Waves and Optics Introduction The energy that generated the wave that moves this surfer was transmitted to the water many miles away. But the mass of water that originally received that energy did not move all those miles to the shore. Rather, it transmitted that energy to the next mass of water, which transmitted it to the next. In waves, energy is transmitted, not matter. There are many different kinds of waves. They include: waves on the surface of water,… …waves on strings,… …waves within the earth,… …sound waves,… …light waves,… …radio waves,… …and x-rays. All of these types of waves transfer energy without transferring matter. This chapter will explore aspects of these waves that are similar to and different from each other. Mechanical & Electromagnetic Waves There are two fundamental types of waves: Mechanical waves and Electromagnetic waves. What makes these two types of waves different are their sources and mediums. Mechanical Waves have matter as their source and matter as their medium. As we have already seen, a guitar string wave has a finger as its source and the string as its medium. The wave is propagated as one particle puts a force on the next particle, causing it to move and continue the wave. Sound is a Mechanical Wave. Sound waves have as their source vibrating matter. In the case of the horn players, their vibrating lips start the sound wave. For a singer, vibrating vocal cords are the source of the sound wave. Sound also travels through solids and liquids. In fact, it travels faster and with less loss the denser the material. Particles are closer together allowing for greater elastic qualities than air. Scratch on your desk or table so that you can just barely hear it. Then put your ear to the desk. You should be able to hear the scratching loud and clear. The Medium of sound is also matter, in this case, air. As one air molecule collides with the next, which collides with the next, sound is propagated through space. On the other hand, sound, like all mechanical waves, cannot travel in a vacuum. Sound needs matter to be transmitted, particles exerting forces on other particles. In a vacuum, there are no particles, so sound cannot be transmitted. Experiments have shown that if you have a bell in a jar, the bell can be heard. But if you evacuate the jar of all air, the bell can no longer be heard. http://www.youtu be.com/watch?v= ce7AMJdq0Gw On the other hand, the bell itself can clearly be seen, air or no air. This means that light is not a Mechanical Wave – it is instead an Electromagnetic Wave. Electromagnetic waves include light, radio waves, microwaves, infrared radiation, ultraviolet light, x-rays and gamma rays. The source of Electromagnetic waves is vibrating charge. In most cases, that means vibrating electrons. Because they do not rely on matter for propagation, Electromagnetic Waves can travel through vacuum. All Electromagnetic Waves travel through a vacuum at the speed of light, ‘c’, where … c = 3.0 x 108 m/s When light does travel through matter, such as water or glass, it travels more slowly. And different waves travel at different speeds. The speed of light, ‘c’, is for Electromagnetic Waves moving through a vacuum only. A WAVE is a vibration or disturbance in space that propagates from one place to another. A MEDIUM is the substance that all SOUND WAVES travel through and need to have in order to move. Two Types of Waves The first type of wave is called Longitudinal. Longitudinal Wave – the displacement of individual particles is in the same direction as the direction of the wave propagation Two Areas of the Wave: Compression- an area of high molecular density and pressure. Rarefaction - an area of low molecular density and pressure. Longitudinal waves are mechanical waves. The second type of wave is called Transverse. Transverse Wave – the displacement is at right angles to the direction of the propagation. Light and Radio Waves are examples. In the animation of the motion of a particle in a transverse wave, the wave moved left to right and the particle moved up and down. Together, they form a plane – the plane of the screen you are viewing this animation on. Direction of Wave Velocity Direction of Motion of the Medium Comparing Longitudinal and Transverse Waves http://www.youtube.com/watch?v=Rbuhdo0AZDU What about a water wave? A cork in water travels both Horizontally and Vertically. In that sense, water has characteristics of both longitudinal and transverse waves. λ = wavelength – distance between wave repeats. T = Period – time required for 1 one wave to pass a fixed point. Units = seconds f = frequency – waves passing a fixed point in a given time period. f = 1/T s-1 1 s Units = Hertz The frequency of a sound wave is perceived as pitch. The higher the frequency, the higher the pitch. For instance a trombone has a lower frequency and pitch than a flute. Low Frequency And Pitch High Frequency And Pitch d (m) d (m) t (s) t (s) The frequency of a light wave is perceived as color. The higher the frequency, the closer to the violet end of the spectrum. The lower the frequency, the closer to the red end. White light is a mixture of all frequencies at once. d (m) d (m) t (s) t (s) Low f Red High f Violet The greater the frequency of a light wave, the shorter the wavelength. d (m) t (s) d (m) t (s) Amplitude = height of the wave There are a number of measurements that can be made on a wave. For mechanical waves, the Amplitude is the maximum distance the medium deviates from its equilibrium position. (click to animate) Amplitude Amplitude The Power/Energy of a wave is proportional to the square of the Amplitude. The greater the Amplitude, the more power in the wave. The greater the Amplitude of an earthquake wave, the more destructive the earthquake. The greater the Amplitude of a sound wave, the greater the volume. The greater the Amplitude of a light wave, the brighter or more intense it is. Speed of a Wave (m/s) v=λ T v = λf Sound waves (longitudinal waves) traveling in air with a speed of 343 m/s. The lowest frequency of sound we can hear is 20 Hz. The highest is 20.0 kHz. Find their wavelengths. v=λ f v=λ f 343m/s = λ 20 s-1 λ = 17.2 m 343 m/s = λ 20,000 s-1 λ = 0.0172 m or … λ = 1.72 cm Factors that influence the speed of sound are density of solids or liquid, and temperature. How long will it take a shout to travel 75.0 m across a football field on an evening when the temperature is 18OC? vsound = 331 m/s + 0.6 m/s (18OC) = 341.8 m/s OC t = d v t = 75.0 m 341.8 m/s t = .219 s Speed in a Solid v = є ρ ε = modulus of elasticity ρ = density = N/m2 = kg/m3 Modulus of Elasticity is the mathematical description of an object or substance's tendency to be deformed elastically (i.e., nonpermanently) when a force is applied to it. What is the speed of sound in a sample of aluminum that has an є of 69 x 109 N/m2 and a density of 2700 kg/m3? v = є ρ v = 69 x 109 N/m2 2700 kg/m3 v = 5055 m/s Speed in a Liquid v = β ρ β ρ = bulk modulus = density = N/m2 = kg/m3 Bulk Modulus of a substance measures the substance's resistance to uniform compression. What is the speed of sound in a sample of ethanol alcohol that has an β of 1.02 x 109 N/m2 and a density of 789 kg/m3 at 20.0OC? v = β ρ v = 1.02 x 109 N/m2 789 kg/m3 v = 1137 m/s Miscellaneous Sound Information Supersonic speed is a rate of travel of an object that exceeds the speed of sound (Mach 1). For objects traveling in dry air of a temperature of 20°C (68°F) this speed is approximately 343 m/s (1,125 ft/s, 768 mph or 1,236 km/h). Speeds greater than five times the speed of sound (Mach 5) are often referred to as hypersonic. A sonic boom is the sound associated with the shock waves created by the supersonic flight of an aircraft. Sonic booms generate enormous amounts of sound energy, sounding much like an explosion. When an object passes through the air, it creates a series of pressure waves in front of it and behind it, similar to the bow and stern waves created by a boat. These waves travel at the speed of sound, and as the speed of the object increases, the waves are forced together, or compressed, because they cannot "get out of the way" of each other, eventually merging into a single shock wave at the speed of sound. This critical speed is known as Mach 1 and is approximately 1,225 km/h (761 mph) at sea level and 20 °C (68 °F). Ultrasound and what we hear … The sound we normally hear is from 20 to 20 000 cycles per second (Hz). Ultrasound means sound that has a higher frequency than our normal hearing. Ultrasound used for medical purposes is from one MHz (one million cycles per second) to 20 MHz. Sea Navigation and Sonar Sonar is a device that uses sound energy to locate objects; measure their distance, direction, and speed; and produce pictures of them. Sonars generally use frequencies in the 20 to 20,000 Hertz range. Sound Waves Sound Waves are a common type of standing wave as they are caused by RESONANCE. Resonance – when a FORCED vibration matches an object’s natural frequency thus producing vibration, sound, or even damage. One example of this involves shattering a wine glass by hitting a musical note that is on the same frequency as the natural frequency of the glass. Natural frequency depends on the size, shape, and composition of the object in question. Because the frequencies resonate, or are in sync with one another, maximum energy transfer is possible. Electromagnetic Spectrum Our eyes can only see certain wavelengths. c=λf f=c λ •What is the frequency of a radio wave that has a length of 1.67 X 1012nm? f=c λ = 3.00 X 8 10 m/s 1670 m f = 1.80 X 105/s Note: changed nm to m: (1 m = 1 x 109nm) (Hz) Find the frequency of a gamma wave that has a length of 0.00300 nm. = f=c λ 3.00 X 108 m/s 3.00 X 10-12 m f = 1.00 X 1020/s (Hz) •Comparing ... long radio wave = 1.80 X 105/s 20 short gamma = 1.00 X 10 /s •What are the wavelengths of these two radio stations? KNYE broadcasts at 95.1 MHz (FM) KDKA broadcasts at 1020 KHz (AM) KNYE = 3.00 X 108 m/s = 3.15 m 6 95.1 X 10 Hz MHz = 106 Hz KDKA = 3.00 X 108 m/s = 294 m 1020 X 103 Hz KHz = 103 Hz • • • • AM broadcast 500 KHz - 1600 KHz FM broadcast 88 MHZ - 108 MHz Cordless phones 46 MHz - 49 MHz TV remote operates in the Infrared region of the EM (electromagnetic) spectra. Wave Hz Meters Water 5 x 10-2 120 7 x 10-7 Red Light 4.3 x 1014 Violet Light 7.5 x 1014 4 x 10-7 • As the frequency goes up: –wavelength decreases. –wave becomes more energetic. Doppler Shift When you hear the sound of an object moving toward you—such as a siren from and emergency vehicle—the pitch of that sound is higher than if the vehicle was standing still. (f is greater and a shorter λ) Likewise, as the vehicle moves away from you, the siren would have a lower pitch. (f is less and a greater λ) Object is moving towards the right. What would happen if the source moved quicker? f would increase and λ would decrease. Doppler Effect on a Moving Observer f’ = ( 1 +/- u ) f v f = normal frequency v = speed of wave f ‘= new frequency u = speed of observer + = observer moving closer to source - = observer moving away from source A street musician sounds the A string of his violin producing a tone of 440 Hz. What frequency does a bicyclist hears as he (A) approaches and (B) recedes from the musician at 11.0 m/s? f’ = ( 1 +/- u ) f v A=+ B=- f’A = 454 Hz f’B = 426 Hz Doppler Effect on a Moving Source 1 1 -/+ u v ( ( f’ = f v = speed of wave u = speed of source - = source moving towards observer + = source moving away from observer A police car’s siren produces a frequency of 1600 Hz when at rest. (air temperature 20OC) What is the frequency detected if the car is moving … A) toward an observer at 25 m/s? B) away from an observer at 25 m/s? v = 331 m/s + 0.6 m/s (20OC) OC v = 343 m/s 1 -/+ u v f’ = ( 1 - f 1_______ 25 m/s 343 m/s f’ = 1726 Hz ( 1 ( ( f’ = 1600 Hz ( 1 + 1_______ 25 m/s 343 m/s f’ = 1491 Hz ( f’ = 1600 Hz Nodes and Antinodes Superposition and Interference Superposition: Combination of two or more waves to form a resultant wave y = y1 + y2 Constructive Interference: When waves combine and the resulting pulse has an amplitude equal to the sum of the amplitude of the individual waves. The net displacement is the sum of the two individual displacements. They pass through each other without being disturbed. Destructive Interference: When the positive displacement of one waves combines with the negative displacement of another to create a net displacement of zero. In and Out of Phase These two waves are in phase, meaning that every point on the top wave is at the same part of the wave cycle as the same point on the bottom wave. The crests line up with crests and the troughs line up with troughs. These two waves are out of phase with each other. None of the points on the top wave are at the same part of the wave cycle as the same point on the bottom wave. These two waves are exactly out of phase with each other. Notice that crests all line up with troughs. This is called being 180º out of phase. Photoelectric Effect The phenomenon of emission of electrons from mainly metal surfaces exposed to light energy (X – rays, UV rays, Visible light and even IR rays) of suitable frequency is known as photoelectric effect. Visible light Photoelectrons Alkali Metals The electrons emitted by this effect are called photoelectrons. The current constituted by photoelectrons is known as photoelectric current. The photoelectric effect demonstrates that light has particle properties. The photoelectric effect demonstrates that light has particle properties. … on the other hand Diffraction When light OR sound is produced by TWO sources a pattern results as a result of interference. Diffraction of Light http://www.youtube.com/watch?v=-mNQW5OShMA&NR=1 Interference Patterns Diffraction … occurs when a wave encounters an obstacle. It is described as the apparent bending of waves around small obstacles and the spreading out of waves past small openings Diffraction – The Central Maximum Suppose you had a source emitting a wave through a small opening or slit. The distance between the slits = d. The distance from the slit spacing to the screen = L. If two waves go through the slit and act constructively, their amplitudes will build and leave a very bright intense spot on the screen. We call this the CENTRAL MAXIMUM. Figure 1 Figure 2 In figure 2 there are several bright spots and dark areas in between. The spot in the middle is the BRIGHTEST and thus the CENTRAL MAXIMUM. These spots are called FRINGES. Each additional fringe is a little less intense. These additional bright spots are known as ORDERS. The first bright spot on either side of the central maximum is called the FIRST ORDER BRIGHT FRINGE. Figure shows the intensity of the orders as we move farther from the bright central maximum. Orders, m Second Order Bright Fringe First Order m=2 Bright Fringe m=1 m, represents the ORDER of the fringe from the bright central. First Order Central Bright Fringe Maximum m=1 Second Order Bright Fringe m=2 Bright Fringes = CONSTRUCTIVE INTERFERENCE. Dark Fringes In the pattern there are dark the DARK REGIONS. These are areas where DESTRUCTIVE INTERFERENCE has occurred. These areas DARK FRINGES or MINIMUMS. First Order Dark Fringe m=1 ZERO Order Dark Fringe m=0 ZERO Order Central Dark Fringe Maximum m=0 First Order Dark Fringe m=1 Dark Fringes = DESTRUCTIVE INTERFERENCE. d If the fringe is further out, Ө changes. X X X X X Ө d X Two slits with a separation of (d) 8.5 x 10-5 m create an interference pattern on a screen (L) 2.3 m away. If the 10th bright fringe above the central fringe is a linear distance of 12 cm from it, what is the wavelength of the light used in the experiment? TanӨ = 0.12 m 2.3 m Ө = 3.0O 10th bright fringe … λ = d sinӨ m λ = (8.5 x 10-5 m) sin3.0 10 λ = 4.4 x 10-7 m or 440 nm Light, Reflection, & Mirrors The Law of “REFLECTION” Law of Reflection … the angle of incidence (incoming ray) equals the angle of reflection (outgoing ray) NORMAL Angles are measured from a perpendicular line to the surface called a NORMAL. The Law of Reflection is always observed (regardless of the orientation of the surface) Two Types of Reflection Specular Reflection – reflection off a smooth shiny surface. Notice the symmetry. Diffuse Reflection – reflection off a rough surface. Notice the Light reflected in sXymmetry. many directions Plane Mirror A flat , plane mirror is mounted vertically. A candle is placed 10 cm in front of the mirror. WHERE IS THE IMAGE OF THE CANDLE LOCATED? Mirror On the surface of the mirror? Same side as the object? Behind the mirror? Object Distance, Do = 10 cm Behind the mirror. mirror Object Distance, Do = 10 cm Virtual Image Image Distance, Di = 10 cm Do = Di, and the heights are equal as well. Virtual Images Virtual images are images that are formed in locations where light does not actually reach. Light does not actually pass through the location on the other side of the mirror; it only appears to an observer as though the light is coming from this location. VIRTUAL IMAGES are ALWAYS on the OPPOSITE side of the mirror relative to the object. Real Image The point of convergence of the bundle of rays emergent from the lens is a real image of that object. For MIRRORS real images always appear on the SAME SIDE of the mirror as the object. Object Image X The characteristics of the image, however, may be different from the original object. These characteristics are: •SIZE (reduced, enlarged, same size) •POSITION (same side, opposite side) •ORIENTATION (right side up, inverted) Spherical Mirrors – Concave & Convex X X X … also called X DIVERGING X X mirror X X X X … also called X CONVERGING X mirror X X X Looking into a X cave X X X A converging mirror is one that is spherical in nature by which it can FOCUS parallel light rays to a point directly in front of its surface. This point is called the FOCAL POINT. To find this point you MUST use light from “infinity” Light from an “infinite” distance, most likely the sun. http://www.youtube.com/watch?v=np8lE Nrge0Q&feature=player_embedded Converging (Concave) Mirror The mirror is spherical. It has a CENTER OF CURVATURE, C. The focal point happens to be HALF this distance. Draw a line through the center of the mirror and call it the PRINCIPAL AXIS. f = C C= f Ray Diagram A ray diagram is a pictorial representation of how the light travels to form an image and can tell you the characteristics of the image. Object C f Principal axis Rule One: Draw a ray, starting from the top of the object, parallel to the principal axis and then through “f” after reflection. Object C f Principal Axis Rule Two: Draw a ray, starting from the top of the object, through the focal point, then parallel to the principal axis after reflection. Object C f Principal Axis Rule Three: Draw a ray, starting from the top of the object, through C, then back upon itself. What do you notice about the three lines? THEY INTERSECT The intersection is the location of the image. Image Characteristics Object C f Principal axis After getting the intersection, draw an arrow down from the principal axis to the point of intersection. … then ask yourself these questions: 1) Is the image on the SAME or OPPOSITE side of the mirror as the object? Same, therefore it is a REAL IMAGE. http://www.youtube.com/watch?v=KVpSCICCD9A&feature=player_embedded 2) Is the image ENLARGED or REDUCED? 3) Is the image INVERTED or RIGHT SIDE UP? The Mirror/Lens Equation There is ANOTHER way to predict image characteristics besides the ray diagram. Use the MIRROR/LENS equation to CALCULATE the position of the image. 1 1 1 = + f do di A concave spherical mirror has a focal length of 10.0 cm. Locate the image for an object distance of 25 cm and describe the image’s characteristics. 1 1 1 1 1 1 = + → = + f do di 10cm 25cm d i di = 16.67 cm What does this tell us? First we know the image is BETWEEN C and f. Since the image distance is POSITIVE the image is a REAL IMAGE. Real image = positive image distance Virtual image = negative image distance What about the size and orientation? Magnification Equation To calculate the orientation and size of the image we use the MAGNIFICATION EQUATION. d i hi M =− = d o ho A POSITIVE magnification, the image is UPRIGHT. A NEGATIVE magnification, the image is INVERTED The magnification value is GREATER than 1, the image is ENLARGED. The magnification value is LESS than 1, the image is REDUCED. The magnification value is EQUAL to 1, the image is the SAME SIZE as the object. Using our previous data … 16.67cm M =− 25cm M = −0.67 x …the image is INVERTED, and REDUCED. The +/- Sign Conventions The sign conventions for the given quantities in the mirror equation and magnification equations are as follows: f is + if the mirror is a concave mirror f is - if the mirror is a convex mirror di is + if the image is a real image and located on the object's side of the mirror. di is - if the image is a virtual image and located behind the mirror. hi is + if the image is an upright image (and therefore, also virtual) hi is - if the image an inverted image (and therefore, also real) A certain concave spherical mirror has a focal length of 10.0 cm. Locate the image for an object distance of 5 cm and describe the image’s characteristics. 1 1 1 1 1 1 = + → = + f do di 10cm 5cm d i d i = -10 cm di M =− = 5cm 2 Characteristics? Upright di = -10 cm M=+2 •VIRTUAL (opposite side) Enlarged Convex Mirrors OBJECT X X X X A convex mirror is sometimes referred to as a diverging mirror due to the fact that incident light originating from the same point and will reflect off the mirror surface and diverge. The diagram shows an object placed in front of a convex mirror. Light rays originating at the object location are approaching and then reflecting from the mirror. Observers must sight along the line of a reflected ray to view the image of the object. Each ray is extended backwards to a point of intersection - this point of intersection of all extended reflected rays is the image location of the object. The image is virtual. Light does not pass through to the image location. It only appears to observers as though all the reflected light from each part of the object is diverging from this virtual image location. Convex Mirrors – Ray Diagrams The revised rules can be stated as follows: Any incident ray traveling parallel to the principal axis on the way to a convex mirror will reflect in such a manner that its extension will pass through the focal point. Any incident ray traveling towards a convex mirror such that its extension passes through the focal point will reflect and travel parallel to the principal axis. 1) Pick a point on the top of the object and draw two incident rays traveling towards the mirror. Draw a ray such that it travels exactly parallel to the principal axis. Draw one ray so that it travels towards the focal point on the opposite side of the mirror. 2) Once these incident rays strike the mirror, reflect them according to the two rules of reflection for convex mirrors. The ray that travels towards the focal point will reflect and travel parallel to the principal axis. The ray that traveled parallel to the principal axis on the way to the mirror will reflect and travel in a direction such that its extension passes through the focal point. 3) Locate and mark the top of the image. The image point of the top of the object is the point where the two reflected rays intersect. Since the two reflected rays are diverging, they must be extended behind the mirror in order to intersect. Draw the extensions until they intersect. The point of intersection is the image point of the top of the object. Both reflected rays would appear to diverge from this point. Characteristics of Convex Mirrors . The diagrams show that in each case, the image is located behind the convex mirror. … a virtual image … an upright image reduced in size (i.e., smaller than the object) Another characteristic of the images of objects formed by convex mirrors pertains to how a variation in object distance affects the image distance and size. A 4.0-cm tall light bulb is placed a distance of 35.5 cm from a convex mirror having a focal length of -12.2 cm. Determine the image distance and the image size. ho = 4.0 cm di = ??? do = 35.5 cm hi = ??? f = -12.2 cm 1 1 1 = + f do di 1/(-12.2 cm) = 1/(35.5 cm) + 1/di -0.0820 cm-1 = 0.0282 cm-1 + 1/di -0.110 cm-1 = 1/di di = -9.08 cm To determine the image height (hi), the magnification equation is needed. hi/ho = - di/do hi /(4.0 cm) = - (-9.08 cm)/(35.5 cm) hi = (4.0 cm) • (9.08 cm)/(35.5 cm) hi = 1.02 cm The negative values for image distance indicate that the image is located behind the mirror. In the case of the image distance, a negative value always indicates the existence of a virtual image located behind the mirror. In the case of the image height, a positive value indicates an upright image. From the calculations in this problem it can be concluded that if a 4.0 cm tall object is placed 35.5 cm from a convex mirror having a focal length of -12.2 cm … … The image will be upright and … 1.02 cm tall and … located 9.08 cm behind the mirror. Convex mirrors always produce images that are upright, virtual, reduced in size, and located behind the mirror. Lenses – The Mirror/Lens Equation CALCULATE the image’s position and characteristics using the same equations used for mirrors. An object is placed 35 cm in front of a converging lens with focal length of 20 cm. Calculate the image’s position relative to the lens as well as the image’s characteristics. 1 1 1 = + f do di di M =− do 1 1 1 = + 20 35 d i di M =− 35 d i = 46.7 cm M = -1.33x This image is REAL (since the object distance is positive) and on the OTHER side of the lens. The image is INVERTED and ENLARGED. Refraction Refraction is based on the idea that LIGHT is passing through one MEDIUM into another. The question is, WHAT HAPPENS? Suppose you are running on the beach with a certain velocity when you suddenly need to run into the water. What happens to your velocity? IT CHANGES! Refraction Fact #1: As light goes from one medium to another, the velocity CHANGES! Suppose light comes from air (vacuum), and strikes a boundary at some angle of incidence measured from a normal line and goes into water. The ratio of the two speeds can be compared. The denominator will ALWAYS be smaller and produce a unitless value greater or equal to 1. This value is called the new medium’s INDEX OF REFRACTION, n. All substances have an index of refraction which can be used to identify the material. Suppose you decide to go spear fishing, but unfortunately you aren’t having much luck catching any fish. WHY? Light BENDS when it reaches a new medium. The object is NOT directly in a straight line path, but rather it’s image appears that way. The actual object is on either side of the image you are viewing. Refraction Fact #2: As light goes from one medium to another, the path CHANGES! Speed of Light Outside a Vacuum What is the speed of light in a sample of water that has an index of refraction of 1.33? v = c n v = 3.0 x 108 m/s 1.33 v = 2.26 x 108 m/s What ALL happens when light reaches a new medium? Some of the light REFLECTS off the boundary and some of the light REFRACTS through the boundary. Angle of incidence = Angle of Reflection Angle of Incidence > or < the Angle of refraction depending on the direction of the light Going from Air to Water The index of refraction, n, for air (vacuum) = 1. The index of refraction for water = 1.33. If you are going from a LOW “n” to a HIGH “n”, the speed DECREASES and the angle BENDS TOWARDS the normal. Going from Water into Air If you are going from a HIGH “n” to a LOW “n”, the speed INCREASES and the angle BENDS AWAY the normal Note: If the angles are EQUAL, then the “n” must be equal for each. The ray will pass straight through. Snell’s Law Snell discovered that the ratios of the index’s and the ratio of the sine of the angles WRT the normal are equal! Snell ' s Law n1 sin θ1 = n2 Sinθ 2 The refractive index of the gemstone, Aquamarine, is 1.577. A ray of light strikes a horizontal boundary of the gemstone with an angle of incidence of 23O from air. Calculate the SPEED of light in Aquamarine. 8 c c 3 x10 m / s n= → vm = = 1.577 vm n vm = 1.90 x 108 m/s Calculate the angle of refraction within Aquamarine. n1 sin θ1 = n2 sin θ 2 (1)(sin 23) = 1.577 sin θ 2 sin 23 θ = sin ( )= 1.577 −1 14.34 degrees Total Internal Reflection There is a special type of refraction that can occur ONLY when traveling from a HIGH “n” medium to a LOW “n” medium. Suppose we are traveling FROM water and going into air. Should the ANGLE OF INCIDENCE get TOO LARGE and the Angle of Refraction EQUALS 90 DEGREES. This special angle of incidence is called the CRITICAL ANGLE, θc, for that material. If the critical angle is EXCEEDED, for that material, the ray will reflect INTERNALLY within the material. We call this idea TOTAL INTERNAL REFLECTION. The angle of incidence EXCEEDS the critical angle for water and the ray reflects according to the law of reflection at the boundary. The Critical Angle How can you calculate the critical angle? … the refracted ray is equal to 90O. n1 sin θ c = n2 sin θ 2 , Sin90 = 1 θc n2 sin θ c = n1 A light ray is traveling in flint glass (n = 1.65) upon striking the boundary, enters air. Calculate the critical angle (Өc) for flint glass. n1 sin θ c = n2 sin θ 2 , Sin90 = 1 n2 sin θ c = n1 1 1 −1 → θ c = sin ( sin θ c = ) 1.65 1.65 θc = O 37.3 Lenses – Application of Refraction There are 2 basic types of lenses. A converging lens (Convex) takes light rays and brings them to a point. A diverging lens (concave) takes light rays and spreads them outward. Converging (Convex) Lens Much like a mirror, lenses also take light rays from infinity and converge them to a FOCAL POINT, f. The difference, however, is that a lens does not have a center of curvature, C, but rather has a focal point on EACH side of the lens. Converging Lenses - Ray Diagrams The rules for ray diagrams are the SAME for lenses as they were for mirrors except you go THROUGH the lens after refraction and instead of going through, C (center of curvature) you go through the actual center of the lens. f f 1. Pick a point on the top of the object and draw three incident rays traveling towards the lens. Draw a ray such that it travels exactly parallel to the principal axis. Draw a second ray so that it passes exactly through the focal point on the way to the lens. Draw the third incident ray such that it travels directly to the exact center of the lens. 2. The ray that traveled parallel to the principal axis on the way to the lens will refract and travel through the focal point. The ray that passes through the focal point on the way to the lens will refract and travel parallel to the principal axis. The ray that traveled to the exact center of the lens will continue in the same direction. 3. The image point of the top of the object is the point where the three refracted rays intersect. All three rays should intersect at exactly the same point. This point is merely the point where all light from the top of the object would intersect upon refracting through the lens. Since this image could be projected on to a screen it is a REAL IMAGE and real images ALWAYS are found on the OPPOSITE side of the lens from the object. Ray Rules For Concave Lenses Any incident ray traveling parallel to the principal axis of a diverging lens will refract through the lens and travel in line with the focal point (… in a direction such that its extension will pass through the focal point). Any incident ray traveling towards the focal point on the way to the lens will refract through the lens and travel parallel to the principal axis. An incident ray that passes through the center of the lens will in affect continue in the same direction that it had when it entered the lens. Diverging Lenses - Ray Diagrams 1. Pick a point on the top of the object and draw three incident rays traveling towards the lens. Draw one ray so that it travels towards the focal point on the opposite side of the lens; this ray will strike the lens before reaching the focal point; stop the ray at the point of incidence with the lens. Draw the second ray such that it travels exactly parallel to the principal axis. Draw the third ray to the exact center of the lens. 2. These incident rays strike the lens and refract. The ray that travels towards the focal point will refract through the lens and travel parallel to the principal axis. The ray that traveled parallel to the principal axis on the way to the lens will refract and travel in a direction such that its extension passes through the focal point on the object's side of the lens. The ray that traveled to the exact center of the lens will continue to travel in the same direction. The three rays should be diverging upon refraction. 3. Locate and mark the image of the top of the object. The image point of the top of the object is the point where the three refracted rays intersect. Since the three refracted rays are diverging, they must be extended behind the lens in order to intersect. Using a straight edge, extend each of the rays using dashed lines. Draw the extensions until they intersect. All three extensions should intersect at the same location. The point of intersection is the image point of the top of the object. The diagrams above show that in each case, the image is … located on the object side of the lens a virtual image an upright image reduced in size (… smaller than the object) Unlike converging lenses, diverging lenses always produce images that share these characteristics. The location of the object does not affect the characteristics of the image. Hence, the characteristics of the images formed by diverging lenses are easily predictable. The diagram below shows five different object locations (drawn and labeled in red) and their corresponding image locations (drawn and labeled in blue). The sign convention for lenses is similar to that for mirrors. Take the side of the lens where the object is to be the positive side. Because a lens transmits light rather than reflecting it like a mirror does, the other side of the lens is the positive side for images. In other words, if the image is on the far side of the lens as the object, the image distance is positive and the image is real. If the image and object are on the same side of the lens, the image distance is negative and the image is virtual. For converging mirrors, ______________, the focal length is positive. Similarly, a converging lens, ____________________, always has a positive f, and a diverging lens has a negative f. The signs associated with magnification also work the same way for lenses and mirrors. A positive magnification corresponds to an upright image. A negative magnification corresponds to an inverted image. Upright and inverted are taken relative to the orientation of the object. A 4.00 cm tall light bulb is placed a distance of 45.7 cm from a convex lens having a focal length of 15.2 cm. Determine the image distance and the image size. ho = 4.00 cm di = ??? do = 45.7 cm hi = ??? f = 15.2 cm 1 1 1 = + f do di 1/(15.2 cm) = 1/(45.7 cm) + 1/di 0.0658 cm-1 = 0.0219 cm-1 + 1/di 0.0439 cm-1 = 1/di di = 22.8 cm hi/ho = - di/do hi /(4.00 cm) = - (22.8 cm)/(45.7 cm) hi = - (4.00 cm) (22.8 cm)/(45.7 cm) hi = -1.99 cm From the calculations in this problem it can be concluded that if a 4.00 cm tall object is placed 45.7 cm from a convex lens having a focal length of 15.2 cm, then the image will be inverted, 1.99 cm tall and located 22.8 cm from the lens. A 4.00 cm tall light bulb is placed a distance of 8.30 cm from a convex lens having a focal length of 15.2 cm. ho = 4.00 cm di = ??? do = 8.3 cm hi = ??? f = 15.2 cm 1 1 1 = + f do di 1/(15.2 cm) = 1/(8.30 cm) + 1/di 0.0658 cm-1 = 0.120 cm-1 + 1/di -0.0547 cm-1 = 1/di di = -18.3 cm hi/ho = - di/do hi /(4.00 cm) = - (-18.3 cm)/(8.30 cm) hi = (4.00 cm) (18.3 cm)/(8.30 cm) hi = 8.81 cm From the calculations in the second sample problem it can be concluded that if a 4.00 cm tall object is placed 8.30 cm from a convex lens having a focal length of 15.2 cm, then the image will be enlarged, upright, 8.81 cm tall and located 18.3 cm from the lens on the object's side (virtual). A 4.00 cm tall light bulb is placed a distance of 35.5 cm from a diverging lens having a focal length of -12.2 cm. Determine the image distance and the image size. ho = 4.00 cm di = ??? do = 35.5 cm hi = ??? f = -12.2 cm 1 1 1 = + f do di 1/(-12.2 cm) = 1/(35.5 cm) + 1/di -0.0820 cm-1 = 0.0282 cm-1 + 1/di -0.110 cm-1 = 1/di di = -9.08 cm hi/ho = - di/do hi /(4.00 cm) = - (-9.08 cm)/(35.5 cm) hi = (4.00 cm) (9.08 cm)/(35.5 cm) hi = 1.02 cm From the calculations in this problem it can be concluded that if a 4.00 cm tall object is placed 35.5 cm from a diverging lens having a focal length of 12.2 cm, then the image will be upright, 1.02 cm tall and located 9.08 cm from the lens on the object's side. Converging lenses play an important role in our lives as our eyes are these types of lenses. Often times we need additional corrective lenses to fix our vision. A camera uses a lens to focus an image on photographic film. The End