PHYSICS 332 LABORATORY REPORT OPTICS Laboratory Name and ID: 1. ……………………………………………………………………………………………… 2. ………………………………………………………………………………………………. Date performed: ……………………..… Time section meets:……………………… Title: Polarization of light Grading Checklist LABORATORY JOURNAL: PREDICTIONS (individual predictions and warm-up completed in journal before each lab session) LAB PROCEDURE (measurement plan recorded in journal, tables and graphs made in journal as data is collected, observations written in journal) BONUS POINTS FOR TEAMWORK LABORATORY REPORT: ORGANIZATION (clear and readable; logical progression from problem statement through conclusions; pictures provided where necessary; correct grammar and spelling; section headings provided; physics stated correctly) DATA AND DATA TABLES (clear and readable; units and assigned uncertainties clearly stated) RESULTS (results clearly indicated; correct, logical, and well-organized calculations with uncertainties indicated; scales, labels and uncertainties on graphs; physics stated correctly) CONCLUSIONS (comparison to prediction & theory discussed with physics stated correctly ; possible sources of uncertainties identified; attention called to experimental problems) TOTAL 1 الصفحة Polarisation Points I. Objects of the experiments The purpose of this laboratory activity is to determine the relationship between the intensity of the transmitted light through two polarizers and the angle, Ø, of the axes of the two polarizers. II. Principles and Theory A polarizer only allows light which is vibrating in a particular plane to pass through it. This plane forms the “axis” of polarization. Unpolarized light vibrates in all planes perpendicular to the direction of propagation. If unpolarized light is incident upon an “ideal” polarizer, only half will be transmitted through the polarizer. Since in reality no polarizer is “ideal”, less than half the light will be transmitted. The transmitted light is polarized in one plane. If this polarized light is incident upon a second polarizer, the axis of which is oriented such that it is perpendicular to the plane of polarization of the incident light, no light will be transmitted through the second polarizer. However, if the second polarizer is oriented at an angle so that it is not perpendicular to the first polarizer, there will be some component of the electric field of the polarized light that lies in the same direction as the axis of the second polarizer, thus some light will be transmitted through the second polarizer (see the bottom figure). The component, E, of the polarized electric field is found by: 𝐸 = 𝐸0 cos ∅ Since the intensity of the light varies as the square of the electric field, the light intensity transmitted through the second filter is given by: 𝐼 = 𝐼0 𝑐𝑜𝑠 2 ∅ where 𝐼0 is the intensity of the light passing through the first filter and Ø is the angle between the polarization axes of the two filters. Consider the two extreme cases illustrated by this equation: • If Ø is zero, the second polarizer is aligned with the first polarizer, and the value of 𝑐𝑜𝑠 2 Ø is one. Thus the intensity transmitted by the second filter is equal to the light intensity that passes through the first filter. This case will allow maximum intensity to pass through. • If Ø is 90º, the second polarizer is oriented perpendicular to the plane of polarization of the first filter, and the 𝑐𝑜𝑠 2 Ø gives zero. Thus no light is transmitted through the second filter. This case will allow minimum intensity to pass through. 2 الصفحة Polarisation • These results assume that the only absorption of light is due to polarizer effects. In fact most polarizing films are not clear and thus there is also some absorption of light due to the coloring of the Polaroid filters. The experiment is set up according to Fig. 1. It must be made sure that the photocell is totally illuminated when the polarization filter is set up. Figure 1 :Experimental setup for investigating the type of polarization of the emergent light (schematically).(a) halogen lamp (d) polarizer (f) analyzer (g) Si photo cell (h) translucent screen Questions 1) The polarization filter (analyzer) is then rotated in steps of 10° between the filter positions +/- 90° and the corresponding photo cell current (most sensitive direct current range of the digital multimeter) is determined. Complete the table? ∅ 90 80 70 60 50 40 30 20 10 0 10 20 30 40 50 60 70 𝑐𝑜𝑠 2 Ø 𝑰 𝑰/𝑰𝟎 2. Write the relationship between the intensity 𝐼 of light transmitted through the analyzer, and the relative angle ∅ between the transmission axis of the analyzer and the plane of polarization of the incident light, intensity 𝐼0 : 3 الصفحة Polarisation 80 90 ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………… 3. Describe in your own words what is mean by and draw a picture of unpolarized light and polarized light. ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………… 4. Plot the graphs of the intensity (I) of transmitted light as a function of the angle (∅) , in radians, between the axis of analyzer and the plane of polarization (second polarizers) of light. What is the shape of the plot of light intensity versus angle? ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………….……… 5. Plot the graphs of and intensity(𝑰/𝑰𝟎 ) as function of 𝑐𝑜𝑠 2 ∅. What is the shape of the plot? Does the curve verify Malus’s law? ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………… 4 الصفحة Polarisation We see in Figure 2 that the intensity reaches a minimum at approximately 1.6 radians (_ 90◦), when the axis of the polariser is perpendicular to the polarisation of the light, as is expected. The maxima occur near where the relative angle is either 0◦ or 180◦. 5 الصفحة Polarisation Figure 3 approximately verifies Malus’ law (equation 2), that the intensity of the transmitted light is proportional to the cosine of the relative angle. Towards the maxima there are deviations such that the graph appears to split into two lines. During the experiment it was found that the value of the intensities at the maxima were not equal, as should have occurred. This is believed to be due to experimental factors such as additional reflections of the laser light on parts of the apparatus, and perhaps some inhomogeneity of the polariser itself. The line of greater slope corresponds to the maximum around the 0◦ point, while the lesser slope corresponds to the 180◦ region. 6 الصفحة Polarisation