negative sign just indicates inversion of the image. However, the object would appear 250 times larger to the human eye compared to the unaided eye. For the second part, one needs to find out the location of image I1 first and add the focal length of the eyepiece. so’ = sofo . (so-fo) = 1.2 x 1.0 (1.2-1.0) = 6 cm Distance between the eyepiece and objective = 6 cm + 0.5 cm = 6.5 cm Answer Example 11.15 The objective and eyepiece of a compound microscope are 18 cm apart. Their respective focal lengths are 7mm and 5 mm. If the final image is formed at infinity, calculate (a) the object location from the objective, (b) the lateral magnification of the objective, and (c) the overall magnification of the system, assuming that the final image is at infinity. 315 (a) The following sketch shows the given information. Object Objective I1 Final image (infinity) so=? Eye piece Eye 17.5 cm 0.5 cm Figure 11.37 Illustration for Example 11.15 Note that the first image I1 is formed so that it is at the focal point of the eye piece, since the final image is at infinity. Using the thin lens equation for the objective, we obtain so = s’ofo s’o-fo where so’ = 17.5 cm and fo=0.7 cm. Substituting, we obtain so = 0.729 cm ~ 7.3 mm Answer (b) The lateral magnification of the objective Mo is given by Mo = -so’ so = -17.5 = -24 Answer 0.729 (c) Mang,e = 25 = 25 = fe 0.5 Moverall = (-24) x 50 = -1200X Answer 50 Clearly, a simple microscope or magnifier will not be able to provide such a high magnification. Objectives are seldom single lenses; rather they are often combinations of a positive and a negative lens cemented together or a triplet of two positive and a negative lens cemented together. Minimization of aberrations (see next section) is the goal behind these. A large size objective will gather more light from the object. This is of significance in the choice of telescope objectives. Similarly, eye pieces (oculars) are comprised usually of two lenses. The eye lens of the eye piece is closest to the eye. The eye lens enlarges the image that it receives. The lens farthest from the eye piece or closest to the objective lens is called the field lens. Its function is to collect as much light as feasible from the objective so that eye lens can magnify it. There are several designs of eye pieces, depending on the type of lens that 316 make up the eye lens and the field lens as well as the relative distance between them. A reticle which may take the form of a graduated scale or cross hairs is an integral part of the eye piece. Its functions are distance gauging and location of center of field3. Some microscope makers traditionally have fixed the optical length as 160mm. The mechanical tube length is slightly larger. Figure 11.38 illustrates these concepts. Fo Final Image Fo’ Fe Eye center of objective system 160 mm (Optical length) Mechanical Tube Length Final image at near point, i.e. 25 cm from eye Figure 11.38 Illustration of optical and mechanical tube lengths The final image formed at the near point of the eye is clearly visible to the observer. If the final image is formed at infinity, it will be blurred. Thus, the real image I1 formed by the objective should not be located exactly at the point Fe, rather slightly to the right of it, at least by a fraction of a millimeter. Objectives and oculars are not only part of microscopes, however are used in other optical instruments such as telescopes and binoculars as well. We will discuss these two instruments next. Telescope While a microscope is used to obtain large magnification of a small object that is close to the observer, a telescope is used to view objects that are very far away, such as heavenly bodies or in surveillance (spy glass). When viewing heavenly bodies, it does not much matter if we view the image upside down or straight side up. The telescope used for such purposes is known as the astronomical telescope. However, for applications where one needs to see an erect image, one needs to use a terrestrial telescope (spy glass) which is the basis behind the construction of a pair of binoculars (as the name implies binoculars have two oculars or eye pieces for obtaining a stereoscopic image of the object). The principle behind a telescope be it astronomical or terrestrial, is to image an object which is practically at infinity. This image will be formed very close to the F’ point of the objective, inverted, real and highly diminished. The next task is to magnify this first image by placing a second positive lens, the eye piece, such that the first image is located 317 just within the focal point of the eye piece. This results in a virtual and magnified image of the original object, which is visible to the observer through the eye piece. The final image, however, is inverted, compared to the original object, which may be a mute point if one is observing far away planets, or lunar eclipse. In Figure 11.39, light is traveling from a far away object and hence the rays are essential parallel. Note that the rays need not be parallel to the optical axis. Let us look at a single ray that strikes the center of the objective lens. The angle, θ that this ray makes with the optical axis is the same as the angle that the ray would make with the unaided eye. Therefore, this angle would be used as a reference when computing angular magnification of the system. Object at ∞ Fo θ θ θ’ Fo’, Fe Human Eye h I1 Fe’ Final Image at ∞ fo fe (Only a few rays are shown for the sake of clarity) Figure 11.39 Astronomical telescope (Keplerian type) The rays are focused to the image point denoted by I1, which is at the Fo’ point of the objective whose focal length is fo. The eye piece with the focal length fe is placed such that the Fe point of the eye piece coincides with the I1 location. In reality, the image I1 of height h, should be slightly to the right side of the Fo’ location so that the final, virtual image, as seen through the eye piece is not at infinity, however at the near point of the observer’s eye. The set of parallel rays that are incident on the observer’s eye make an angle θ’ with the eye. Since the image makes the same angle with the observer’s eye, the angular magnification of the system can be defined as Mang = θ’ θ From Figure 11.37, we notice that and θ’ = h fe (small angle approximation tan θ’ ~ θ’) θ = h fo (small angle approximation tan θ ~ θ) 318 Substituting these, we obtain, Mang = fo fe It is customary to designate that the final image is inverted with respect to the original image by using a negative sign, as Mang = -fo fe Thus, compared to the unaided eye, the far away object will appear (fo/fe) times bigger to the eye. Using an objective focal length large compared to the eye piece will result in a large angular magnification. This approach will work when one observes the object through the eye piece. If one needs a hard copy photograph of the heavenly body, then, a photographic film is placed slightly to the right side of Fe’ such that a real image is obtained on the film, which can later be developed and magnified. Example 11.15 An astronomical telescope has an objective with focal length of 75 cm and an eye piece focal length of 3 cm. (a) Find the angular magnification of the system, and (b) show how you would convert this to a terrestrial telescope. (a) Mang = -75/3 = -25X Answer The far away object looks 25 times larger through the telescope than to the unaided eye, (b) For design purposes, choose a third lens of focal length 3 cm also. The goal here is to minimize the total telescope tube length and still achieve an erect final image. Refer to Figure 11.40 Objective Inverting lens Eye piece Object at infinity Final image 75 cm 6 cm 6 cm Eye 2.68 cm 25 cm (near point) Figure 11.40 Converting an astronomical telescope to a terrestrial telescope Note that the third lens, denoted as the inverting lens, flips the first image upside down without magnifying it. The eye piece is placed so this image is located, say a few millimeters or so within the eye piece’s focal point (3.2 mm to be precise) so that the final erect and enlarged image will be formed at the near point of the eye, rather than at 319 infinity. The angular magnification of the system will be slightly lower than 25X. By adjusting the eye piece one can obtain a clear final image. A second method of inverting the image is by the use of a prism or two prisms. This concept was discussed in Chapter 5. The prism(s) would now provide the image inversion, replacing the third lens. Two such terrestrial telescopes placed side by side result in a pair of binoculars that many of us are familiar with. We know that the images we see (players in a ball game, or musicians in an opera) are erect. The two telescopes side by side provide the stereoscopic effect. Additionally, many binoculars, telescopes and microscopes have built in apertures or pupils in them, sometimes adjustable, so that the pupil size of the system can match the pupil size of the human eye. In most cases, the objective limits the light into the system and thus serves as the aperture. With the right aperture, one can see the image not only clearly and entirely, but also at the best possible level of illumination. Clearly, the human pupil is much larger at night time than during the day time. Therefore, optical systems used for night time viewing should have an adjustable or a different pupil size to optimize night viewing of objects. Binocular specifications often go by M x d, where M denotes the magnification and d the effective objective diameter in millimeters. The telescope discussed in this section which employs positive lenses for both the objective and the eye piece is called a Keplerian telescope. One could also use a negative lens for the eye piece. This is known as a Galilean telescope. It should also be mentioned (see also Chapter 5) that a concave mirror may be used to gather light from the object and focus it. This is the principle behind the Newtonian telescope3. There are many other forms of telescopes also not discussed here. Beam Expander (Collimator) We know that the laser beam, despite it being more directional than conventional light sources, has a beam divergence, which results in the beam diameter expanding the farther the beam is from the laser cavity. The procedure to calculate the beam divergence and spot diameter are discussed in the chapters on pulsed lasers and wave optics. If one needs to obtain a larger diameter beam from a laser, yet maintain the parallel nature of the rays, a collimator (beam expander) is used. Collimated light essentially has a parallel beam with minimum divergence. Collimators are also called beam expanders. Collimating telescopes are useful in aligning and adjusting optical systems when an object located at infinity is desired. The Keplerian system3,5 employs two positive lenses. We will continue using the terms objective and eyepiece although when using lasers one should not look through he eyepiece. If a nearly parallel beam of light such as from a laser is incident upon the objective lens at points A and B in Figure 11.41. The radius (more specifically, the 1/e2 radius) of the beam is denoted as ro, the subscript ‘o’ referring to the objective. Following the rules of geometric optics, the beam will be focused at point C. If the eyepiece is placed in such a 320 manner that its focal point coincides with point C, then the beam, after striking the eye piece at points D and F will become parallel to the right side of the objective lens. Objective A ro B Eye Piece D C E re F fo fe Figure 11.41 Keplerian collimator. Consider triangles ABC and CEF. They are similar triangles (refer to geometric principles at the beginning of Chapter 5) because the corresponding angles are equal. Therefore, AB = EF BC CE or or ro fo re ro = = re fe fe fo The ratio of the radii (or diameters) of the beams equal the ratio of the focal lengths3,5. Two observations arise. The beam is expanded and collimated if the eye piece focal length is larger than the objective focal length. Secondly, if the original beam had a beam divergence of θo, then the beam divergence θe of the exiting beam from the eye piece can be shown to have a lower divergence6 modified by the same ratio, inversely through, as θe θo fo fe If the laser power were to be large to begin with the irradiance at the point C can be very high. In order to avoid this, a negative lens may be used as the objective resulting in a Galilean system3,5, as shown in Figure 11.42. = If ro is the radius of the nearly parallel beam that is incident on the negative lens at points A and B, then per the principles of geometric optics, the rays of light will appear to emerge from point C, the objective’s secondary focal point. This is denoted by the rays AD and BF. If, however, a positive lens eye piece whose focal length fe is larger in magnitude than the objective lens’ focal length us placed such that the eye piece’s primary focal point coincides with the point C, then the rays emerging from points D and E will be essentially parallel and larger in diameter than the original beam. 321 D re A G ro E C B F fo fe Figure 11.42 Galilean collimator. As before, we see that triangles CAG and CDE are similar to each other leading to CG AG = CE DE or re = fe ro fo 2 The ratio of the 1/e radii (or diameters) of the beams equal the ratio of the focal lengths3,5. Note that only the magnitude of the objective lens focal length is to be used, not its negative sign. The beam divergence of the exiting beam is still given by θe θo fo fe where only the magnitude of fo is to be used. = Note that the principle behind the collimator is exactly the same as the astronomical telescope. The irradiance of the expanded and collimated beam in either collimator will essentially be lower. Example 11.15 Calculate (a) the beam diameter, and (b) the beam divergence for the following systems: (1) a Keplerian collimator with objective lens focal length of 5 cm and eye piece focal length of 15 cm, and (2) a Galilean collimator with objective focal length of -3 cm and eye piece focal length of 15 cm. The incident beam has a diameter of 2 mm and a beam divergence of 15 miiliradians (mrads) in either case. (1) Keplerian Diameter of the exiting beam = 2 mm x (15/5) = 6 mm Answer Divergence of the exiting beam = 15 mrad x (5/15) = 5 mrad Answer (2) Galilean Diameter of the exiting beam = 2 mm x (15/3) = 10 mm Answer Divergence of the exiting beam = 15 mrad x (3/15) = 3 mrad Answer 322 Example 11.16 Design a Keplerian collimator such that it expands the beam 5X. The collimator length shall not exceed 50 cm. Let the focal length of the objective = fo Let the focal length of the eye piece = fe Therefore, fe = fo The length, fo + fe = 50 cm 5 Substituting the first equation in the second and solving, we obtain fo=8.33 cm and fe=41.7 cm Answer Example 11.17 Design a Galilean collimator such that it expands the beam 5X. The collimator length shall not exceed 25 cm. Let the focal length of the objective = fo Let the focal length of the eye piece = fe Therefore Therefore fe fo = 5 fe- fo fo = 25 cm = 6.25 cm in magnitude (negative lens); fe = 31.25 cm Answer Exercise 11.2A 1. Calculate the angular magnification of a 50 mm focal length positive lens. 2. Calculate the angular magnification of a 10 D focal power positive lens. 3. Calculate the overall magnification of a home made compound microscope with an objective of focal length 1.2 cm and eye piece focal length of 0.5 cm for an object placed 1.5 mm beyond the focal point of the objective. Calculate the distance between the two lenses. 4. Calculate the overall magnification of a home made compound microscope with an objective of focal length 2 cm and eye piece focal length of 1 cm for an object placed 0.5 mm beyond the focal point of the objective. Calculate the distance between the two lenses. 5. The objective and eyepiece of a compound microscope are 15 cm apart. Their respective focal lengths are 1 cm and 5 mm. If the final image is formed at infinity, calculate (a) the object location from the objective, (b) the lateral magnification of the 323 objective, and (c) the overall magnification of the system, assuming that the final image is at infinity. 6. The objective and eyepiece of a compound microscope are 18.5 cm apart. Their respective focal lengths are 1.5 cm and 1.0 cm. If the final image is formed at infinity, calculate (a) the object location from the objective, (b) the lateral magnification of the objective, and (c) the overall magnification of the system, assuming that the final image is at infinity. 7. An astronomical telescope has an objective with focal length of 100 cm and an eye piece focal length of 3 cm. (a) Find the angular magnification of the system, and (b) show, using a sketch, how you would convert this to a terrestrial telescope. 8. An astronomical telescope has an objective with focal length of 100 cm and an eye piece focal length of 5 cm. (a) Find the angular magnification of the system, and (b) show, using a sketch, how you would convert this to a terrestrial telescope. 9. Calculate (a) the beam diameter, and (b) the beam divergence for the following systems: (1) a Keplerian collimator with objective lens focal length of 3 cm and eye piece focal length of 15 cm, and (2) a Galilean collimator with objective focal length of -3 cm and eye piece focal length of 9 cm. The incident beam has a diameter of 2 mm and a beam divergence of 30 miiliradians in either case. 10. Calculate (a) the beam diameter, and (b) the beam divergence for the following systems: (1) a Keplerian collimator with objective lens focal length of 3 cm and eye piece focal length of 18 cm, and (2) a Galilean collimator with objective focal length of -6 cm and eye piece focal length of 9 cm. The incident beam has a diameter of 2 mm and a beam divergence of 4.2o in either case. 11. Design a Keplerian collimator such that it expands the beam 3X. The collimator length shall not exceed 50 cm. 12. Design a Keplerian collimator such that it expands the beam 5X. The collimator length shall not exceed 30 cm. 13. Design a Galilean collimator such that it expands the beam 3X. The collimator length shall not exceed 50 cm. 14. Design a Galilean collimator such that it expands the beam 5X. The collimator length shall not exceed 30 cm. 11.3 Aberrations1,3,4 Up to this point, we had considered idealized situations with thin lenses and small angles so that the small angle approximation is inherent in some of the underlying formulas. 324 This enabled the mathematics to be simplified tremendously so that one could use algebra-based, rather than calculus-based approach to solve problems. Lens aberration causes the image formed to have a different shape and location compared to the predictions of diffraction at the lens aperture. Consequently, a light distribution in the image is obtained which is different from diffraction phenomenon. Refer to the chapter on wave optics for a clearer understanding of diffraction. For the present, we will look at how the small angle approximation helps simplify ray tracing procedures and the consequences of the approximation. In calculus, the sine of an angle θ can be expanded as an infinite series, as sin θ = θ - θ 3 + θ 5 – θ 7 + θ 9 – θ 11 + …….. 3! 5! 7! 9! 11! where the “!” symbol represents factorial. For example, 3! = 3 x 2 x 1 =6, 4! = 4 x 3 x 2 x 1 = 24, 5! = 5 x 4 x 3 x 2 x 1 = 120… and n! = n (n-1) (n-2) (n-3)…4.3.2.1., in general for any integer n. Note that 0! = 1. In the series expansion for sin θ, the angle θ is always expressed only in radians. (1 radian = 57.296o approximately). When the angle θ is small, higher powers of the angle become even smaller. Therefore, with little error one can eliminate all higher powers of θ, leading to sin θ ~ θ ~ tan θ which is the small angle approximation discussed in Chapter 5. We will see how the approximation is used to derive the lens maker’s equation and the thin lens equation. Referring to Figure 11.43, consider a curved surface of radius r of refractive index n’ adjacent to a medium of refractive index n. O is an axial object point and I is the axial image point. θ φ1 X s n n’ Y h θ’ Z P φ2 φ3 r C I s’ Figure 11.43 Derivation of refraction equation for a single surface. 325 Consider an arbitrary ray of light making an angle θ with the normal. The normal itself is shown by the dotted line, and as required, it passes through C, the center of curvature of the curved surface. (Refer to Chapter 5). The angle of refraction in the second medium is θ’. h represents the vertical height of the point of incidence above the optical axis. The foot of the perpendicular is at P on the optical axis. The object distance is denoted by s, and the image distance by s’, as usual. Applying Snell’s law at the interface nsin θ = n’sin θ’ Equation a If the oblique ray XY from the axial object X to the interface between the two media is such that the height h is small in relation to s, s’ or r, then the small angle approximation can be used both for θ and for θ’. Accordingly, sin θ = θ and sin θ’ = θ’ The Snell’s law (Equation a) simplifies to n’ = sin θ = θ n sin θ’ θ’ Equation b Equation c Equation d Again because h is small compared to s, s’ or r, the angle φ1is also small; the same is true of φ2 and φ3. Using the principle that the exterior angle in a triangle is equal to the sum of the two interior opposite angles (see geometric principles at the beginning of Chapter 5), we have φ1 + φ2 (consider triangle XYC) θ= Equation e φ2 = θ’ + φ3 (consider triangle IYC) and Equation f or θ’ = φ2 - φ3 Equation g Dividing Equation e by Equation g, we obtain φ1 + φ2 = θ φ2 - φ3 θ’ Equation h Substituting for the left hand side of Equation h from Equation d, we obtain φ1 + φ2 = n’ Equation k φ2 - φ3 n From triangles XYP, YPC and YPI and applying the small angle approximations, and realizing that the points Z and P are very close to each other, we obtain tan φ1 = h = Equation m φ1 s 326 tan φ2 = φ2 h = Equation n r tan φ3 = h = Equation p φ3 s’ Substituting for the angles φ1, φ2, and φ3 from Equations m, n and p into Equation k, we obtain h + h = n’ s r Equation q h - h n r s’ Canceling h from the left hand side and multiplying both the numerator and the denominator of the left hand side by ss’r and simplifying, we obtain rs’ + ss’ ss’ – rs = n’ n Equation r n’(ss’-rs) Equation s Cross multiplying, we obtain n(rs’ + ss’) = Dividing both sides by ss’ and simplifying, we obtain n(r + 1) = n’(1 – r) s s’ Distributing via expansion of brackets on both sides, nr + n = n’ n’r s s’ or Equation t Equation u r(n s + n’) s’ = n’ – n Equation v n s + n’ s’ = n’ – n r Equation w or Equation w explains the refraction phenomenon through a single curved surface when light passes from medium of refractive index n to medium of refractive index n’ and very close to the optical axis. This equation is valid for paraxial rays, i.e., rays nearly parallel to the optical axis and close to it so that the small angle approximations hold. Lenses are comprised of two such spherical surfaces. Thus Equation w can be applied twice, one to each of the spherical surfaces shown in Figure 11.44. 327 Surface 1 X Surface 2 C2 C1 s1 t I2 I1 s2’ s2 s1’ Figure 11.44 Derivation of Lens maker’s equation. Consider an axial object point X located on the optical axis of a biconvex lens. The first surface is convex with its center of curvature at C1, and the second surface is concave with its center of curvature at C2. The object distance is s1 from surface 1. After refraction at the first surface, the intermediate image will be located at I1, and at a distance of s1’ from surface 1. The thickness of the lens is marked as t. The object for the second surface is the image I1, which is located at s2 from surface 2. After refraction at the second surface, the final image is located at the axial image point I2, and located at a distance s2’ from surface 2. We can now apply Equation w derived earlier to each surface as n s1 + n’ s1 ’ = n’ – n r1 Equation x n’ s2 + n s2 ’ = n’ – n = n – n’ - r2 r2 Equation y The minus sign in Equation y before r2 is to indicate that it is a concave surface, consistent with the rules developed in Chapter 10. Also, for surface 2, light is traveling from a medium of index n’ to a medium of index n. For a thin lens, t ~ 0 and therefore s1’=-s2. The minus sign here denotes a virtual object for surface 2. Making this substitution and adding Equations x and y, n - n’ + n’ + n s1 s2 s2 s2’ = (n’-n)[1 – 1 ] r1 r2 Equation z 328 This further simplifies when the first medium is air (i.e., n=1) to 1 + 1 = (n’-1) [ 1 – 1 ] Equation aa s1 s2 ’ r1 r2 Using a rule of ray tracing from Chapter 10, we know that when the object is located at infinity (i.e., s1-> ∞), the incident rays will be parallel and the image will be located at the secondary focal point for a positive lens (i.e., s2’->f). Making these substitutions, we obtain 1 f = (n’-1) [ 1 – 1 ] r1 r2 Equation bb Equation bb is the lens maker’s equation that we had used extensively in Chapter 10. Now, using the lens’ maker’s equation, we will derive the thin lens equation as follows. Consider an object AB of height h standing on the point A on the optical axis in Figure 11.45. The primary focal point of the positive lens is shown by F, the secondary focal point by F’. Also, two locations, which are at twice the focal distance from the lens are shown by the points 2F and 2F’, respectively, on the optical axis. Using Rules 1, 2, and 3 of ray tracing discussed in Chapter 10, the image is obtained at A’B’ with height h’. While the derivation here and the previous ones in this section use positive lens as an example, the arguments are equally valid for negative lens as well. The student is encouraged to derive this relationship for negative lens. B h 2F A L (lens) P F s Q F’ R s’ 2F’ A’ h’ Rule 2 B’ Rule 1 Rule 3 Figure 11.45 Application of similar triangles principle to derive the thin lens equation. Being a thin lens, we again represent it by a vertical line. Consider triangles ABQ and A’B’Q. They are similar for the following reasons. /BAQ = /B’A’Q = 90o /AQB = /A’QB’ (vertically opposite angles) Obviously making the third respective angles also equal in each of the triangles. Since we proved that the triangles are similar, the ratios of corresponding sides must be equal, i.e., 329 A’B’ = A’Q -AB AQ or h’ = -s’ = M (lateral magnification) h s Triangles ABF and FQR are similar /BAF = /FQR = 90o /BFA = /QFR (vertically opposite angles) Obviously making the third respective angles also equal in each of the triangles. Since we proved that the triangles are similar, the ratios of corresponding sides must be equal, i.e., AF = FQ AB QR or s-f = h Equation (cc) f -h’ Triangles A’B’F’ and QPF’ are similar /F’A’B’ = /PQF’ = 90o /PF’Q = /B’F’A’ (vertically opposite angles) Obviously making the third respective angles also equal in each of the triangles. Since we proved that the triangles are similar, the ratios of corresponding sides must be equal, i.e., QF’ = F’A’ PQ A’B’ f = h Equation (dd) s’-f -h’ The right hand sides of Equations cc and dd are equal. Therefore, the left hand sides must also equal, i.e., . s-f f = f s’-f Equation (ee) Cross multiplying and expanding the brackets, we obtain f2 = ss’ –sf -s’f + f 2 Equation (ff) or Equation (gg) f = ss’ (s + s’) Inverting 1 = s + s’ f ss’ or 1 = 1 + 1 f s s’ This is the derivation of the thin lens equation Equation (hh) Equation (kk) 330 Clearly, the thin lens equation arises from the small angle approximation, which states that the angle in radians, its sine and tangent are equal. This justified our using a straight line to represent a lens. Clearly, if the ray of light is far away from the optical axis, the curvature of the lens becomes significant, leading to inaccurate prediction of image location. Several types of monochromatic aberrations, i.e., aberrations caused even if monochromatic light is used for imaging, will be discussed first. (a) Spherical aberration: This is related to the spherical nature of the lens which we just discussed. The rays very close to the optical axis known as paraxial rays will follow the thin lens equation predictions. However, as shown below, rays which are farther away come to a focus farther away than the paraxial rays. This is called spherical aberration, illustrated in Figure 11.46 below. N1 P1 L Fn’ Fp’ P2 N2 Figure 11.46 Illustration of spherical aberration in a positive lens. Parallel rays travel from left to right and strike a positive lens, L. Paraxial rays are marked by P1 and P2, for example, whereas, non-paraxial rays (ones that are parallel to the optical axis, however are far away from the optical axis) N1 and N2, for example. The paraxial rays are focused at point Fp’ whereas the non-paraxial rays at Fn’. This does not mean that there are only two distinct focal points, since the focal points lie on a range, and could be anywhere between Fp’ and Fn’, the two extremes. Clearly, the Fp’ and Fn’ are exaggerated for clarity. Thus if the lens is the objective of a telescope, the image could lie anywhere in a range caused by spherical aberration. Similarly, for a negative lens, we observe the same phenomenon illustrated in Figure 11.47. If the lens is thick the spherical aberration is exacerbated. A thick lens with a short focal length suffers more severely from spherical aberration than a thin lens with a long focal length. When spherical aberration is present, an axial point image of an axial point object can never be realized. Instead the image will be within a circle of a fixed radius. Clearly, if the rays farthest away from the optical axis contribute to spherical aberration, they can be eliminated via the use of an aperture close to the left of the lens, allowing 331 only rays close to the optical axis to pass through the system. While this minimizes the spherical aberration, it would also reduce the intensity of light in the optical system. N1 Fp’ P1 Fn’ P2 N2 Figure 11.47 Illustration of spherical aberration in a negative lens. Also, in some optical instruments such as a telescope, the objective may be placed with the higher curvature side facing the object (for example, the convex side of a planoconvex lens), to minimize spherical aberration. If the flat side of a plano-convex lens faces the object, this leads to severe spherical aberration. A compound lens, which is combination of a positive and a negative lens with the same refractive index, is also used in some optical instruments to minimize spherical aberration. If cost were no factor, a lens with a constantly changing radius of curvature, most probably hand crafted, would minimize spherical aberration. (b) Coma: Frequently one images an object that has a spatial extent, rather than a point object. With spherical aberration, a point axial object can at best be imaged as a small circle, outside of which the beam diverges. If, however, the object has off –axis points (i.e., a tall object standing on the optical axis), then the off-axis points will be imaged to resemble a coma “,” or a comet with a tail. Even if the lens is optimized for minimum spherical aberration, coma will still be present. Fortunately, however, a plano-convex lens with the curved side facing the object also results in minimum coma, besides minimum spherical aberration. An example of a coma is shown in Figure 11.48. The rays through the center (ray Y, for example) will focus to a point if the spherical aberration is minimized. However, rays farthest away will exhibit coma in the image plane (the plane of F’). (c) Astigmatism. This is the phenomenon when the image appearance is optimized in one direction, for example, the vertical, the appearance (or focusing) in the perpendicular direction (for example, horizontal) is poor, and vice versa. Again, this is caused by rays that are incident on the lens farthest away from the optical axis. Thus, there exists a circle of 332 least confusion in which the image seen provides the best compromise for mutually perpendicular directions. Use of multiple lenses, with different curvatures is a way to minimize astigmatism due to cancellation effect. X1 Y Parallel rays from object F’ plane X X2 Y image appearance (example) Figure 11.48 Coma. (d) Curvature of field There exist two curved surfaces in which the image is focused in mutually perpendicular directions. These two surfaces meet on the optical axis at which point the astigmatism is negligible. Even if astigmatism is corrected for at other points on the curved surface, the fact is that the image plane is not a plane, but rather curved. This is known as the curvature of field. The human retina is curved, clearly an attempt to focus light from objects clearly. Use of stops can minimize the curvature of field effect in optical instruments. (e) Distortion This is an aberration which manifests itself as relatively disproportionate distances among points on the image compared to the same corresponding points in the object. It is caused by variable lateral magnification based on the distance of the object point from the optical axis. Thus, a straight line object that does not pass through the optical axis will appear as a curve in the image. Different types of distortions are illustrated in Figure 11.49. No distortion Barrel distortion Figure 11.49 Distortion. Pincushion distortion 333 Use of a combination of positive and negative lenses with opposing types of distortion effects will minimize distortion aberration. While a single lens seldom exhibits distortion problems, use of a stop to minimize, say spherical aberration, introduces distortion. Properly designed middle stop in a system will minimize distortion. Again, since the retina is curved, it accommodates the curvature of field of the image plane, and also is practically distortion free. It is essential for the eye piece of optical instruments to be distortion free. Such eye pieces are known as orthoscopic since the image proportions are the same as the object proportions. This concludes the discussion of monochromatic aberrations which are present even if one wavelength of light is used for imaging. However, when white light is used in an optical system, and this is the case with many optical instruments, this leads to chromatic aberration, which is discussed below. (f) Chromatic aberration1-4: As we learned in Chapter 5, the refractive index depends on the wavelength of light. We also recognized that when white light enters glass, if the faces are not parallel to each other, this leads to dispersion. In the case of a lens, the different wavelengths that make up white light undergo dispersion, i.e., refracted to different extents. Figure 11.50 illustrates this concept for a single, positive lens. Red White light Violet Violet Red Green Figure 11.50 Chromatic aberration. Since the refractive index of violet is slightly larger than red for typical glasses and lens materials, it will be refracted most, whereas red wavelength will be refracted the least. In between are the other colors. Only green is shown in the figure for clarity. Thus, light that was white originally is focused to different points along the optical axis, provided other types of distortion mare minimized. You might have observed a rainbow hue when trying to image an object illuminated by white light. The human eye is most sensitive to the yellow portion of the spectrum and thus one would automatically adjust the focusing to accommodate clarity for yellow light. In general, different wavelengths would thus result in different focal points or image locations, both in the axial direction and in the lateral direction. This phenomenon is called chromatic aberration. It also results in different magnifications for different wavelengths. Similarly, with a negative lens, when white light is incident in it, the violet wavelengths will diverge more strongly than red wavelengths. Realizing the opposing effect that chromatic aberration can produce in positive and negative lens, a lens that is composed of 334 a positive and a negative lens cemented together will have the chromatic aberrations canceled out. Such a lens is called an achromat or achromatic doublet. Using a combination of flint and crown glasses, an achromat can be constructed with minimum chromatic aberration, yet can serve overall as a positive lens. An example of an achromat is shown in Figure 11.51. Figure 11.51 An achromat. In building optical systems, one needs to be aware of the possible aberrations. While a single lens is often enough for imaging purposes, the system soon becomes complex as many additional lenses are introduced to overcome the different aberrations. Realize that greater the number of lenses, the light intensity is reduced somewhat. Antireflection coatings (see the chapter on Wave Optics) are used on lenses to minimize Fresnel reflections. 11.4 Suggested Laboratory Experiments7 A. Stops Use a single positive lens and an object about 5 mm tall, illuminated by an expanded He:Ne beam. Choose an object location such that a clear, slightly diminished image is obtained. Do not move the object or lens from this point onwards. (i) Use a front stop with an opening approximately half the lens diameter. Place it 2 cm to the left of the lens. Place a power meter at the image location. Do not move the power meter from now on. Read the power. Move the stop away from the lens in 2 cm increments, taking a power reading at each point. Draw a graph of stop-to-lens distance on the horizontal axis versus measured power on the vertical axis. What are your conclusions? (ii) Use a rear stop with an opening approximately half the lens diameter. Place it 2 cm to the right of the lens. Place a power meter at the image location. Do not move the power meter from now on. Read the power. Move the stop away from the lens in 2 cm increments, taking a power reading at each point. Draw a graph of stop-to-lens distance on the horizontal axis versus measured power on the vertical axis. What are your conclusions? B. Optical Instruments: Astronomical Telescope (Keplerian type): Use a millimeter graph paper as an object affixed to a frosted glass screen and illuminated by an incandescent lamp from the back. Use an achromat as an objective; determine using a screen the location of the inverted image. The object should be far away from the objective that the image will be highly diminished. Place an eye piece lens so that the screen is within its focal length. Remove the screen; adjust the eye piece 335 location to observe a large, inverted image of the graph paper through the eye piece. Count the number of millimeters on the original graph that correspond to one millimeter on the image seen through the eye piece. This may be tricky since this involves use of both eyes: one looking at the object, one at the virtual image. The number of millimeters counted on the object that corresponds to one millimeter of the image provides the angular magnification of the system C. Aberrations: Use a large positive lens for this experiment (10 cm or larger in diameter). Illuminate an object using an expanded He:Ne beam. Place a center stop (a glass plate whose center is coated black will serve this purpose). Determine the focal length using the procedure from Chapter 10. Replace the center stop by an edge stop (an edge stop can be a glass plate coated black so that it allows only a small portion at the center to let light through). Determine the focal length using the procedure from Chapter 10. Compare the focal lengths obtained from the two stops. The difference provides an indication of spherical aberration. Repeat these experiments using white light source. Comment on the results. • • • • • Chapter Summary The limiting aperture is that physical element that controls the amount of light into an optical system. This is sometimes accomplished by the use of a front stop, a middle stop, a rear stop (or a field stop). The element that is the limiting aperture need not be the stop. It could be any element in that system. To determine which element it is, image every element through the optics to the left of it (assume always that light travels from the object from left to right). That element, the image of which subtends the smallest angle at the foot of the object is considered the limiting aperture that limits the amount of light from the object to the image plane. Once the limiting aperture is correctly identified either through ray tracing or trigonometric calculations, ray tracing is performed through the system starting from the head of the object. The chief ray travels through the center of the entrance pupil (the image of the limiting aperture through the preceding optics), the center of the limiting aperture itself, and the center of the exit pupil (the image of the limiting aperture through the succeeding optics). The marginal rays travel at the extremities of the entrance pupil, the limiting aperture and the exit pupil. The intersection of the chief ray and the two marginal rays provides the location of the image of the object’s head. The object’s foot is imaged, as always, as a point on the optical axis, in line vertically with the image’s head. A simple magnifier uses a single positive lens. The angular magnification is given by 25/f, where f is the focal length in centimeters. This assumes that the magnified, erect and virtual image is obtained at the near point of the human eye, which is traditionally assumed to be 25 cm (10”) from the eye. In order to obtain higher magnifications, two lenses need to be used, and the set up is called a compound microscope. The lens closest to the object which gathers light from the object is called the objective. The lens closest to the observer is known as the eye piece or ocular. The overall system magnification of a 336 • • • • • • compound microscope is given by the product of the angular magnification of the eye piece and the lateral magnification of the objective. The enlarged, inverted and virtual image of the original object is best viewed at the near point of the observer’s eye. An astronomical telescope also uses an objective and an eye piece, although somewhat differently. The object is essentially at infinity. The objective gathers the light from the object and forms a small, inverted and real image close to its secondary focal point. By placing the ocular (eye piece) so that the first image is within the eye piece’s focal point, an enlarged, inverted, and virtual image is obtained best visible at the near point of the observer’s eye. The astronomical telescope can be made in two versions. While both use a positive lens for the objective, the Keplerian type uses a positive lens for the eye piece, whereas the Galilean type uses a negative lens. In either case, the astronomical telescope can be converted to a terrestrial telescope by appropriately placing a third lens between the objective and ocular for inverting the image. A prism can also be used to accomplish this. Two such terrestrial telescopes employing prisms can be used to form a pair of binoculars. The principles behind an astronomical telescope can be used to construct a laser beam expander (also called collimator). Again, a combination of two positive lenses whose distance from each other is the sum of the focal lengths will result in a Keplerian collimator. Using a negative lens as an objective and a positive lens as the eye piece such that the former has a smaller magnitude of focal length, a Galilean collimator can be constructed. In either case, the ratio of the diameters of the expanded beam to the original beam is given by the ratio of the focal length magnitudes of the eye piece to the objective. The divergence of the expanded beam is reduced inversely by the same ratio. Aberrations are caused by departures from ideal, simplified equation predictions for image location and size. The small angle approximation resulted in the lens maker’s equation and the thin lens equation. When the assumptions are violated, aberrations result. Monochromatic aberrations occur for single wavelength light. Spherical aberration results from curved nature of the lens surfaces and is most severe for non-paraxial rays (rays farther away from the optical axis). Off-axis rays, even if parallel to each other, will result in a comet shaped image with a tail. This aberration is called coma. Astigmatism is an aberration where clarity of vertical and horizontal features of the image occurs at different image plane locations, and a compromise, the circle of least confusion has to be found. The image is seldom confined to a plane; rather, there exists a curvature of field. Distortion is an aberration that manifests itself as different lateral magnifications to different points on the object. Most aberrations can be minimized via the use of multiple lenses nearly canceling one another’s distortion and use of an appropriate aperture. Chromatic aberration occurs due to the wavelength dependence of the refractive index of lens materials. This results in the lower wavelengths focusing closer to the lens and the longer wavelengths farther away, for a positive lens. Use of an achromat doublet minimizes chromatic aberration. 337 End of Chapter Exercises 1. A positive lens of focal length 10 cm and 5 cm in diameter has a rear stop 1 cm from it. The rear stop is 4 cm in diameter. (a) Calculate the angles subtended by the image of each element at the foot of the object 1 cm tall object standing 6 cm to the left of the lens. (b)Which element is the aperture? (c)If the object were 2 cm tall, which element would now be the aperture? 2. Identify the aperture for the following system shown in Figure 11.52 (not to scale): f1= 20 mm L1, diameter=20 mm f2=25 mm L2, diameter=20 mm STOP Object 3 mm high dia.=10mm 35 mm 6 mm 14 mm Figure 11.52 Illustration for problem 2. 3. A simple magnifier has a power of 10D. For an object placed within its focal point, and the image formed at the near point, calculate (a) the angular magnification, and (b) the lateral magnification. 4. The objective and eyepiece of a compound microscope are 18.5 cm apart. Their respective focal lengths are 2.5 cm and 0.5 cm. If the final image is formed at infinity, calculate (a) the object location from the objective, (b) the lateral magnification of the objective, and (c) the overall magnification of the system, assuming that the final image is at infinity. 5. Design a Keplerian collimator such that it expands the beam 5X. The collimator length shall not exceed 42 cm. 6. Why are multiple lenses used in optical systems? Give all the reasons that apply. 7. Of the following lenses, which would be most suitable to minimize both spherical aberration and coma: (a) an equi-biconvex lens, (b) a plano convex lens with the flat side facing the light, (c) a plano convex lens with the curved side facing the light, (d) a positive meniscus lens. 8. An eye piece free of distortion is called ____________. 338 Group Discussion Questions These require the students to form small groups of three or four and discuss each question. Some questions may require you to do a search on the Internet and find the answers. 1. Why are stops used in imaging? Name the stop that all of us carry on us. 2. Describe a situation when a lens is the limiting aperture, the entrance stop and the exit stop. 3. Why are the marginal rays so called? 4. Describe the use of negative lenses in optical instruments. 5. Find someone who wears eye glasses (not contacts). Determine the focal length, the focal power and the type of correction the glasses provides. (Caution: do not assume that eye glasses have a uniform and unique radius of curvature.) 6. Discuss the types of aberrations that you have encountered in every day life experience with optics and optical instruments. References: Jenkins,F. and H. White. “Fundamentals of Optics.” Singapore: McGraw Hill, 1981. 2 Welford, T. “Optics.” Oxford: Oxford University Press, 1988. 3 Bureau of Naval Personnel. “Basic Optics and Optical Instruments.” New York: Dover Publications, Inc., 1969. 4 Sears, F.W, M. W. Zemanksy and H.D. Young. “College Physics.” Reading: AddisonWesley Publishing Company, 1991. 5 Melles Griot Product Information, http://mellesgriot.com/ 6 Laser Beam Products Information, http://www.lbp.co.uk/ 7 Gottlieb, H. “Experiments Using a Helium Neon Laser.” Bellmawr: Metrology Instruments, Inc. (1981) 1 339 Chapter 12 Introduction to Wave Optics 12.1 Huygen’s Principle 12.2 Interference 12.3 Diffraction 12.4 Suggested Laboratory Experiments Chapter Summary End of Chapter Exercises Chapter Objectives After completing the chapter, the student should be able to (i) describe Huygen’s principle (ii) explain the phenomenon of interference of light waves. (iii) explain the phenomenon of diffraction Key words: Huygen’s Principle, interference, diffraction – Fresnel and Fraunhofer, near field, far field, spherical waves, planar waves, Airy disk, ARC 12.1 Huygen’s Principle A point source of light would radiate light waves traveling in all directions, radially outward. This results in spherical waves. Realize that a point source is an idealized concept because the source has zero dimensions, yet it radiates light! A light source such as a fluorescent lamp is an extended source, since it is made up of infinite number of point sources. However, an extended source, which is far away, can act as a point source. It is the distance of the source from the viewer as well as the wavelength of light that determines if a light source can be treated as a point source or extended source. A point source emanating spherical waves is shown in Figure 12.1. The waves travel in all three dimensions, and, hence, the name spherical waves. However, when one looks at waves some distance away from the source, the waves have lost most of their curvature. This is evident from the diagram above, when for the fourth set of waves from the source illustrated, the radius from the origin is so large enough that the circular shape of the wave is not evident and the waves look more linear. While a point source generates spherical waves, after a certain distance, the waves become planar, i.e. become parallel to each other. A case in point is the sun, which, by virtue of its distance from the earth can be considered a point source. While the waves emanating from the sun are spherical in nature because the sun is a sphere, by the time they reach the earth, we essentially consider the beam of light to be collimated, i.e., rays of light parallel to each other. Collimated light thus obtained, or from a laser, produce planar wave fronts, as illustrated in Figure 12.2. Huygen’s principle states that when planar wave fronts strike an obstacle such as an aperture or an obstacle edge, those points of contacts themselves act as point sources for 340 secondary light waves to be created1,2. While rectilinear propagation theory of light discussed in detail in Chapter 5 is sufficient to explain the phenomena of reflection and refraction, only wave theory of light, combined with Huygen’s principle, can explain unusual phenomena such as interference and diffraction, which are discussed in detail in the next two sections. Point Source Figure 12.1 Spherical waves from a point source. Collimated light Spherical wave fronts Planar wave fronts Figure 12.2 Spherical waves versus collimated light. Huygen’s principle is illustrated in Figure 12.3. Slit/aperture Planar waves Spherical waves Figure 12.3 Illustration of Huygen’s priciple. Edges of obstacles serve as secondary point sources of light as well, producing new spherical wave fronts. This phenomenon can be appreciated when one looks at an 341 incandescent bulb with a circular shade behind it, hanging from the ceiling, as shown in Figure 12.4. If light traveled only in straight lines, then one would see a single, perhaps slightly larger, shadow of the shade on the ceiling. In reality, one does observe a few circular shapes of “shadows” delineated by light and dark areas as illustrated below. These are known as fringes, and, are clearly caused by the edge of the lamp shade acting as an obstacle to light originating from the lamp. Ceiling Fringes Lamp with shade Figure 12.4 Diffraction patterns caused by the edge of the lamp shade. 12.2 Interference Two light waves are shown in the diagrams in Figure 12.5. They are both in phase and traveling in the same direction. The net effect of these two waves can be obtained by adding the two waves, shown in Figure 12.6. Different amplitudes are chosen, on purpose, of each wave. Wave #1 10 5 0 0 -5 -10 10 20 30 40 50 60 70 Wave #2 (in phase) 15 5 -5 0 10 20 30 -15 Figure 12.5 Waves in phase. 40 50 60 70 342 In this case, since the peaks and ebbs of the two waves line up in space, they constructively add, leading to larger amplitude for the combined wave, i.e., all points on the combined wave are enhanced, compared to the original individual waves. Combined Waves 1 and 2 25 15 - -15 -25 0 10 20 30 40 50 60 70 Figure 12.6 Interference of waves in phase. On the other hand, if the waves are completely out of phase, i.e., the peak of one wave coincides with the ebb of the second wave, or vice versa, the waves destroy each other to some extent when added, as shown in Figure 12.7. Wave #1 10 5 0 0 -5 10 10 20 30 Wave#2 40 50 60 70 (out of phase) 15 5 -5 0 10 20 30 40 50 60 70 -15 Figure 12.7 Waves out of phase. The amplitude of the added wave is much lower than each individual wave’s, as shown in Figure 12.8. This phenomenon of constructive and destructive interference is present in wave motion. The waves can be mechanical (as in sound waves or vibrations in a string or rope) or electromagnetic, as in light waves. The illustrations shown above are for two extreme cases of constructive and destructive interference. In reality, the two waves can 343 be out of phase by any amount, and, one would have a net result that corresponds to an in-between situation between constructive and destructive. Combined waves 1 and 2 5 0 0 10 20 30 40 50 60 70 -5 Figure 12.8 Interference of waves out of phase. How does one create two light waves that are out of phase using a coherent source? A coherent source, such as a laser, by definition, produces waves that are spatially and temporally in phase. A scientist, by name Young, demonstrated interference phenomenon using his famous double slit experiment, shown in Figure 12.9. As we know from Chapter 4, one of the properties of the laser is coherence. However, the laser was not invented until the 1960’s. How did Young obtain coherent light? The answer is simple. All light sources are coherent to a certain extent, i.e., the distance from the source up to which point light remains essentially coherent (known as the coherence length) varies depending on the light source. The coherence length of lasers can be up to several tens of meters and much higher. However, using a lamp which has a coherence length of a few centimeters, it is possible to place the slit within the source’s coherence length. That is how Young got away with it! A slit by definition is narrow in one dimension and long in the other dimension. Basically, it is a narrow rectangular opening. The width is very small, typically fractions of a millimeter, and of the order of magnitude as the wavelength of light. The slit length is of the order of a few millimeters to centimeters, which is much larger than the wavelength of light. Even though coherent light is incident on the slit from the left side as shown in the diagram below, each slit, per Huygen’s principle, acts like a secondary source of light, creating spherical waves. Thus, while light incident on the slit may be composed of planar waves, the light exiting the slits are spherical waves, and can interfere with each other1-3. In Figure 12.9, spherical or planar waves are traveling from left to right. The waves are coherent. There is no phase difference among the waves as they strike the double slits marked S1 and S2 because the slits are at the same distance from the source. Immediately upon striking the slits, as per Huygen’s principle, spherical waves are created on the right side of the slits. The waves generated from each slit will interfere with each other, provided the slits are reasonably close to each other. When the peak of a wave from S1 interferes with the peak of a wave from S2, then constructive interference happens. When the peak of a wave from one of the slits interferes with the ebb (valley) of a wave from 344 the other slit, destructive interference occurs. Thus, a series of constructive and destructive interferences occur, leading to bright and dark spots in the intensity of light. Screen P Double slit Spherical or θ S1 d B C Planar wave fronts y θ A K S2 L Q Figure 12.9 Young’s double slit experimental set up. When one places a screen PQ to capture the light at a distance infinitely far away from the slits, one would see dark and bright lines on it caused by interference. Consider an arbitrary location on the screen, and call it B. Connect B with the two slits, resulting in two straight lines S1B and S2B, respectively. Connect the point B with C, the point midway between the two slits. The line CA establishes the horizontal, i.e., the optical axis. The line CB makes an angle θ with the optical axis. Draw a perpendicular from S1 to S2B. S1K is now perpendicular to S2B and making the same angle θ with S1S2. If d is the distance between the two slits, then, looking at triangle S1S2K, S2K = dsinθ Realize that for the arbitrary location B, S2K is the additional distance that a wave from slit 2 will have to travel, compared to a wave from slit 1. Now, we can decide whether light hitting point B will have a bright (constructive interference) or a dark (destructive interference) spot by considering the quantity dsinθ . We know that if two waves have a path difference ofλ, 2λ , 3λ , 4λ , etc., the waves essentially look alike, however, are displaced by a whole number of wavelengths. These waves will constructively interfere with each other and produce a bright spot. On the other hand, if waves have a path difference of λ/2, 3λ/2, 5λ/2, 7λ/2, etc., then, they undergo destructive interference, and a dark spot will be observed. Thus, for any arbitrary location B on the screen, if a bright fringe is to be obtained, dsinθ= mλ with m=0, +1, +2, +3, +4, etc. 345 where m is an integer. The positive integers indicate locations on the screen above the optical axis where a bright fringe is observed, while negative integers denote locations below the optical axis. For any arbitrary location B on the screen, if a dark fringe is to be obtained dsinθ= mλ with m=+1/2 , +3/2, +5/2, +7/2, etc. where m is now a half integer. The positive half integers indicate locations on the screen above the optical axis where a dark fringe is observed, while negative half integers denote locations below the optical axis. m is also called the order of the fringe. From these relationships, we see that at a point corresponding to the optical axis, we notice a bright fringe. Then, alternate dark and bright fringes are obtained on either side of the central bright fringe. One would have expected to see a dark fringe at the point A since it appears to be in the “shadow” region of either slit. However, due to interference, just the opposite is observed. The location of the bright and dark fringes can be easily obtained as follows. From Chapter 5, we use the small angle approximation of sinθ = θ = tan θ. This is allowed since the distance L is relatively large compared to any other dimension. Realize that ym = L tanθ ym = L mλ d where ym =distance of the mth fringe from the central bright fringe d =distance between the two slits λ =wavelength of light L =distance from slits to screen If one uses integral values for m (i.e., m=0, 1, 2, 3...) one obtains locations of the bright fringes, i.e. y0, y1, y2, y3, etc. If one uses half-integral values for m (i.e. m=1/2, 3/2, 5/2, etc.), one obtains locations of the dark fringes, i.e. y1/2, y3/2, y5/2, etc. The fringes are symmetric with respect to the center line. Remember, the angle mentioned in these equations must be in radians, not in degrees. The small angle approximation holds only when angles used are measured in radians. The opening of the slit must be of the order of magnitude as the wavelength of light. Its exact value is not as important and does not appear in the equations. The location of fringes would depend upon the slit spacing, slit to screen distance and the wavelength of light, in addition to the fringe order. The resultant pattern on the screen would appear as shown in Figure 12.10. The figure is highly exaggerated for clarity and only a few fringes are shown. 346 y1/2 Center line y-2 Figure 12.10 Interference fringes from a double slit. Note that the fringe location of any order fringe, bright or dark, is measured from the center line (i.e. of the central bright fringe) to the center line of that fringe. The distance between any two successive bright fringes is equal to the distance between any two successive dark fringes and is given by Lλ/d. Each bright fringe is of the same intensity as its neighboring bright fringe. However, depending upon the situation, one usually sees a few fringes clearly and the others appear too faint to be clearly discernible. Thus, the fringe spacing varies directly as the slit-to-screen distance and wavelength of light and inversely as the slit spacing. Example 12.1 An argon ion laser with output wavelength of 514.5 nm illuminates a double slit 50 µm wide and 0.25 mm spacing. The fringes are seen on a screen 2 m from the slit. Calculate (a) the distance between adjacent bright fringes, (b) the distance between adjacent dark fringes, (c) the location of the third bright fringe, (d) the location of the fourth dark fringe, (e) the angle corresponding to the first bright fringe and (f) the angle corresponding to the second dark fringe. (a) and (b): the distance between successive bright or dark fringes is the same and is given by Lλ/d = 2 m x 514.5 x 10-9 m 0.25 x 10-3 m = 0.00412 m = 4.12 mm Answer (c) m=3 y3 = 2 m x 3 x 514.5 x 10-9 m 0.25 x 10-3 m = 12.3 mm Answer 347 (d) ym = L mλ d with m=7/2 y7/2 = 2 m x (7/2) x 514.5 x 10-9 m 0.25 x 10-3 m = 14.4 mm Answer (e) dsinθ= mλ with m=0, +1, +2, +3, +4, etc. Here m=1 for the first bright fringe θ1= sin-1(mλ/d) = sin-1 (1 x 514.5 x 10-9/0.25 x 10-3) = 0.00206 radians = 2.06 milliradians = 0.118o Answer (f) dsinθ= mλ with m=+1/2 , +3/2, +5/2, +7/2, etc. Here m=3/2 for the second dark fringe θ1= sin-1(mλ/d) = sin-1 (1.5 x 514.5 x 10-9/0.25 x 10-3) = 0.00309 radians = 3.09 milliradians = 0.177o Answer Example 12.2 Using a certain type of laser, it is observed that adjacent bright fringes are 1.90 mm apart on a screen 1.5 m away from a 0.5 mm spacing double slit. What is the wavelength of the laser? Fringe spacing = Lλ/d Rearranging, we obtain λ = (Fringe spacing) x d L = 1.90 x 10-3 x 0.5 x 10-3 1.5 = 633 nm Answer Exercise 12.2A 1. Can you obtain interference fringes with just a single slit? Explain. 2. It is found that ten successive bright fringes are found to occupy 5.89 mm distance when measured accurately with a micrometer. If the screen is 1 m away from a double slit illuminated by a sodium lamp giving our 589 nm wavelength yellow light, what is the slit spacing? 348 3. A He-Ne laser with output wavelength 633 nm illuminates a double slit with 0.05 mm opening and 0.25 mm spacing. How far the screen should be placed so that the third bright fringe is located exactly 20 mm from the central bright fringe? 4. A sodium lamp with output wavelength of 589 nm illuminates a double slit 50 µm wide and 0.25 mm spacing. The fringes are seen on a screen 1.5 m from the slit. Calculate (a) the distance between adjacent bright fringes, (b) the distance between adjacent dark fringes, (c) the location of the third bright fringe, (d) the location of the fourth dark fringe, (e) the angle corresponding to the first bright fringe and (f) the angle corresponding to the second dark fringe. 5. Using a certain type of laser, it is observed that adjacent bright fringes are 1.90 mm apart on a screen 75 cm away from a 0.25 mm spacing double slit. What is the wavelength of the laser? 6. A He:Ne laser with output wavelength of 633 nm illuminates a double slit 50 µm wide and 0.50 mm spacing. The fringes are seen on a screen 2.0 m from the slit. Calculate (a) the distance between adjacent bright fringes, (b) the distance between adjacent dark fringes, (c) the location of the third bright fringe, (d) the location of the fourth dark fringe, (e) the angle corresponding to the first bright fringe and (f) the angle corresponding to the second dark fringe. 7. Using a 488 nm wavelength output laser, calculate the fringe spacing of adjacent dark fringes. The screen is 75 cm away from a 0.50 mm spacing double slit. 8. If the fourth dark fringe is 3.40 mm away from the central bright fringe located on a wall 1 m away from a double slit of 0.5 mm spacing, what is the wavelength of the light used? Thus far, we have considered monochromatic light sources. The observed interference pattern will exhibit dark and bright hues of the color of the laser light. What would happen to the fringes if one uses white light? Recall the formula ym = L mλ d Since white light is composed of 400nm-700nm wavelength range, upon interference, light undergoes dispersion. Within each bright fringe order (i.e. with m fixed), the violet light will be seen relatively closer to the central bright fringe compared to the red light. Thus, using collimated white light, one would see a few rainbow patterns, the color separation becoming more obvious with higher order fringes. This explains how one sees a rainbow pattern on a thin gasoline film floating on a puddle of water. The thin film results in two light waves interfering with each other, one reflected off the air-gasoline interface and the second off the gasoline-water interface. Since the incident light is white, the interference pattern manifests itself as a rainbow hue. Several fringes can be observed, some straight and some curved, depending on the evenness of the film. Interference caused by reflected beams is illustrated in Figure 12.11. 349 Incident light Reflected waves Thin film Substrate Figure 12.11 Interference from a thin film. The top film has to be thin enough that the light reflected off the substrate interface has sufficient intensity to interfere with the reflected light off the thin film surface. If the incident light is monochromatic, the reflected waves from the two surfaces create an interference pattern of bright and dark fringes. By measuring the fringe spacing, and knowing the wavelength of the light source used, one can calculate the unknown thickness of the film. This principle is called interferometry. Interferometers find applications in the semiconductor industry and other industries, where a coating is either deposited or being etched. This results in the coating thickness changing with time, and, consequently, the resulting interference pattern. By looking at the interferogram, one would know the end point of the process, i.e. when to shut off the coating or etching tool. Note that unlike the double slit experiment where the path difference increases with the angle for the double slit, it decreases with the angle for the thin film. The student is encouraged to experiment with air wedges that can be created using simple microscope slides to understand why and provide an explanation. Anti-reflection coatings (ARC): We know that dielectric materials such as glass reflect a portion of the light back, which, at normal incidence, is about 4%, for typical glass with refractive index of 1.5. It is desirable to minimize this Fresnel reflection. Coating lens and window surfaces with a material such as magnesium fluoride (MgF2) with refractive index of about 1.38 at 589 nm wavelength would minimize Fresnel reflection via interference. Antireflection (AR) Coating Dielectric material Reflected light waves 1 2 d Figure 12.12 Interference from an anti-reflection coating film. 350 In Figure 12.12, if the thickness of the AR coating is such that the phase difference between the two reflected waves, (1) one off the AR coating surface and the second (2) off the dielectric material surface, is exactly one-half the wavelength of light inside the coating material, then the reflected waves undergo destructive interference and cancel each other out. This results in practically zero net Fresnel reflection. Since the wave 2 travels an additional distance equal to twice the thickness of the AR coating material, for destructive interference to occur, we demand that1 2d = λARC 2 d = λARC 4 where λARC refers to the wavelength of light inside the ARC material, which, based on discussions from Chapter 5 is given by λARC = λair n where n is the refractive index of the ARC material. This formulation also holds for coating curved surfaces such as lenses. The ARC is deposited after the window or lens is ground to the required curvature and flatness specifications. Magnesium fluoride (MgF2) is a common ARC material. Example 10.3 What should the ARC coating thickness be for a glass lens to be used in conjunction with He:Ne laser light (633 nm wavelength)? Assume that the refractive index of ARC is 1.378 at the laser wavelength. λARC = λair n = 633 nm/1.378 = 459 nm It is not a surprise that the red light is now bluish in color inside the ARC! d = λARC 4 = 459/4 = 115 nm Answer Clearly, the optic has to be coated precisely with 115 nm of ARC thickness and uniformly to take maximum advantage of destructive interference. 351 A material whose refractive index lies somewhere in the vicinity of the geometric mean of the refractive indices is considered optimum. Geometric mean of two numbers is obtained by multiplying the two numbers and taking the square root of the product. ARC’s are also used in applications where standing waves are a problem. For illustrative purposes only, let us look at a mechanical wave such as the one generated by a rope tied to a pole. If one tugs on the rope slightly and release the pressure, sine waves can be created that appear to travel from your hand up to the pole. When it reaches the pole, the wave is reflected back. The superposition of the incident and reflected wave leads to a standing wave pattern as illustrated in Figure 12.13. Illustration of standing waves 1 0.8 0.6 0.4 0.2 0 0 10 20 30 40 50 60 70 - 0.2 - 0.4 - 0.6 - 0.8 -1 Post Figure 12.13 Standing waves. The incident wave is shown using solid line and the reflected wave using dashed line. At any instance in time, to an outside observer, it would appear as though there are two waves, both stationary. Imagine now, that instead of a mechanical wave, we have a light wave that is incident on a reflective surface after traversing through a film. Standing waves would be generated, with the wave pattern as shown above. The intensity of light is related to the square of the amplitude. Thus, no matter if the wave is located above or below the central reference line, the square is always positive, leading to light intensity distribution in the film as shown in Figure 12.14. If the film is photoresist material used in the semiconductor industry for imaging applications, the light intensity inside the film will not be uniform, rather wavy as illustrated above. This is not a desirable situation since the film undergoes photochemical reactions just like a photographic film, and the rate of reaction inside the film would also vary in a wave-like fashion. Thus, areas corresponding to peaks in light intensity will undergo complete reaction or even overexposure, whereas areas that receive minimum light intensity will undergo incomplete reaction or underexposure. When the film is developed out, the development process will result in wavy or striation patterns, rather than uniform and straight features. Use of ARC on the reflective surface would suppress the standing waves, restoring more uniform light intensity throughout the film. 352 Light intensity profile due to standing waves 2 1.8 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0 0 10 20 30 40 50 Film 60 70 Reflective substrate Figure 12.14 Intensity variation due to standing waves. Thin films of air can also bring about interference patterns or fringes. Two microscope glass slides flush with each other, however, separated at one end with a wedge (for example, by a piece of masking tape) can create linear fringes when illuminated normally and viewed normally with respect to the slides. On the other hand, if a curved surface such as a lens sits on a flat surface such as a glass plate, one obtains circular fringes, known as Newton’s rings. By measuring the distance between adjacent bright or dark fringes, and knowing the degree of curvature of one surface, one can calculate the degree of curvature (or flatness) of the second surface. A few examples of the fringe patterns obtained for various situations are shown in Figure 12.15 4. If a monochromatic light source such as a sodium lamp or a laser is used, it is easier to discern the dark and fright fringes and make accurate measurements of the spacing. On the other hand, if a white light source is used, the bright fringes would exhibit a dispersion (rainbow) pattern making it difficult to measure the spacing. In the figures below, the bottom piece, B is perfectly flat, to the degree of flatness one can obtain in real world. Master flats are sold with flatness specifications of a tenth wavelength, a twentieth of a wavelength, etc. Greater the degree of flatness, higher is the cost. The test piece is marked A. Checking flatness and curvature is another area where the phenomenon of interference is used. A whole set of reference interferograms are available to test convex, concave, cylindrical, saddle and spheroidal surfaces4. By checking the observed pattern against the reference, one can arrive at the type of curvature and the magnitude. 353 Top view of the fringes Test piece in contact with the master A B A and B both “perfectly” flat Straight fringes A Newton’s Rings B is “perfectly” flat; A is curved B Curvature= 1 fringe spacing A B Figure 12.15 Typical interferogram patterns. 12.3 Diffraction Clearly, interference occurs only if there are two light waves, originally from the same source, however, exhibiting a phase difference due to their path difference. Let us go back to the rectilinear light propagation theory and examine Figure 12.16 where light travels through an opening. If the opening size is relatively large in comparison to the wavelength of light, for example, aperture size of a few millimeters in diameter or higher, one can expect the light to progress more or less in a straight line as it passes through the aperture opening2. Collimated light Shadow regions Bright region Large size aperture Figure 12.16 “Large” aperture follows geometric optics considerations. When collimated light with planar wave fronts is used, one obtains more or less predictable bright and “shadow” regions on a screen. This is because the aperture size 354 (order of millimeters or bigger) is much larger than the wavelength of light (order of few hundreds of nanometers). When the size of the aperture or opening through which light passes is reduced, so that it approaches its wavelength (or several times the wavelength), a phenomenon called diffraction takes place. Instead of traveling in straight lines as geometric optics suggests, light spreads, and the size of the image obtained on a screen is much larger than the size of the opening with accompanying fringe pattern of light and dark areas. Light spreads in whichever direction it is constricted. For example, with a horizontal slit, light with planar wave fronts spreads in the vertical direction and with a vertical slit, light spreads in the horizontal direction, as shown in Figure 12.17-18. Incoming light Vertical slit Light spreads horizontally Figure 12.17 Diffraction from a vertical slit. Incoming light Horizontal slit Light spreads vertically Figure 12.18 Diffraction from a horizontal slit. Diffraction, in simplistic terms, is bending of light around an obstacle that constricts it. In a lighter vein, and philosophically speaking, it appears as though light is free-spirited. When unconstrained, so that obstacles are out of its way, light is viewed as traveling in a straight line unperturbed, until it encounters a different medium (refraction.) The moment the path of the light is constricted by an obstacle, caused by a small size opening, whose dimensions approach the wavelength of light, the wave nature of light manifests itself and light bends as if protesting or misbehaving! Diffraction occurs when a portion of the wave front, is obstructed or constrained in some way. During this process, when a portion of wave front in altered in phase, or amplitude, 355 it results in an intensity distribution called the diffraction pattern. Thus, the diffraction phenomenon is a deviation from geometric optics. Diffraction is distinguished from interference as follows: interference considers essentially two wave fronts of different phases whereas diffraction is a single wave phenomenon. Both are grounded on Huygen’s principle which states that when light waves encounter an obstacle, the multitude of points that makes up the obstacle or the opening themselves serve as sources of new, secondary, spherical wave fronts. While diffraction is a single wave effect, using Huygen’s principle, it can be modeled as multi-wave effect. Depending upon the location of the source, the opening and the screen, two types of diffraction phenomena are observed. If the source-to-slit distance (denoted by Z), or the slit-to-screen distance (denoted by Z’), or both, are small, then Fresnel diffraction results. If the source-to-slit distance and the slit-to-screen distance are both infinite, then, Fraunhofer diffraction results. The words “small” and “infinite” need to be clarified. The ratio of the square the aperture size to the wavelength of light is used as a guide to determine what type of diffraction will take place. If the characteristic dimension of the aperture is d, and the wavelength of light is λ, then, if 1,2,5,6 Z >> d2 and Z’ >>d2 λ λ then, Fraunhofer, or far field diffraction will occur. On the other hand, if Z ~ d2 or Z’ ~ d2 λ λ then, Fresnel, or near field diffraction will occur. While the symbol “~” means “of the order of,” Fresnel diffraction will also occur if either Z or Z’ is less than d2/λ . In between these two extreme cases of Fresnel and Fraunhofer diffraction, there exists an uncertain area where either theory is only approximate. Some authors arbitrarily chose a value of 100 or so to delineate, while others are satisfied with a value of 10, based on the approximations to the governing mathematical equations. This is for simplicity of calculations, since ideally Z and Z’ must be infinite for far field diffraction to occur. Notice that we have been concentrating more on far field diffraction than near field diffraction. This is because the mathematics associated with Fraunhofer diffraction is much simpler than Fresnel diffraction, and, therefore, we will examine only far field diffraction during the remainder of this section. Fresnel theory, while mathematically complex, can provide correct results for both near and far field, the much simpler Fraunhofer approach is used for far field calculations. Before presenting the relevant equations for far field diffraction, a comparison of the near field and far field diffraction is in order and is presented in Table 12.16. 356 Far field diffraction (Fraunhofer) Table 12.1 Near field diffraction (Fresnel) 1. Both Z and Z’ are essentially infinite. 1. Either Z or Z’ is finite 2. Planar wave fronts arrive both at the aperture and the screen 2. Spherical wave fronts arrive at either the aperture or the screen or both 3. The size of the diffraction pattern observed on the screen changes uniformly and proportionately when Z’ is changed. There is no change in the diffraction pattern shape. 3. When Z’ is altered, the diffraction pattern changes both in shape and size. The aperture itself can be a long, narrow slit, a rectangle, a square, a hexagon, an octagon, or a circle. One obtains various far field diffraction patterns in each case. Example 12.4 A 100 µm diameter wire (2 cm in length) is in the path of a CO2 laser beam (λ=10.6 µm. Estimate Z and Z’ for far field diffraction conditions to be satisfied. The length of the wire is much larger than the diameter such that the diffraction is introduced only the wire diameter dimensions. Let us calculate the ratio d2 = (100 x 10-6)2 = 943 µm ~ 1 mm. Thus, Z or Z’ >> 10 cm would suffice. Answer λ 10.6 x 10-6 Any time a laser is used as light source, one does not need to be overly concerned about far field calculations for Z, since we have a collimated beam to begin with, and only planar wave fronts arrive at the aperture. Only Z’ needs to be calculated to verify Fraunhofer conditions. Alternatively, if laboratory space is limited, when one uses a non-collimated source as shown in Figure 12.19, place a positive lens so that the source is at the focal length of the lens1,2. This results in collimated light emerging from the lens. Again, for this case, calculation of Z is unnecessary. Only Z’ needs to be calculated to verify Fraunhofer conditions. Example 10.5 A sodium lamp (589 nm wavelength) illuminates a circular aperture 75 µm in diameter. Calculate the lamp to aperture distance and the aperture to screen distance so that Fraunhofer diffraction patterns will result. d2 = (75 x 10-6)2 = 5.6 x 10-9 m2 357 Ratio = 5.6 x 10-9 m2 = 0.0096 m = 9.6 mm 589 x 10-9 Collimated light Lamp Spherical waves Planar wave fronts obtained Lens focal length Figure 12.19 Obtaining far field conditions using a positive lens. Clearly, a one-meter distance for both Z and Z’ will satisfy Fraunhofer conditions, since it far exceeds the ratio calculated above. These distances are realizable easily on a laboratory bench. However, this is not always true. When one comes up with values that are several meters in length, then there are two practical approaches. (1) By placing the source S at the primary focal point of a lens A and by placing a second lens B, and focusing on a screen, the requirements for both Z and Z’ to be large are automatically met2. This is because the source and the screen essentially move to infinity, as illustrated in Figure 12.20. S FA A Slit B Screen FB Figure 12.20 A practical arrangement to obtain far field diffraction in limited space (2) If on uses a laser as a source, then the requirement for Z is met with little difficulty. This is because the divergence of a laser beam is rather small, and collimated light comes out of a laser. Exercise 12.3A In each problem below, estimate the source-to-aperture distance and aperture-to-screen distance for far field diffraction to occur: (Note: For rectangles use length x width dimensions in place of d2.) 358 Problem number 1 2 3 4 5 6 7 8 Wavelength of source Aperture details 514.5 nm 632.8 nm 589 nm 488 nm 589 nm 589 nm 589 nm 589 nm Circle, 100 µm in diameter Square, 100 µm on the side Rectangle, 100 µm x 150 µm Long, narrow slit with 125µm opening Square, 200 µm on the side Circle, 150 µm in diameter Long, narrow slit with 125µm opening Rectangle, 200 µm x 300 µm Light passing through a long, narrow slit will undergo diffraction, and if far field conditions are satisfied, the diffraction patterns, as well as the intensity distribution in the fringes will be as shown in Figure 12.21. The central maximum will be the brightest, and the intensity distribution quickly drops off on either side. The second sets of fringes are equally spaced from the central maximum, and, each has less than 5% of the intensity of the central maximum. The third bright fringe has only about 1.7% of the maximum intensity, the fourth has less than 1%, and so on. Thus, the central maximum has about 84% of the total intensity of the light incident on the slit. This is good news in one sense, since much of the light is still located close to the center line of the slit. However, the width of the central bright fringe is very different from the slit width because of diffraction. To calculate the width of the central bright fringe, we proceed as follows. Consider a narrow, long, slit of width b, located at a distance Z’ from on the screen1,2, as shown in Figure 12.22. In Figure 12.22, AB is the narrow slit of width b illuminated from the left by planar light waves. The screen is located in the far field diffraction regime, and the intensity pattern of the diffraction fringes appear on the screen, such that at Y0, the centerline of the central maximum occurs. At Y1, a dark fringe occurs; Y2 is the center of the second maximum; Y3 is the second dark fringe location, etc. A similar pattern is observed above the center line at Y0, and is not shown in the figure for the sake of clarity. Two lines are drawn connecting Y1 with A and B. BC is the perpendicular drawn from the point B to AY1. If the distance Z’ is practically infinity, then angle /ABC also equals θ, the angle of inclination of Y1 from Y0. Using arguments similar to the derivation of the equations for Young’s double slit experiment, it can be assumed that the additional path difference AC should equal λ, the wavelength of light. This is because the wavelets emanating from points A and C cause destructive interference at point Y1, implying that sources A and B generate a phase difference, each one-half wavelength compared to the dashed center line. Thus, 359 AC = λ = bsin θ, the condition that explains a dark fringe at Y1, which separates the central maximum fringe from the second bright fringe. When Z’-> ∞, an ideal requirement for the far field (Fraunhofer) condition, the small angle approximation can be used, i.e., sin θ = θ Thus, combing these last two equations, we obtain the half-angle of divergence as θ =λ b Diffraction pattern: Intensity distribution 1 0.8 Intensity 0.6 0.4 0.2 0 -10 -5 0 5 Distance from center line, arbitrary units 10 Spreading due to diffraction Long, narrow slit Planar wave fronts Figure 12.21 Diffraction pattern intensity from a long, narrow slit. 360 Slit Screen Planar Waves A C b θ B Intensity profile Z’--> ∞ Y0 Y1 Y2 Y3 Figure 12.22 Diffraction pattern fringe location from a long, narrow slit. This is a relationship of the half-angle for the central maximum. Remember that for the small angle approximation to work, the angle must be expressed in radians only (see Chapter 5.) For Z’ distance which satisfies the Fraunhofer condition, however, not quite infinite, the half height of the central maximum Y1Y0 is given by Z’λ b The width of the central bright fringe is then given by 2Y1Y0. Y1Y0 = Z’θ = Unlike Young’s double slit experiment, where one calculated the location of different order bright or dark fringes, for the case of diffraction, one is more interested in the width of the central maximum, since that would tell us how far the light has spread, resulting in the distribution of about 84% of the original intensity. Example 12.6 Light from an argon-ion laser of 488 nm wavelength passes through a long, narrow single slit of 75 µm opening. If the Fraunhofer patterns are observed on a screen 1.5 meter behind the slit, calculate (a) the half-angle of the central maximum, and (b) the half-width of the central bright fringe. (a) θ =λ (b) b θ = 488 x 10-9 = 75 x 10-6 Y1Y0 = Z’θ = 6.51 milli-radian Answer 1.5 m x 6.51 milli-radian = 9.76 mm Answer 361 Thus, while geometric optics would predict that one would have a bright spot of about 75 µm width (i.e., slit width), due to diffraction caused by the wave nature of light, we now have a bright spot width of 9.76 mm, about 130 times bigger, on the screen 1.5 meters away from the slit! Thus far we have looked at a long, narrow slit, where diffraction patterns are observed only in the direction that the light was constricted. If light were to be constricted in two directions, then we will have a rectangular slit, and diffraction patterns will be observed both in the horizontal direction and the vertical direction, as shown in Figure 12.23. Again, the far field conditions will have to be satisfied first. If the slit size is a unit wide in the horizontal direction and b units wide in the vertical direction2,7,8, the half-widths (extending the arguments posed for a single, narrow slit discussed earlier) are given by Z’λ a Z’λ b X1Y0 = Y1Y0 = Rectangular slit with b > a a Y1 b Y0 X1 Planar waves Screen Z’ View of fringe patterns on the screen (Additional fringes will be observed along the directions of arrows , shown). Figure 12.23 Diffraction pattern fringe location from a rectangular slit. Only the central fringe is shown in Figure 12.23 for the sake of clarity. For a square slit with a=b, then Y1Y0=X1Y0, and the diffraction pattern would appear identical in both the horizontal and the vertical directions. If we were to use an aperture, shaped in the form of a regular polygon with four or more sides, then, by extrapolation of the arguments one could expect the diffraction pattern to spread in many different directions from the center Y0, as illustrated in Figure 12.24. A star-burst pattern would be visible on the screen. 362 Figure 12.24 Diffraction patterns from a single, multisided aperture. Extend your imagination. What happens to a polygon (i.e., a multi-sided figure) when one increases the number of sides, say, to twenty or more? The figure would look more like a circle. Thus, if we were to use a circular aperture, light would be diffracted all 360o, resulting in rings. While they look like Newton’s rings discussed under the section on interference, they are created solely by diffraction by the edges of a circular aperture. The central bright circular spot is known as the Airy disk, in honor of the British astronomer Sir George Airy. The diffraction pattern would look as shown in Figure 12.25 9,10: Planar wave fronts θ D Screen Circular aperture Airy disk (~84% of light intensity) Radius R Z’ Figure 12.25 Diffraction patterns from a single, circular aperture. In this figure, D is the diameter of the circular aperture, R the radius of the Airy disk, θ the half-angle to the edge of the Airy disk, and Z’ the distance of the screen from the aperture such it satisfies the Fraunhofer condition. Again, the distance Z’ is so large compared to D such that small angle approximation can be used for the half angle. or tan θ ~ θ = R-(D/2) ~ R Z’ Z’ R = Z’θ since (D/2) << Z’ 363 Unfortunately, unlike the case of the narrow, rectangular, slit, one cannot equate θ to λ/D. One needs to know integral calculus and differential equations to solve this problem. Therefore, without proof, the result for the half-angle is presented as2,7,10 θ = 1.22λ D Therefore, R = 1.22λZ’ D When one focuses collimated light using a positive lens of aperture diameter D and focal length f, the radius of the focused spot is given by a similar expression as R = 1.22λf D Notice that the focal length replaces Z’ from the previous equation. The diffraction limited spot size can now be calculated. This has implications in imaging applications such as in photolithography used in semiconductor manufacturing. Unlike the predictions of geometric optics which insists that parallel rays of light will be focused to a point of zero dimension, diffraction limited optics provides us an equation to predict the spot size. The dramatic increase in the Airy disk radius as one decreases the aperture diameter is shown in Figure 12.26 for a fixed Z’ of 5 meters and using 633 nm light to illuminate the aperture. Airy disk radius, R, mm 633 nm light, Z'=5 m 200 150 100 50 0 0 50 100 150 200 250 Aperture diameter, µm Figure 12.26 Variation of Airy disk radius with aperture diameter. 364 When two distant objects are viewed by a telescope, one could tell them apart as two different entities (i.e. resolving power of the telescope), only if the angular separation at the viewing optics exceeds the half angle given by2,7,10 θ = 1.22λ D This is called Rayleigh’s resolution criterion. In this case, D normally stands for the diameter of the objective lens of the telescope. Example 12.7 A He:Ne laser (λ = 633 nm) illuminates a rectangular slit 100 µm x 200µm in size. What is the size of the central maximum bright spot seen on a screen 3.5 m away from the slit? Let us first check for far field conditions. d2 is equivalent to = (100 x 10-6) x (200 x 10-6) = 2 x 10-8 m2 Ratio = d2 / λ = 2 x 10-8 m2 / 633 x 10-9 m = 0.0316 m Actual Z’ of 3.5 m is 3.5/0.0316 = 111 compared to the ratio, which is large enough to approximate far field conditions. It does not much matter which way the slit is oriented. We use the following equations to solve for the half-widths in either direction as X1Y0 = Y1Y0 = X1Y0 = = X1Y0 = Z’λ a Z’λ b 3.5 x 633 x 10-9 100 x 10-6 22.2 mm -9 3.5 x 633 x 10 6 200 x 10 = 11.1 mm The actual dimensions of the central maximum are obtained by doubling these values: i.e., spot size on the screen = 44.3 mm x 22.2 mm Answer Thus, it is clear that smaller the opening size in any direction, larger the spot size on the wall for the same direction! The spot is about 50,000 times larger in area than the aperture! Example 12.7 A He:Ne laser (λ = 633 nm) illuminates a circular slit 100 µm in diameter. (a) What is the diameter of the Airy disk seen on a screen 3.5 m away from the slit? (b) What is the beam divergence? Let us first check for far field conditions. 365 d2 = (100 x 10-6)2 = 1 x 10-8 m2 Ratio = 1 x 10-8 / 633 x 10-9 = 0.016 Actual Z’ is 3.5/.016 = 222 compared to the ratio - > satisfies Fraunhofer requirement. (a) R = 1.22λZ’ D = 1.22 x 633 x 10-9 x 3.5 100 x 10-6 0.027 m radius = 54.1 mm diameter Answer = (b) θ = 1.22λ D =1.22 x 633 x 10-9 100 x 10-6 = 7.72 milli-radians Answer We know that with a single, narrow slit, a diffraction pattern of bright and dark fringes can be obtained. We also know that with a double slit, an interference pattern of bright and dark fringes can be obtained. In fact, both phenomena come into play with a double slit. Inside each diffraction maximum (i.e., bright fringe), one would observe bright and dark fringes caused by interference. The interference fringes are most clearly seen inside the diffraction central maximum and become less pronounced when one looks at the second, third, or higher diffraction maxima. Clearly, this is because nearly 84% of the light is concentrated in the central maximum for the diffraction pattern and, hence, it is easier to discern the interference pattern within. What would happen if one uses a three-slit system, a four-slit system, etc? The patterns will become such that the interference patterns within each diffraction maximum will disappear depending upon the ratio of slit width to slit separation. Distinct bright fringes are observed, one corresponding to each diffraction order, as shown in Figure 12.27. The reasoning is as follows. The interference principal maxima narrows as the number of slits is increased. The brightness of the secondary interference maxima decreases drastically. This results in the primary maxima alone being visible. A diffraction grating may also be used, which may be thought of as consisting of infinite number of slits. The grating is basically a transparent slide with fine engravings on it, thousands of lines per centimeter. Light incident on one side of the grating undergoes diffraction and splits into a few modes. Depending upon the wavelength of light and the line spacing, a few modes may be feasible. Unlike a single slit diffraction pattern, the light is scattered over a large angle. A grating may also act as a reflection grating where the light reflected off the grating is diffracted. A compact disc or a digital video disk serves as a reflection grating. The grating equation is very similar to Young’s double slit interference equation. It simply states1,2,9 dsinθ m= mλ, m=0, +1, +2, +3… 366 where d=spacing of the grating λ= wavelength of light θ=angle of the bright fringe location from the center line, and m=order of the bright fringe. This equation is valid when far field condition is met and for normal incidence of light on the grating surface. Grating m=2 θ2 m=1 θ1 m=0 Planar waves m=-1 m=-2 Figure 12.27 Diffraction grating. Example 12.8 A diffraction grating has 6000 rulings per centimeter. It is illuminated by 514.5 nm wavelength laser light. Identify the location of all the diffraction fringes possible. Clearly the zeroth order fringe lies along the optical axis of the laser. d = 1/6000 cm = 0.000167 cm = 1.67 x 10-6 m Rearranging the grating equation, we obtain θ m=sin-1(mλ/d) For m=1 (or -1), θ 1=sin-1(1λ/d) = sin-1 (514.5 x 10-9m/ 1.67 x 10-6 m) = 17.9o Answer For m=2 (or -2), θ 2=sin-1(2λ/d) = sin-1 (2 x 514.5 x 10-9m/ 1.67 x 10-6 m) = 38.0o Answer For m=3 (or -3), θ 3=sin-1(3λ/d) = sin-1 (3 x 514.5 x 10-9m/ 1.67 x 10-6 m) = 67.5o Answer For m=4 (or -4), θ 4=sin-1(4λ/d) = sin-1 (4 x 514.5 x 10-9m/ 1.67 x 10-6 m) = sin-1 (1.23), which is not feasible. Thus, a total of seven bright fringes will appear corresponding to m=0, +1, +2, and +3 only. If the screen is really large, then the third order fringes may easily be captured on the screen. If not, higher order fringes will be seen on the wall perpendicular to the screen, since the angles are becoming larger. 367 The maximum number of diffraction fringes obtainable for select wavelengths as a function of the number of grooves per centimeter of grating is shown in Figure 12.28. Number of Maximum diffraction order possible fringes (Max. 1000 order) 400nm 500nm 600nm 700nm 100 10 1 100 1000 10000 Grooves per cm 100000 Figure 12.28Diffraction grating: Maximum orders possible. Greater the number of rulings per unit length, less the number of diffraction fringes possible; higher the wavelength, less the number of fringes. These observations are a direct consequence of the diffraction equation. Reflection gratings find applications in determining the wavelength of an unknown source of light or analyzing and recording its spectrum. Many diffraction gratings are manufactured with a blaze to provide maximum reflection of a definite wavelength of light in a certain direction11. In the discussion thus far, we clearly noted the wavelength dependence in the equations. Since monochromatic light was used for example calculations, we could easily obtain either the half-angle or the width of the diffracted pattern. If one uses polychromatic light, needless to say, dispersion occurs. Different colored light waves are diffracted to different degrees leading to dispersion. A compact disc or a digital video disk is a good example of a reflection grating, since it is composed of thousands of lines. When white light is incident on a disk, by angling the disk properly, one sees a rainbow pattern as result of dispersion caused by diffraction. This principle can be used to create a home made spectroscope12. A spectrometer works on the principle of the diffraction grating. As discussed in Chapter 2, the spectrometer gathers light from a source through a slit. When the polychromatic light hits a diffraction grating, different wavelengths that make up the light are diffracted to different angles. The diffracted light strikes a graduated scale, and the different visible lines may be read off the scale. It is unwise to use a source such as a mercury arc lamp that emits in the ultraviolet and view it through a spectrometer, unless a UV filter is used. A diffraction grating or a prism can also be used to separate out a particular wavelength, out of many emitted by a multi-line laser. 368 Exercise 12.3B 1. Light from a laser of 633 nm wavelength passes through a long, narrow single slit of 0.25 mm opening. If the Fraunhofer patterns are observed on a screen 2 meters behind the slit, calculate (a) the half-angle of the central maximum, and (b) the width of the central bright fringe. 2. Light from a laser of 532 nm wavelength passes through a long, narrow single slit of 0.10 mm opening. If the Fraunhofer patterns are observed on a screen 2 meters behind the slit, calculate (a) the half-angle of the central maximum, and (b) the width of the central bright fringe. 3. If the entire width of the central maximum was measured to be 29.5 mm, on a screen 2.5 m away from a 0.1 mm wide slit illuminated from a source, what is the wavelength of light used? 4. If the half width of the central maximum was measured to be 9.8 mm, on a screen 2 m away from a 0.1 mm wide slit illuminated from a source, what is the wavelength of light used? 5. A He:Ne laser (λ = 633 nm) illuminates a rectangular slit 100 µm x 100µm in size. What is the size of the central maximum bright spot seen on a screen 2 m away from the slit? 6. A He:Ne laser (λ = 633 nm) illuminates a rectangular slit 100 µm x 125µm in size. What is the size of the central maximum bright spot seen on a screen 2.5 m away from the slit? 7. A laser (λ = 514.5 nm) illuminates a rectangular slit 100 µm x 100µm in size. What is the size of the central maximum bright spot seen on a screen 2.5 m away from the slit? 8. A laser (λ = 514.5 nm) illuminates a rectangular slit 100 µm x 125µm in size. What is the size of the central maximum bright spot seen on a screen 2.5 m away from the slit? 9. A He:Ne laser (λ = 633 nm) illuminates a circular slit 200 µm in diameter. (a) What is the diameter of the Airy disk seen on a screen 5 m away from the slit? (b) What is the beam divergence? 10. A laser (λ = 488 nm) illuminates a circular slit 150 µm in diameter. (a) What is the diameter of the Airy disk seen on a screen 4 m away from the slit? (b) What is the beam divergence? 11. A diffraction grating has 10,000 rulings per centimeter. It is illuminated by 514.5 nm wavelength laser light. Identify the location of all the diffraction fringes possible. 369 12. A diffraction grating has 10,000 rulings per centimeter. It is illuminated by 633 nm wavelength laser light. Identify the location of all the diffraction fringes possible. 13. A diffraction grating has 2,500 rulings per centimeter. It is illuminated by 633 nm wavelength laser light. Identify the location of all the diffraction fringes possible. 14. A diffraction grating has 5,000 rulings per centimeter. It is illuminated by 633 nm wavelength laser light. Identify the location of all the diffraction fringes possible. 12.4 Suggested Laboratory Experiments9 a) Interference: Young’s Double Slit Experiment 1. Set up an optical axis. 2. Follow safety precautions associated with using a He:Ne laser. 3. Use a slide which has a double slit. Ensure that the laser illuminates both the slits. Mask any other double slits that may be part of the slide. Note the slit opening and slit spacing in Table 12.2. 4. Obtain the interference pattern on a piece of white paper (screen) affixed to a wall about 2 meters (6 feet) away. 5. Remove the slide. Mark the location of the He:Ne laser beam on the screen. Replace the double slit. 6. With respect to the center line of the laser, mark the center of each bright fringe on the screen using a pencil. Make sure that bright fringes on either side of the center line are covered. Concentrate only on the set of bright fringes clustered together around the center line. These are interference fringes embedded in the central maximum diffraction fringe. Ignore the other diffraction fringes. 7. Calculate the fringe locations based on theory. ( ym = L mλ m=1,2,3…) d Enter in Table 12.2 as also the measured fringe locations. 8. Explain any differences observed. Switch over to a different double slit with a different slit spacing. Repeat steps 4 through 7. Enter answers in Table 12.2. b) Diffraction: 1. Choose the following slits: a long, narrow, slit, a rectangular slit and a circular aperture. Note the opening sizes for each in Table 12.3. 370 Slit spacing #1. Slit spacing #2. Table 12.2 Fringe location (measured) For m=1 For m=2 For m=3 For m=-1 For m=-2 For m=-3 Fringe location (calculated) For m=1 For m=2 For m=3 For m=-1 For m=-2 For m=-3 Fringe location (measured) For m=1 For m=2 For m=3 For m=-1 For m=-2 For m=-3 Fringe location (calculated) For m=1 For m=2 For m=3 For m=-1 For m=-2 For m=-3 2. Perform far field calculations for Z’ and ensure that the Z’ lies in the Fraunhofer regime. For most typical laboratory He:Ne lasers, if the slit is placed about 30 cm from the laser source, the far field condition of Z will be satisfied based on the laser cavity geometry. 3. Observing all safety precautions, turn the laser on. 4. Locate the beam on a piece of white paper (screen) affixed to a wall. 5. Place the long, narrow slit on a slide holder. 6. Verify step 2. Observe the diffraction pattern on the screen. Measure the width of the central maximum. Calculate the half-width. Enter the answer in Table 12.3. 7. Replace the long, narrow slit with a rectangular aperture. Repeat step 6, remembering to measure both dimensions. 8. Replace with a circular aperture (a pinhole). Repeat step 6. 9. Calculate, in each case, the half-width of the central maximum predicted by theory. Enter them in Table 12.3. Explain any differences observed with measurements. 10. Replace the slit with a diffraction grating with 5000 rulings per centimeter. Calculate the grating spacing. Predict the angular locations of all feasible order diffraction fringes. Do you observe the diffraction fringes that the theory predicts? 371 Type of slit/opening Long, narrow b= Rectangular, b= , a= Circular, D= Predicted halfwidth Table 12.3 Measured halfwidth Vertical Airy Disk R= Predicted halfwidth N/A Measured halfwidth N/A Vertical Horizontal Horizontal Airy Disk R= N/A N/A Chapter Summary • • • • • • • • • Huygen’s principle simply states that when planar waves encounter an obstacle, each point on the obstacle will serve as sources for secondary, spherical waves. When spherical waves travel over relatively long distances, they lose their identity as spherical wave fronts; rather, they appear to be planar wave fronts. Collimated light has planar waves. Laser light, and light emerging from a positive lens, with a point source placed at the lens’ focal point, are examples. When waves that are out of phase combine, constructive and destructive interference take place, resulting in bright and dark spots (or fringes) of light intensity. Young’s double slit experiment was a classical demonstration that with a single coherent source, a path difference can be introduced between waves emerging from a double slit illuminated by the source. This results in interference patterns or fringes. The slit openings should be of the order of the wavelength of light. The principle of interference is used in designing anti-reflection coatings (ARC) to minimize Fresnel reflection off dielectric surfaces, such as lenses and windows. Another manifestation of the wave nature of light is diffraction. When light passes through an opening whose size is of the order of the light wavelength, light bends. The spreading of light is called diffraction. Secondary spherical waves created at the slit or aperture produce a diffraction pattern of bright and dark fringes. The mathematics being cumbersome, the effect of near field (Fresnel) diffraction was not considered in this chapter. Only cases where the source-to-slit distance and the slit-to-screen distance are both much higher than the ratio of slit area to light wavelength were considered. This would result in far field (Fraunhofer) diffraction for which the mathematical theory is much simpler. The vast majority of the light intensity is concentrated in the central maximum. The intensity distribution of the neighboring fringes rapidly falls off. For the case of a circular aperture, the fringes appear as annular rings, and the central maximum is a circle, known as the Airy disk. Diffraction occurs in a more pronounced fashion in the direction in which light is constrained to a greater extent. 372 • A diffraction grating is the mathematical limit of infinite number of slits illuminated by planar waves. With polychromatic light, a grating can be used to separate the individual wavelengths of the incoming light. End of Chapter Exercises 1. A He-Ne laser with output wavelength 633 nm illuminates a double slit with 0.1 mm opening and 0.50 mm spacing. How far the screen should be placed so that the third bright fringe is located exactly 20 mm from the central bright fringe? 2. A sodium lamp with output wavelength of 589 nm illuminates a double slit 100 µm wide and 0.25 mm spacing. The fringes are seen on a screen 2 m from the slit. Calculate (a) the distance between adjacent bright fringes, (b) the distance between adjacent dark fringes, (c) the location of the third bright fringe, (d) the location of the fourth dark fringe, (e) the angle corresponding to the first bright fringe and (f) the angle corresponding to the second dark fringe. 3. What should the ARC coating thickness be for a glass lens to be used in conjunction with He:Ne laser light (488 nm wavelength)? Assume that the refractive index of ARC is 1.379 at the laser wavelength. 4. What should the ARC coating thickness be for a glass lens to be used in conjunction with YAG laser light (1064 nm wavelength)? Assume that the refractive index of ARC is 1.383 at the laser wavelength. 5. Estimate the source-to-aperture distance and aperture-to-screen distance for far field diffraction to occur, given wavelength of light source is 532 nm and the aperture is a square 75 µm on the side. 6. Estimate the source-to-aperture distance and aperture-to-screen distance for far field diffraction to occur, given wavelength of light source is 700 nm and the aperture is a circle of diameter 125 µm. 7. If the entire width of the central maximum was measured to be 40 mm, on a screen 2.5 m away from a 0.075 mm wide slit illuminated from a source, what is the wavelength of light used? 8. If the half width of the central maximum was measured to be 5 mm, on a screen 2 m away from a 0.2 mm wide slit illuminated from a source, what is the wavelength of light used? 9. A He:Ne laser (λ = 633 nm) illuminates a rectangular slit 80 µm x 150µm in size. What is the size of the central maximum bright spot seen on a screen 3 m away from the slit? 373 10. A laser (λ = 488 nm) illuminates a rectangular slit 100 µm x 100µm in size. What is the size of the central maximum bright spot seen on a screen 2.7 m away from the slit? 11. A diffraction grating has 1000 rulings per centimeter. It is illuminated by 488 nm wavelength laser light. Identify the location of all the diffraction fringes possible. 12. A diffraction grating has 2500 rulings per centimeter. It is illuminated by 589 nm wavelength collimated light. Identify the location of all the diffraction fringes possible. Group Discussion Questions These require the students to form small groups of three or four and discuss each question. Some questions may require you to do a search on the Internet and find the answers. 1. How would the diffraction grating equation differ if light were not normally incident on the grating surface? 2. What is resolving power of a grating? How does it compare with the resolving power of other optical instruments? 3. Using a spy glass (i.e., a terrestrial telescope) with an objective lens diameter of 5 cm and focal length 20 cm, how close should two persons be standing with respect to each other, if they are both 20 km from where you are with your spy glass? Use Rayleigh’s diffraction resolution criterion. Assume a light wavelength of 550 nm. References 1 Sears, F.W, M. W. Zemanksy and H.D. Young. “College Physics.” Reading: AddisonWesley Publishing Company, 1991. 2 Jenkins,F. and H. White. “Fundamentals of Optics.” Singapore: McGraw Hill, 1981. 3 University of Winnipeg Physics Department, http://theory.uwinnipeg.ca/physics/light/node9.html 4 Optical Techniques for Measuring Flatness, Product Literature from Edmund Scientific Co., Barrington, NJ, Science Kit & Boreal Laboratories, Tonawanda, NY. 5 M. Ciofini, , A. Labate , R. Meucci and , Peng-Ye Wang, Experimental evidence of selection and stabilization of spatial patterns in a CO2 laser by means of spatial perturbations, Optics Communications 154 1998 307–312 6 Kent University http://www.lci.kent.edu 7 Educational Technology Center, Northeastern University, http://www.atsweb.neu.edu/physics/j.sage/w03/diffract.pdf 8 Wolfram Research, http://scienceworld.wolfram.com/physics/FraunhoferDiffractionRectangularAperture.htm l 9 Gottlieb, H. “Experiments Using a Helium Neon Laser.” Bellmawr: Metrology Instruments, Inc., 1981 10 Welford, T. “Optics.” Oxford: Oxford University Press, 1988. 11 Product Literature from Edmund Scientific on reflection gratings.. 12 Science Toys web site, http://www.scitoys.com/ 374 Appendix I Useful Conversions h=6.625 x 10-34 J.s c=3 x 108 m/s 1 eV = 1.602 x 10-19 J To convert FROM E(J) E(eV) E(J) k(cm-1) E(eV) k(cm-1) TO E(eV) E(J) k(cm-1) E(J) k(cm-1) E(eV) Conversion Table Multiply Divide E(J) E(eV) E(J) k(cm-1) E(eV) k(cm-1) Length 1 cm 1 in = = 0.394 in 2.54 cm 1 cm 1 ft = = 0.0328 ft 30.5 cm 1m 1 ft = = 3.28 ft 0.305 m 1 mi = 1 mi = 1 km = 5280 ft 1.61 km 0.621 mi 1 cm2 = 1 in2 = 0.155 in2 6.45 cm2 1 cm2 = 1 ft2 = 1.08 x 10-3 ft2 930 cm2 1 m2 1 ft2 = = 10.8 ft2 9.30 x 10-2 m2 1 kg 1 lb = = 2.20 lb 0.454 kg Area Other by 1.602 x 10-19 1.602 x 10-19 5.03 x 1022 5.03 x 1022 8060 8060 375 Appendix II Scientific System and Metric Prefixes Positive exponents (or powers) of ten (some examples): 10 1 = 10 x 1 = 10 2 10 = 10 x 10 = 100 10 3 = 10 x 10 x 10 = 1000 4 10 = 10 x 10 x 10 x 10 = 10,000 10 5 = 10 x 10 x 10 x 10 x 10 = 100,000 10 6 = 10 x 10 x 10 x 10 x 10 x 10 = 1000,000, etc. Negative exponents (or powers) of ten (some examples): 10 -1 = 0.1 x 1 = 0.1 -2 10 = 0.1 x 0.1 = 0.01 10 -3 = 0.1 x 0.1 x 0.1 = 0.001 10 -4 = 0.1 x 0.1 x 0.1 x 0.1 = 0.0001 -5 10 = 0.1 x 0.1 x 0.1 x 0.1 x 0.1 = 0.00001 10 -6 = 0.1 x 0.1 x 0.1 x 0.1 x 0.1 x 0.1 = 0.000001, etc. Examples: 13,200 = 1.32 x 104 and 0.00266 = 2.66 x 10-3 Prefix name(symbol) value (power of 10) Value (decimal) milli (m) 10-3 0.001 -2 centi (c) 10 0.01 deci (d) 10-1 0.1 base unit (none) 100 1 1 deca (da) 10 10 hecto (h) 102 100 3 kilo (k) 10 1000 The following metric engineering prefixes are commonly used arranged from small to large values: Prefix name Prefix symbol Value (powers of ten) Value (decimal) atto a 10-18 0.000 000 000 000 000 001 femto f 10-15 0.000 000 000 000 001 -12 pico p 10 0.000 000 000 001 nano n 10-9 0.000 000 001 -6 micro µ 10 0.000 001 milli m 10-3 0.001 0 base unit none 10 1 kilo k 103 1000 6 mega M 10 1,000,000 giga G 109 1,000,000,000 12 tera T 10 1,000,000,000,000 peta P 1015 1,000,000,000,000,000 18 exa E 10 1,000,000,000,000,000,000 376 Appendix III Calculator Review (General) It is important to learn how to use the calculator correctly. Most mistakes result from operator error rather than due to a faulty calculator. Human beings have a tendency to blame inanimate objects for their own mistakes. You have often heard the saying, “it was a computer error.” Computers and calculators seldom make mistakes on their own. The quality of the results from a machine is directly dependent on the quality of the data entry, i.e. how carefully the operator enters the information and follows the right steps. The other piece of advice to the student is not to depend on the calculator solely but to use mental or paper/pencil calculations to verify that the derived answer is reasonable. Review the calculator manual before starting to use it. The steps given in this section apply most closely to Texas Instruments calculator TI-503Xa® and may not apply to all calculators. For example, some calculators require you to enter the number first followed by the operation symbols, while others require just the opposite procedure. In any case, if you own a calculator that performs the following functions, it would suffice as far as solving problems in this book: Key/ symbol + x ÷ Function Procedure Addition (Press this key between entries of numbers, press = key) Subtraction (Press this key between entries of numbers, press = key) Multiplication (Press this key between entries of numbers, press = key) Division (Press this key between entries of numbers, press = key) +/- Change sign key. x2 Square key x Square root yx y to the x LOG (log) LN (ln) 1(or x-1) x ex logarithm (base 10) logarithm (base e) reciprocal 10x 10 to the b c EE (or exp) a exponential If a negative number is to be entered in any of the operations, enter its magnitude first, then use +/key Enter number first, press x2 key; display shows the square Enter number first, press x key, display shows square root Enter value of y, press yx key, then enter value of x, press = Enter the number, press LOG (log) key Enter the number, press LN (ln) key Enter number, press the key inverse of LN key; enter number; press 2nd, then LN key Inverse of LOG key; enter number; press 2nd, then LOG key fractions This enables fractions to be entered exponent of 10 This enables scientific and engineering notations to be entered 377 Key/ symbol Function Procedure DRG Degrees/radians/grad Enables calculator to compute in any one of the three modes. Pressing 2nd DRG lets one convert from one form to another DMS Degrees/minutes/seconds. Enables conversion between decimal angles and degrees/minutes/seconds format π pi (3.14….) Enters value automatically when pressed sin sine of an angle Enter value of angle, press sin key. cos cosine of an angle Enter value of angle, press cos key. tan tangent of an angle Enter value of angle, press tan key. arc sine of an angle Enter value, press 2nd, then sin key. sin-1 -1 cos arc cosine of an angle Enter value, press 2nd and then cos key. tan-1 arc tangent of angle Enter value, press 2nd and then tan key. x> <y Needed for polar to rectangular conversion or vice versa R-P or P-R Needed for polar to rectangular conversion or vice versa RCL or STO Recall memory or store memory buttons; many calculators allow values to be stored in different memory cell; in most cases press STO followed by 1 to store in cell 1; press RCL 1 to get the number back into display from memory. As we work with specific sections more details and examples will be presented regarding calculator use. For now we will work numerical problems involving basic arithmetic operations, square, square root and signed numbers. Example 1 Simplify: -3.62 + 4.85 – 6.92 + 2.00 Entry operation Display shows Enter 3.62 3.62 Press +/- to change sign -3.62 Press + -3.62 Enter 4.85 4.85 Press 1.23 Enter 6.92 6.92 Press + -5.69 Enter 2 2 Enter = -3.69 Answer Mental check by rounding off each number to the nearest integer: -4+5-7+2 = -4, thus the answer is in the ball park and possibly no gross entry errors such as a wrong decimal place were made. 378 Example 2 Simplify: 3.22 x (-5.25) x 1.01 ---------------------------(-2.56) x (1.66) x (-5.85) Entry operation Enter 3.22 Press x Enter 5.25 Press +/Press x Enter 1.01 Press ÷ Press “(” Press 2.56 Press +/Press x Enter 1.66 Press x Enter 5.85 Press +/Press “)” Press = Display shows 3.22 3.22 5.25 -5.25 -16.905 1.01 -17.07405 0 2.56 -2.56 2.56 1.66 -4.2496 5.85 -5.85 24.86016 -0.686803705 Answer Answer verification (i) sign : Three negatives will yield a net negative for the answer - > checks (ii) magnitude: 3 x 5 x 1 is 15 divided by 3 x 2 x 6, which is 36. 15÷36 is approximately 0.4, decimal location checks; realize that value is very approximate. Example 3 ____ Compute √1.322 + 1.672 Entry operation Display shows Enter 1.322 1.322 Press x 1.1497825 Press + 1.1497825 Enter 1.672 1.672 Press = 2.8217825 Answer (It would be more appropriate to round off the answer to 2.822, i.e. three decimal places.) Example 4 Find the reciprocal of 4 379 Entry operation Enter 4 Press 1/x (or ÷ followed by =) Display shows 4 0.25 Answer Example 5 Calculate –3[-2 + 42 {13-22} –7{11-4}-10] –11 If your calculator is sophisticated that you can type in the above exactly as it looks, you will get the right answer. Type 3, +/-, x, “(“ , 2, +/-, + 42 x “(“ 13 – 22 “)” – 7 x “(“ 11 – 4 “)” – 10 “)” – 11 = 1306 Answer It is important not to miss a single bracket and to have the same total number of left “(“ brackets as right “)” brackets. Example 6 Find the square of 2.2 Entry operation Enter 2.2 Press x2 button Display shows 2.2 4.84 Answer Example 7 1.2 x 10-4 x 3.2 x 10-8 Entry operation Enter 1.2 EE 4 followed by +/Press x key Enter 3.2 EE 8 followed by +/Press = Display shows 1.2-04 1.2-04 3.2-08 3.84-12 implying the answer is 3.84 x 10-12 380 Appendix IV Basic Geometric and Trigonometric Definitions and Identities Principle 1 If a straight line stands on another straight line, the sum of the adjacent angles is equal to 180o. D C A B Line CD stands on line AB. /DCA + /DCB = 180o Principle 2 If two straight lines intersect, the vertically opposite angles are equal. AB and CD intersect at point P. Then, /APC = /DPB and /DPA = /CPB D A P C B Principle 3 If a straight line intercepts a pair of parallel lines, the alternate angles are equal and the corresponding angles are equal. E A K C B D L F AB and CD are parallel to each other. EF intercepts AB and CD at points K and L, respectively. Then, /EKB = /KLD; /BKL = /DLF; /EKA = /KLC; /AKL = /CLF (corresponding angles are equal) /BKL = /KLC; /AKL = /KLD (alternate angles are equal) Also, /BKL + /KLD = 180o and /AKL + /KLC = 180o Principle 4. The sum of the three angles of any triangle equals 180o. In the following principles, when two triangles are congruent, they are exactly identical in all respects, i.e. the three sides and three angles of one triangle are equal to the three 381 sides and three angles of the second triangle. In other words, one can draw out one triangle on a piece of paper cut out its outline and align it on the second triangle. Then the triangles will match up perfectly. Principle 5 Two triangles are considered congruent, if two sides and the included angle in one triangle are equal to two sides and the included angle in the second triangle. Principle 6 Two triangles are considered congruent, if any side and any two angles in one triangle are equal to any side and any two angles of the second triangle. Principle 7 Two triangles are considered congruent if each side in one triangle is equal to each side in the second triangle. Principle 8 Two right triangles are considered congruent if the hypotenuse and any other side of one triangle equal the hypotenuse and any other side of the second triangle. Principle 9 Two triangles are considered to be similar to each other, if each angle in one triangle equals each angle in the second triangle. Note that while congruent triangles will be similar to each other, similar triangles are not necessarily congruent with each other. Similar triangles have proportionate sides. Principle 10 Pythagorean Theorem B c a C A b Pythagorean Theorem states that c2 = a2 + b2 or _______ c = √a2 + b2 Principle 11 Multisided enclosed figures are called polygons. The triangle is the simplest polygon with three sides. A quadrilateral has four sides, a pentagon has five sides, hexagon has six sides, octagon has eight sides (STOP sign shape), etc. If the sides of a polygon are equal to each other, and the interior angles are equal to each other, then we have a regular polygon. An equilateral triangle and a square are simple examples of regular polygons. 382 Quadrilateral Parallelogram Rhombus (unequal sides and angles) (opposite sides equal) (all sides equal) (opposite sides parallel) Trapezium Rectangle Square (angles equal, opposite sides equal) (angles equal, all sides equal) Trigonometry. Definitions based on right triangle B c a A C b sin A = a/c sin B = b/c cos A = b/c cos B = a/c tan A = a/b tan B = b/a csc θ = 1/sinθ sec θ = 1/cos θ cot θ = 1/tan θ sin A = cos B cos A = sin B tan A = cot B cot A = tan B (The Greek letter θ is used to designate an angular measurement, pronounced “theta.") In general, sin θ = cos (90o - θ ) cos θ = sin (90o -θ ) tan θ = cot (90o-θ ) cot θ = tan (90o-θ ) Values of trigonometric functions for selected angles Function 0o 30o 45o 60o 90o θ= 0 ½ 1 sin θ √2/2 (0.707) √3/2 (0.866) 1 ½ 0 cos θ √3/2 (0.866) √2/2 (0.707) 0 1 tan θ √3/3 (0.577) √3 (1.73) ∞ Trigonometric Identities: sin2θ + cos2θ = 1 1 + tan2θ = sec2 θ 1 + cot2θ = csc2 θ 383 Appendix V Additional Useful Formulas in Fiber Optics1-4 Refer to Chapter 6 also for pertinent terminology. Dispersion: The following two formulas provide modal dispersion per meter of cable length: Δtm = ncoreε . (step index fiber) c (1-ε ) Δtm = where ε = ncore ε 2 (graded index fiber) 8c 1 – ncladding ncore The following two formulas provide dispersion values per km of cable length: Δtmat = dispersive coefficienta x spectral width of sourceb Δtwg = waveguide coefficienta x spectral width of sourceb a The coefficients are usually specified in ps/(ns.km fiber length). b The spectral width of the source usually has units of ns, specified at 3dB values (typically tens of ns for LEDs and fractions of ns to ns for lasers). Light sources: For an LED surface emitting source, the fraction of light from the source that enters the fiber is given by: Power Fraction = ncore2-ncladding2 The source bandwidth (BW) is related to the rise time of the source using the following approximation BW = 1 (2.857 x Rise time) (see Chapter 9 for definition of rise time). References: 1 Sterling, D. Jr., “Technician’s Guide to Fiber Optics.” Albany: Delmar Publishers, 2000. 2 Meardon, S.L.W. “The Elements of Fiber Optics.” Englewood Cliffs: Prentice Hall, 1993. 3 Bell College School of Science and Technology website http://floti.bell.ac.uk/MathsPhysics/system.htm 4 Corning Cable Systems web site http://www.corningcablesystems.com/web/college/fibertutorial.nsf/ofpara?OpenForm 384 Answers to Odd Numbered Problems Exercise 1.1A 1. 989 3. 3.14 5. –32.2 7. –0.0245 17. –0.025 19. 870,000 25. –32.17 27. –0.02450 Exercise 1.1B 1. 8.758 x 10-3 11. 2.24 x 104 21. 2.56 x 10-3 9. 873,000 21. 988.8 29. 872,500 11. 990 13. 3.1 15. –32 23. 3.145 3. –4.5236 x 101 5. 9.7823 x 102 7. 2.3665823 x 105 9. –6.8925 x 102 13. 1.6 x 10-19 15. -5.875 x 10-5 17. 3.256875 x 103 19. 1.928 x 103 Exercise 1.1C 1. 8.76 x 10-3 11. 2.24 x 104 19. 1.93 x 103 3. –4.52 x 101 5. 9.78 x 102 7. 2.37 x 105 9. –6.89 x 102 13. 1.6 x 10-19 15. -5.88 x 10-5 17. 3.26 x 103 21. 2.56 x 10-3 Exercise 1.1D 1. 1.0 mW 15. 3.3 MW 27. 3.75 µs 3. 10 mm 5. 1 µm 17. 2.6 µW 29. 10 ng 7. 32 mW 19. 100 fs 9. 100 mg 11. 2.2 THz 21. 3.32 PHz 23. 33 MHz Exercise 1.1E 1. 0.01 ms = __10_____µs 3. 3300 kg = 5. 1.2 kg = _1.2 x 106_ mg 7. 0.00003 s= 9. 2320 g = _2.32 x 106_mg 11. 2320 k$ = ___0.33___ kW 15. 0.0085 W = 13. 3.3 x 102 W = 17. 236,000 Hz = _0.236____MHz 19. 6.023 x 103 ms = 21. T=100 ps to ν = __10 GHz__ 23. ν =125 THz to T= 25. T=22000 ns to ν =__45.5 kHz__ 27. f=0.1 Hz to T = 29. T=3.3 ms to ν = __303 Hz__ Exercises 1.2A 1. yes 1.2 1 0.8 0.6 0.4 0.2 0 0 1 2 3 __3.3 x 106 g __30____ µs ___2.32____M$ ___8.5___mW ___6.023___s __8 fs_ __10 s_____ 13. 1 THz 25. 25 µs 385 3 no 18 16 14 12 10 8 6 4 2 0 0 2 4 6 5. yes 12 10 8 6 4 2 -4 0 -2 0 -2 2 4 6 -4 7. yes 1 0.8 0.6 0.4 0.2 0 -2 0 2 4 6 386 9. yes 14 12 10 8 6 4 2 0 -6 -4 -2 -2 0 2 4 -4 -2 0 -1 0 -2 2 4 11. yes -6 -3 -4 -5 -6 -7 -8 -9 13. yes -4 -2 20 15 10 5 0 -5 0 -10 2 4 6 387 15.no 200 150 100 50 0 -4 -3 -2 -1 0 17. No, unless the data collected exhibited considerable scatter (i.e. error.) 500 400 300 200 100 0 0 100 200 300 400 Exercises 1.2B 1. 200 150 100 50 0 -50 50 150 250 350 388 3. 10 8 6 4 2 0 -4 -2 0 2 4 0 2 4 5 20 15 10 5 0 -4 -2 -5 7. 60 40 20 0 -20 0 1 2 3 4 5 389 9. 1 0.5 0 -180 -80 20 120 -0.5 -1 11. 4 3.5 3 2.5 2 1.5 1 0.5 0 0 2 4 6 13. 500 400 300 200 100 0 0 0.5 1 1.5 2 390 15. The corresponding range is 1.3 to 2.3 3 2.5 2 1.5 1 10 12 14 16 18 20 17. (0,1) 3 2.5 2 1.5 1 0.5 0 -1 -0.5 0 0.5 0 1 1 19. (0,8) 8 7 6 5 4 3 2 1 0 -2 -1 21. (-2,-4) and (0,0) 2 391 0 -5 -4 -3 -2 -1 0 -5 -10 -15 -20 23. 1 0.8 0.6 0.4 0.2 0 0 200 400 600 800 1000 25. Image goes to infinity 30 20 10 0 -10 1 6 11 16 -20 25 20 15 10 5 0 0.001 27. see above. 0.006 0.011 392 Exercise 1.3A Problem Answer Problem Answer Problem Answer Problem Answer Problem Answer 1 -5 9 1 17 y=1 25 m=3/5, b=-32/5 33 y=-2x - 8 3 5 11 infinity 19 y=1.5x + 2.2 27 m=-1, b=-1 35 y=-2 5 0.5 13 2 21 y=1000x- 5000 29 m=2/7, b=-4 37 y=-3x – 2 41. y = 2.0182x - 1 120 100 80 d 60 40 20 0 0 10 20 30 t 40 50 60 43. y = 0.0759x + 0.0533 16 14 12 10 I 8 6 4 2 0 0 50 100 150 V 200 250 7 13/3 15 -5 23 m=2, b=3 31 y=-1.5x+ 0.5 39 y=-3.9x+ 7.67 393 Exercise 1.4A Problem Answer Problem Answer 1a ______ √u2 + 2as 7 F/P Problem Answer 15a (V/πh)1/2 1b v2-u2 2a 9 ___ 2π√l/g 15b V/(πr2) End of Chapter 1 Exercises Problem 1 Answer 7.74 x 10-3 Problem 9 Answer 7.7 mW Problem 17 Answer 100 ms Problem 25 Answer 1.6 fs 3a u = [s- ½ at2 ] t 11 m = 0.625k 3 -1.59 x 10 11 1.0 mm 19 2 µs 35 m=2, b=1 3b a = 2(s-ut) t2 13a 1 [R – 1 – αT] T2[ R0 ] 13c R0=R/[1+αT+βT2] 5 9.29 x 102 13 100 nm 21 5.3 x 106 g 37 m=7/2, b=3 41. y=-2 (Figures for problems 27, 29, 31, 33 on next page) 27. 2 1 0 -1 0 -2 -3 -4 -5 -6 -7 2 4 6 8 5 RT-R1-R2 13b 1 [R – 1 – βT2] T [ R0 7 2.36 x 105 15 38 mW 23 2 GHz 39 y=-x-5 394 29. 70 60 50 40 30 20 10 0 0 1 2 3 4 5 31. 250 200 150 y 100 50 0 0 50 100 150 x 200 250 300 395 33. 40 30 20 s' 10 0 -10 0 5 10 15 20 25 30 50 60 -20 -30 -40 s 43. 80 70 60 50 40 30 20 10 0 d(nm) 0 d = 1.5011eV + 0.5451 10 Chapter 2 Exercise 2.1A Problem Photon 20 30 E(eV) λ (nm) 40 λ (µm) λ (Angstrom) 1 3 5 7 9 11 13 15 17 19 blue IR IR IR x-ray UV IR Violet UV Violet/UV 477 2200 901 24,800 12.5 157 1064 405 106 400 0.477 2.2 0.901 24.8 0.0125 0.157 1.064 0.405 0.106 0.4 4770 22,000 9010 248,000 125 1570 10,640 4050 1060 4000 ν(THz) unless o/wise specified 629 136 333 THz 12.1 24 PHz 1.91 PHz 282 741 2.83 PHz 750 THz T (fs) unless o/wise specified 1.59 7.33 3.00 82.7 41.7 as 523 as 3.55 1.35 0.353 fs 1.33 396 Exercise 2.3A Problem 1 3 5 7 9 11 13 15 17 19 λ (nm) 477 564 901 2840 800 193 1064 2470 1330 400 ν (THz) 629 532 333 THz 106 375 1.55 PHz 282 122 226 750 THz E(J) 4.17x10-19 3.52x10-19 2.21x10-19 7 x 10-20 2.48x10-19 1.03x10-18 1.87x10-19 8.05x10-20 1.5 x 10-19 4.97x10-19 E(eV) 2.60 2.2 1.38 0.437 1.55 6.41 1.17 0.502 0.936 3.10 k (cm-1) 21,000 17,700 11,100 3520 12500 51,800 9,400 4050 7520 25,000 Exercise 2.3B 1. 1.95 mW 3 2 s 5. 6.20 nm Exercise 2.4A 1. 509 THz, 3.37x10-19 J, 2.10 eV, 17,000 cm-1 1.96 fs 3. 250 THz, 1.66 x 10-19 J, 1.03 eV, 1.2 µm, 8330 cm-1 5. 1.37 eV, 332 THz, 903 nm, 11,100 cm-1, 3.01 fs 7. 283 THz, 3.53 fs, 1.88 x 10-19J, 1.17 eV, 9430 cm-1 9. T= 5.2 fs, 192 THz, 1.27 x 10-19 J, 0.795 eV, 1.56 µm, 6400 cm-1 11. 22.6 THz, 44.2 fs, 13.3 µm, 753 cm-1, 0.0936 eV End of Chapter 2 Exercises ν THz k(cm-1) E(J) E(eV) 1. 356 nm 843 28,100 5.58 x 10-19 3.48 288 THz 9,600 1.91 x 10-19 1.19 3. 1.04 µm 5. 496 nm 604 THz 20,200 4.01 x 10-19 2.5 264 THz 8810 1.75 x 10-19 1.09 7. 1.14 µm 9. 589 nm 509 THz 17,000 3.37 x 10-19 2.10 11. 6500 A 462 THz 15,400 3.06 x 10-19 1.91 13. 1050 W 15. 1.08 billion km 17. No. It would provide only 3.31 eV whereas the threshold is 4.78 eV, λ Chapter 3 Exercise 3.3A 1. 70.7 mW/cm2. 3. (a) 14.3 MW (b) 3.57 MW/cm2 (c) 2 W 5. 1.59 kW/cm2 7. 25.5 kW/cm2 9. 1.23 x 105 mW/cm2. 11. 85,000 µW/cm2. End of Chapter 3 Exercises 1. c 2 f 3 g 4. a 5 b 6 e 7 d 3. Based on ANSI 1993 standards: Laser Class Power Class 1 below 0.4 µW Class 2 below 1 mW Class 3a 1 mW – 5 mW Class 3b 5 mW – 500 mW Class 4 above 500 mW (Note that the ANSI 2000 standards consider Class 3 to encompass 1-500 mW power, with the notations of Class 3(a) for visible and Class 3(b) for non-visible lasers. 5. 223 W/cm2 7. 1.11 mW/cm2 9.284 mJ/cm2 11. 6.35 W 397 Chapter 4 End of Chapter Exercises 1. 20 18 16 14 12 10 8 6 4 2 0 100 1000 10000 Wavelength (nm) 3. OC has an aperture that lets the output beam. 5. 4.51 eV 7. pumping or excitation 9. 84.4 eV. 11. 4.62 W Chapter 5 Exercise 5.2A 1. 37o 37o 53o 53o 3. 22o 22o 5 40o 50o 50o 65o 55o 65o 398 7b 7. a Object Object C F Image F Image 9a. 9b Image Image F F R R 11. 7 13 (a) C 37o o 35 o C 37 35o 15. Use similar triangles principle A Person 6’ M A’ Mirror 3’ Image 6’ N B B’ Triangle A’MN and triangle A’AB are similar. Since object and image distances from mirror are equal for plane mirrors, AM = A’M or A’M=one half of AA’. Since corresponding sides are proportional for similar triangles, MN is one half of AB, i.e. one needs only a 3’ mirror for the 6’ person to see his/her entire image. The point M does not have to line up with A horizontally. MN can be anywhere within the limits of AB. 399 Exercise 5.3A 1. (a) 2.08 x 108 m/s (b) 1.80 x 108 m/s (c) 1.50 x 108 m/s (d) 1.10 x 108 m/s 3(a) 13.2o (b) 22.5o (c) 28.1o (d) 37.9o 5 (a) 32.5o (b) 64.2o (c) and (d) not feasible since angle of refraction in second medium cannot exceed 90o. 7. see graph below. n = 1/slope = 1.64 sin(angle of refraction) y = 0.6181x 0.45 0.4 0.35 0.3 0.25 0.2 0.15 0.1 0.05 0 0 0.2 0.4 0.6 0.8 sin(angle of incidence) Exercise 5.3B 1 (a) 1.02 mm (b) 2.70 mm (c) 3.98 mm (d) 9.79 mm (3) 59.1o Exercise 5.3C 1. 3.01 mm 52o 3. 1 Air λ=710 nm 29.8o 2 8 v=1.89x10 m/s 3 v=1.70x108 m/s 4 v=1.43x108 m/s 26.5o 22.0o 1 Exercise 5.4A 1 (a) 44.0o (b) 38.1o (c) 34.4o (d) 26.4o (e) 19.5o (f) 11.5o 52o Glass (n=1.585) λ=448 nm Glass (n=1.767) λ=402 nm λ=337 nm Gem (n=2.105) Air 400 3. 3 cm 30 o 18.4 o 7 cm 30 o 3.67 cm 10 cm 5. 5 cm 75o 75o 5 cm 30.7o 8.42 cm 30.7o 3.91 cm 15 cm End of Chapter 5 Exercises 1. Angles 2, 5 and 6 are 75o each; angles 1, 3, 4 and 7 are 105o each 3. 3(a) Object -------------------c------------f-----------------------------------------------------------Image (b) Object Image - ------------f----------------------------------------------------------------C---- 401 5. 3.12 mW 7. (a) 16o, (b) 27.5o, (c) 34.7o and (d) 47.9o. 9. 4.76” 11. (a) 33.9o (sapphire @ 589 nm)) (b) 44.2o (calcium fluoride @589 nm)) (c) 16.9o (silicon at 3000 nm) (d) 40.8o (crown glass @ 400 nm) Chapter 6 Exercise 6.3A 1(i) (a) Multi (b) Step (c) plastic. (ii) (a) Multi (b) either (c) glass, (iii) (a) single (b) step (c) glass, (iv) (a) multi (b) either (c) glass, (v) multi (b) step (c) plastic, (vi) (a) single (b) step (c) glass. 3. 5 5. 657. 7. glass/glass Exercise 6.4A 1 (a) 80.6o (b) 14.1o (c) 0.243 3. (a) 80.6o (b) 14.1o (c) 0.244 5. (a) 80.7o (b) 14.2o (c) 0.246 7. (a) 48.8o (b) 61.3o (c) 0.877 Exercise 6.5A 1. 1.06 dB 3. Pout=889 µW, 2.27 dB/km. 5. 13.0 dB, 3.94 dB/km 7. Pout=19.9 mW, 0.99 dB 9. Pout=6.17 mW, 0.636 dB/km Exercise 6.6A 1. 0.561 dB 3. 0.915 dB 5. 0.884 dB 7. 4.08 dB 9. 1.58 dB 11. 5.67 dB End of Chapter Exercises 1(i) (a) Multi (b) Step (c) plastic. (ii) (a) Multi (b) either (c) glass, (iii) (a) Single (b) step (c) glass, (iv) (a) multi (b) either (c) glass, (v) (a) single (b) step (c) glass (vi) (a) multi (b) step (c) plastic. 3. 737 modes 5. 76.7o, 20.1o, 0.344. 7. Pout=2.67 mW, 2.27 dB/km 9. 14 dB, 4.24 dB/km 11. 0.730 dB 13. 18.4 dB Chapter 7 Exercise 7.3A 1. 34.6o 3. 61.1o 5. 43.3o 7. 39.2o 9. 37.0o 11. 50.8o Exercise 7.4A 1. 1.1 mW 3. 1.07 mW, 53.6% 5. 50.8o 7. 22.8o End of Chapter Exercises 1. 46.2o 3. 48.8o 5. 57.8o 7. 40.6o 9. 41.2o 11. 41.1o 13. 52o, 52o, 1.28 15. 62o, 62o, 1.88 17. 3.36 mW 19. 709 µW 21. 45o 23. 30o Chapter 8 Exercise 8.1 A 1. 554 µW, T=11.1%, 88.9% lost 3. 1.4 mW, T=42.7%, 65.4% lost 5. P=3.22 mW, T=64.4%, 35.6% lost 7. 0.462 cm 9. 4.93 cm 11. 2.30 in-1 13. 0.893 cm-1. Exercise 8.2 A 1. 1.52 3. 0.522 5. 75.9% 7. 5.62% 9. Seven, as follows (1) OD=0.3, T=50.1% (2) OD=0.6, T=25.1%, (3) OD=0.9, T=12.6% (4) OD=1, T=10% (5) OD=1.3, T=5.01% (6) OD=1.6, T=2.51% and (7) OD=1.9, T=1.26% 11. Use 0.301 OD + 0.477 OD together. Exercise 8.3 A 1. 0.00959 cm-1 3. 81.8%, 18.2% Exercise 8.3 B 1. 13.5% 3. 0.0101, 0.0175, 0.00765, all in km-1 units 5. 1.25, 11.3, 31.3, all in m-1 units 402 End of Chapter Exercises 1. 1.05 mW, 12.2% and 95.7% 3. 1.39 cm 5. 0.805 in-1 7. 0.886 9. 66.1%. 11. Seven, as follows: (1) OD=0.3, T=50.1% (2) OD=0.5, T=31.6%, (3) OD=0.7, T=20.0% (4) OD=0.8, T=15.8% (5) OD=1.0, T=10.0% (6) OD=1.2, T=6.31% and (7) OD=1.5, T=3.16% 13. 0.00406 cm-1 15. 22.3% Chapter 9 Exercise 9.1A 1. (i) 12.7 (ii) 63.6. 3. 207 µradians Exercise 9.2A 1 (a) 7 Hz (b) 143 ms (c) 1.40 x 10-7 (d) 7.5 MW 3. (a) 500 W (b) 250 Hz (c) 800 µs (d) 400 mJ 5. Rise time=5.5 µs Fall time=7.5 µs Pulse width = 4 µs T=46 µs PRF=21.7 kHz Duty cycle = 0.087. 7. 0.008 9. (a) 1.67 kHz (b) 600 µs 11. 100 mJ. End of Chapter Exercises 1. 12.2 and 72.9 3. 448 µradians 5. (a) 450 W (b) 250 Hz (c) 1.33 ms (d) 600 mJ 7. 0.01 9. (a) 3.33 kHz (b) 300 µs 11. 1 J Chapter 10 Exercise 10.1A 1. 24 mm, 41.7 D 3. 48 cm, 2.08 D 5 . 17.8 cm, 5.63 D 7. ∞, 0. 9. 10 cm, 10 D 11. -14 cm, -7.14 D. 13. ∞, 0 (i.e., a flat piece of glass, not a lens). Exercise 10.2A 1. s’=60 mm, real, M= -5, inverted, Image height = -5 mm. 3. f=5 cm, real, M=-1, inverted, Image height = 1 cm. 5. f=-20 cm, virtual, M=0.5, erect, Image height=1 cm. 7. s=6.25 cm, s’=25 cm, real, M=-4, inverted. 9. s=15 cm, s’=-3.75 cm, virtual, erect, Image height= 5 mm 11. s’=-3.33 cm, virtual, M= 0.667, erect, Image height = 4.67 mm 13. s=26 mm, virtual, M = 0.5, erect, Image height = 0.5 mm 15. s’=10 cm, real, M = -1, inverted, Image height = 5 mm 17. s’=-7.5 cm, virtual, M = 0.5, erect, Image height = 12.5 mm 19. f=minus infinity, M = 1, erect, Image height = 1.2 cm (This is a glass plate, not a lens! You would see a faint, virtual image of the object in the plate). Exercise 10.3A 1. Regular ray tracing method: 403 6 5 4 3 2 1 0 0 5 10 5 10 15 20 25 30 Oblique method: 6 5 4 3 2 1 0 0 15 20 25 30 3. Regular method: 6 5 4 3 2 1 0 0 5 10 15 20 404 Oblique method: 6 5 4 3 2 1 0 Plane through F’ 0 5 10 15 20 5. Regular ray tracing: 20 15 10 5 0 0 50 100 150 Oblique method: 20 15 10 5 Plane thru’ F’ 0 0 50 100 150 405 Exercise 10.4A 1. Final image is located 60 mm to the right of the second lens, erect, real and 16 mm in height. 3. Final image is located 54.5 mm to the right of the second lens, inverted, real and about 1 mm in height. 5. Final image is located 32.8 cm to the right of the third lens, inverted, real and about 1.5 cm in height. Exercise 10.4B 1. Regular ray tracing method: 4 3.5 3 2.5 2 1.5 1 0.5 0 L1 0 10 L2 20 30 40 50 Oblique method for second lens. 4 3.5 3 2.5 2 1.5 1 0.5 0 0 10 20 30 40 50 406 3. Regular ray tracing method: L1 L2 2 1.5 1 0.5 0 0 10 20 30 40 Oblique method for L2: L1 Oblique ray for L2 4 3.5 3 2.5 2 1.5 1 0.5 0 0 10 F2 plane L2 Construction ray Final Image 20 30 40 5. Regular ray tracing method: First image Final image Second image L2 L3 L1 4 3.5 3 2.5 2 1.5 1 0.5 0 0 10 20 30 40 50 60 407 Oblique method: L1 Oblique ray for L2 4 3.5 3 2.5 2 1.5 1 0.5 0 0 10 20 L2 L3 F3’ plane F2’ plane Obliq. for L3 30 40 50 60 End of Chapter Exercises 1. f=80 mm, P=12.5 D 3. f=-8.88 cm, P=-11.3 D 5. s’=20 mm, real, M=-1, inverted, image height = 1 mm 7. f=-20 cm, virtual, M=0.5, erect, image height= 5 mm 9. Regular ray tracing 4 3.5 3 2.5 2 1.5 1 0.5 0 0 5 10 15 20 25 408 Oblique method: 4 3.5 3 2.5 2 1.5 1 0.5 0 0 5 10 15 20 25 11. Regular ray tracing 4 3.5 3 2.5 2 1.5 1 0.5 0 0 10 20 30 40 Oblique method: 4 3.5 3 2.5 2 1.5 1 0.5 0 0 10 20 30 40 13. Final image located 35.8 mm to the right of the second lens, inverted, real and about 0.4 mm high. 409 15. 6 5 4 3 2 1 0 0 10 20 30 40 50 17. Final Image @ 5.9cm to the right of L2 2 1.5 1 0.5 0 0 10 20 30 40 50 60 410 Chapter 11 Exercise 11.1A 1. Object Exit Pupil Lens Image Ent. Pupil/Aperture 60 50 40 mm 30 20 10 0 0 5 10 15 20 25 30 cm 3. Object Aperture/Ent. Pupil Lens Image Exit Pupil 60 50 40 mm 30 20 10 0 0 2 4 6 cm 8 10 411 5. Object Aperture/Ent. Pupil Lens Image Exit Pupil 60 mm 50 40 30 20 10 0 0 10 20 30 cm 7. Object Image Lens Ent Pupil Aperture and Exit Pupil 60 50 40 mm 30 20 10 0 0 5 10 cm 15 412 9. Object Image Lens Ent. Pupil Aperture/Exit Pupil 60 50 mm 40 30 20 10 0 0 5 10 cm Exercise 11.1B 1. θL1=15.9o, θs=4.61o, θL2=4.57o, therefore L2 is the aperture. 3. θL1=18.4o, θs=7.1o, θL2=4.23o, therefore L2 is the aperture. 5. θL1=14.4o, θs=20.6o, θL2=56.3o, therefore L1 is the aperture. Exercise 11.2A 1. 5X 3. 400X, 11.3 cm 5. 1.07 cm, 13.5, 675X 7. 33.3X 9. 10 mm, 6 mrad, and 6 mm, 10 mrad 11. fo=12.5 cm, fe=37.5 cm 13. fo=-25.0 cm, fe=75 cm End of Chapter Exercises 1. (a) Angle subtended by image of lens at the foot of the object = 22.6o Angle subtended by image of stop at the foot of the object = 17.4o (b) The stop is the aperture (c) The stop is still the aperture; it does not depend on how tall the object is because the foot of the object is located at the same distance as before. 3. Mang=2.5, M=3.5 5. fo=7 cm; fe=35 cm 7. (c) a plano-convex lens with the curved side facing the light Chapter 12 Exercise 12.2 A 1. One can look at diffraction occurring from a single slit as interference caused by spherical wave fronts emanating per Huygen’s principle from different locations of the slit opening including the slit edges. Thus, the diffraction patterns caused by a single slit are because of interference. 3. 2.63 m 5. 633 nm 7. 732 µm. Exercise 12.3 A 1. >> 2 m 3. >> 2.6 m 5. > 7 m 7. >> 2.7 m Exercise 12.3 B 1. (a) 2.53 mradians (b) 10.1 mm. 3. 590 nm 5. 25.3 mm square 7. 25.7 mm square 9. (a) 38.6 mm (b) 3.86 mradians 11. m=1, angle=31o is the only fringe possible other than m=0. 13. In addition to the zeroth order fringe, we have (i) 9.11o, (ii) 18.5o, (iii) 28.3o, (iv) 39.3o, (v) 52.3o, and (vi) 71.7o, as locations of six additional fringes. End of Chapter Exercises 1. 5.26 m 3. 88.5 nm. 5. >> 1.1 m 7. 600 nm 9. 47.5 mm x 25.3 mm 11. In addition to the zeroth order fringe, twenty additional fringes exist at 2.80o, 5.60o, 8.42o, 11.3o,...and 77.4o 413 View publication stats