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Chapter 24 The Wave Nature of Light How many times does the incident light reflect from this set of three mirrors? 0% re e 0% th on e 0% tw o 1. one 2. two 3. three Definition: Refraction Refraction is the movement of light from one medium into another medium. Refraction cause a change in speed of light as it moves from one medium to another. Refraction can cause bending of the light at the interface between media. Index of Refraction 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑙𝑖𝑔ℎ𝑡 𝑖𝑛 𝑣𝑎𝑐𝑢𝑢𝑚 𝑛= 𝑠𝑝𝑒𝑒𝑑 𝑜𝑓 𝑙𝑖𝑔ℎ𝑡 𝑖𝑛 𝑚𝑒𝑑𝑖𝑢𝑚 𝑛= 𝑐 𝑣 When light slows down… …it bends. Let’s take a look at a simulation. http://interactagram.com/physics/optic s/refraction/ The Lawnmower model fast medium slow medium fast medium slow medium Speed changes but not direction Speed and direction change Ray is bent toward normal Refraction occurs when there is a change in wave Frequency Amplitude Speed All of these 0% ft he ee d se 0% A ll o Sp de 0% m pl itu A ue nc y 0% Fr eq 1. 2. 3. 4. n1 < n 2 When the index of refraction increases, light bends toward the normal. FST SFA Problem Light enters an oil from the air at an angle of 50o with the normal, and the refracted beam makes an angle of 33o with the normal. a) Draw this situation. b) Calculate the index of refraction of the oil. c) Calculate the speed of light in the oil Problem Light enters water from a layer of oil at an angle of 50o with the normal. The oil has a refractive index of 1.65, and the water has a refractive index of 1.33. a) Draw this situation. b) Calculate the angle of refraction. c) Calculate the speed of light in the oil, and in the water Critical Angle of Incidence • If light passes into a medium with a greater refractive index than the original medium, it bends away from the normal and the angle of refraction is greater than the angle of incidence. • If the angle of refraction is > 90o, the light cannot leave the medium. • The smallest angle of incidence for which light cannot leave a medium is called the critical angle of incidence. Calculating Critical Angle n1sin(θ1) = n2sin(90o) n1sin(θc) = n2sin(90o) sin(θc) = n2/n1 θc = sin-1(n2/n1) Total Internal Reflection Video: prism and fiber optic Total Internal Reflection Video: tank of water Problem A. What is the critical angle of incidence for a gemstone with refractive index 2.45 if it is in air? B. If you immerse the gemstone in water (refractive index 1.33), what does this do to the critical angle of incidence? Why is the sky blue? The Visible Spectrum and Dispersion Wavelengths of visible light: 400 nm to 750 nm Shorter wavelengths are ultraviolet; longer are infrared The Visible Spectrum and Dispersion The index of refraction of a material varies somewhat with the wavelength of the light. The Visible Spectrum and Dispersion This variation in refractive index is why a prism will split visible light into a rainbow of colors. Formation of Rainbows Since each color Ray of sunlight (all colors) has a different wavelength, they Red light each refract slightly Normal differently into and Blue light out of the water droplet thus separating from white to individual colors Water Droplet Close-up The Visible Spectrum and Dispersion Actual rainbows are created by dispersion in tiny drops of water. Diffraction The bending of a wave around a barrier. Diffraction of light combined with interference of diffracted waves causes “diffraction patterns”. The phenomenon of diffraction involves the spreading out of waves past openings which are on the order of the wavelength of the wave. Ripple Tank • Diffraction around obstacles in a ripple tank. • Diffraction and interference in a ripple tank. Interference of waves Two sound waves interfere each other destructive constructive d1 d2 d 1 d 2 n ( n 1 / 2 ) n 0,1,2,.... (constructive) (destructive) Interference – Young’s Double-Slit Experiment If light is a wave, interference effects will be seen, where one part of wavefront can interact with another part. One way to study this is to do a double-slit experiment: Interference – Young’s Double-Slit Experiment If light is a wave, there should be an interference pattern. Interference – Young’s Double-Slit Experiment The interference occurs because each point on the screen is not the same distance from both slits. Depending on the path length difference, the wave can interfere constructively (bright spot) or destructively (dark spot). Interference – Young’s Double-Slit Experiment We can use geometry to find the conditions for constructive and destructive interference: (24-2a) (24-2b) 24.3 Interference – Young’s Double-Slit Experiment Condition for a maximum: d sinθ = mλ For small angles, we may assume sin θ ≈ θ THEN tanθ ≈ sinθ ≈ θ ≈ x/L (where x is the distance between the Central maximum and the next bright fringe (one λ difference) Therefore: d(x/L) ≈ mλ → x ≈ mλL / d (for small angles, screen is far away, L>>d) Double slit interference Between the maxima and the minima, the interference varies smoothly as shown in the intensity profile: **How does interference pattern change with λ or with d? ** Physics Classroom two-slit diffraction two-slit diffraction Interference – Young’s Double-Slit Experiment Since the position of the maxima (except the central one) depends on wavelength, the firstand higher-order fringes contain a spectrum of colors. Interference – Young’s Double-Slit Experiment: Remember, as λ increases, θ increases as well (if all else remains the same). Therefore, red light is diffracted at a larger θ than violet light. Diffraction Grating The maxima of the diffraction pattern are defined by (24-4) single slit diffraction Single slit diffraction applet single slit diffraction http://hyperphysics.phyastr.gsu.edu/hbase/phyopt/sinsl it.html Monochromatic light passed through a double slit produces an interference pattern on a screen a distance 2.0m away. The third-order maximum is located 1.5cm away from the central maximum. Which of the following adjustments would cause the third-order maximum instead to be located 3.0cm from the central maximum? 0% di ... ... ng th e ee n sc r al vi H U si ng a sc r a si ng 0% ... e th U lin g Tr ip 0% ee n .. w ... 0% d. th e g bl in 5. 0% ou 4. Doubling the distance between slits Tripling the wavelength of the light Using a screen 1.0m away from the slits Using a screen 3.0m away from the slits Halving the distance between slits D 1. 2. 3. Respons e Grid FRQ Laser light is passed through a diffraction grating with 7000 lines per cm. Light is projected onto a screen far away. An observer by the diffraction grating observes the first order maximum 25° away from the central maximum. a) What is the wavelength of the laser? b) If the first order maximum is 40cm away from the central maximum on the screen, how far away is the screen from the diffraction grating? c) How far, measured along the screen, from the central maximum will the second-order maximum be? Interference by Thin Films Another way path lengths can differ, and waves interfere, is if the travel through different media. If there is a very thin film of material – a few wavelengths thick – light will reflect from both the bottom and the top of the layer, causing interference. This can be seen in soap bubbles and oil slicks, for example. Interference by Thin Films The wavelength of the light will be different in the oil and the air, and the reflections at points A and B may or may not involve phase change. Thin Films Why do you see a rainbow pattern on a soap bubble? It’s a result of thin film interference. The colored pattern is a result of reflection from both the front and back surfaces of the bubble and different bubble thickness There are two effects: 1) An inversion of the wave incident on the first surface of the bubble 2) A path length difference between wave reflected from 1st surface and wave reflected from 2nd surface Thin Films There is a path length difference ABC: Results in constructive interference if ABC = 2t = mλn Results in destructive interference if ABC = 2t = (m+1/2) λn t λn = λ/n wavelength of light material In addition, depending on index of refraction of the materials, an additional phase change may exist: • If n2 > n1 the reflected wave has a phase change of (½ cycle, equivalent to a ½ path length difference) (like reflection from a string with a fixed end, medium 2 is denser) • If n2 < n1 the reflected wave has a phase change of 0 (equivalent to a 0 path length difference) (like reflection from a string with a free end, medium 2 is less dense) Thin Films Intuitive explanation of phase change at a interface (picture below is for a soap bubble) gravity Thin Films 1) Colored pattern is result of varied thickness of the bubble: Gravity causes the bubble to be thinner at the top and thicker at the bottom. 2) The thinnest part of film (top) is dark (destructive interference), Thin film interference Air Soap film, thickness t Water Extra path length = 2t • Film thickness: t ~ (0.5 – 1 m) • Interference between light reflected off the top and bottom surfaces can either be constructive (bright band) or destructive (dark band) • Constructive interference occurs for a certain wavelegth depending on the local film thickness bright color appears at that value of Problem solving strategy: 1. 2. 3. 4. Count the phase changes. A phase change occurs for every reflection from low to high index of refraction The extra distance traveled by the wave in the thin film is twice the thickness of the film. The wavelength in the film is λn = λ/n where n is the index of refraction of the film’s material. Use 2t = mλn . If the light undergoes zero or two phase changes, then m is a whole number for constructive interference (half integers for m give destructive interference). For one phase change, the conditions are reversed – whole numbers give destructive interference, half integers , constructive. Practice Problem A soap bubble appears green (λ = 540nm) at the point on its front surface nearest the viewer. What is the smallest thickness the soap bubble film could have? Assume n=1.35. Anti-reflective / Nonreflective coating Nonreflective coatings are commonly found on high quality optical equipment including camera lenses, telescopes, and eye glasses Less light reflection means more light transmission A thin film is placed on the surfaces of the lens The film thickness is such that light (designed to apply to just wavelength) reflected from it destructively interferes. No coating can apply to all wavelengths. The amount of reflection depends on the difference in index of refraction between the two materials Other wavelengths will have partial destructive interference. Wavelengths farthest from the center will have the most reflection (why the camera lens looks purplelish) Anti-reflective / Nonreflective coating Practice Problem What is the thickness of an optical coating of MgF2 whose index of refraction is n=1.38 and is designed to eliminate reflected light at wavelengths (in air) around 550 nm when incident normally on glass for which n=1.5? 2Michelson Interferometer The Michelson interferometer is centered around a beam splitter, which transmits about half the light hitting it and reflects the rest. It can be a very sensitive measure of length.