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25 Pages

### Chapter 35c

Course: PHY 2048 PHY, Spring 2009
School: University of Florida
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Word Count: 1715

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35: Chapter Interference ! ! ! ! ! Interference in thin films reflection and transmission. HITT question. Michelsons Interferometer. Youngs Interference Experiment. Intensity in Double-Slit Interference. December 4, 2009 Ch. 35: Interference - Part C 1 The Index of Refraction ! The index of refraction n for any medium is defined as the ratio of the speed of light in vacuum and the speed of light in the...

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35: Chapter Interference ! ! ! ! ! Interference in thin films reflection and transmission. HITT question. Michelsons Interferometer. Youngs Interference Experiment. Intensity in Double-Slit Interference. December 4, 2009 Ch. 35: Interference - Part C 1 The Index of Refraction ! The index of refraction n for any medium is defined as the ratio of the speed of light in vacuum and the speed of light in the medium. n= ! c (index of refraction) The speed, wavelength and frequency of light in a medium with refraction index n are: c (speed) = n n = n (wavelength) c fn = = =f n December 4, 2009 (frequency) 2 Ch. 35: Interference - Part C Wavelength and the Index of Refraction The phase difference between two light waves can change if the waves travel through different materials having different indexes of refraction. ! The two light waves travel initially in air (n~1), and are initially in phase, then travel distance L in media with refraction indexes n1 and n2, and then emerge in air. ! What is the phase difference in terms of n2>n1 wavelengths as the two waves emerge in air? ! N1 = L n 1 = L L n1 and N2 = L L n 2 = L n2 N2 N1 = December 4, 2009 n2 n1 = L ( n2 n1 ) 3 Ch. 35: Interference - Part C Reflection Phase Shifts ! ! Refraction at an interface does not cause phase shifts. Reflection at an interface can cause phase shifts depending on the indexes of reflection on the two sides of the interface. Reflection Off lower index Off higher index Reflection Phase Shift 0 0.5 wavelength Higher means half rule. December 4, 2009 Ch. 35: Interference - Part C 4 Interference from Thin Films ! Light waves (ray i) are incident on a thin film of thickness L and index of refraction n2 . ! ! ! ! The angle of incidence is ~0o. The ray r1 is reflected from the front (left) surface of the film. The ray r2 is reflected from the back surface of the film. The region ac of the film will appear bright or dark to the observer viewing it almost perpendicularly, depending on whether the reflected rays r1 and r2 interfere constructively or destructively. Ch. 35: Interference - Part C 5 December 4, 2009 Reflections from a Thin Film in Air ! Light waves (ray i) are incident on a thin film of thickness L and index of refraction n2 . ! air air ! ! ! n1<n2, n2>n3 The angle of incidence is ~0o. Ray r1 is reflected from the front surface of the film, with a 0.5 n2 phase shift. Ray r2 is reflected from the back surface of the film no phase shift. Rays r1 and r2 have the same path length in air, but r2 reflects twice in medium 2, giving path length difference of ~2L (~0o). Ch. 35: Interference - Part C 6 December 4, 2009 Reflections from a Thin Film in Air Reflection phase shifts Path length difference Thin film index r1 0.5 wavelength 2L n2 r2 0 n1<n2, n2>n3 In phase: 2L n 2 odd number + 0.5 = integer or 2 L = 2 n2 + 0.5 = integer+0.5 or 2 L = integer Ch. 35: Interference - Part C Out of phase: December 4, 2009 2L n2 7 n 2 Transmission through Thin Layers Light waves (ray i) are fall perpendicularly on a thin film of thickness L and index of refraction n2 . ! Ray r3 is refracted from medium 2 to air 1.33 air medium 3. ! Ray r4 is reflected from the back surface of the film into the film, and reflected from the front surface into the film. There is no phase shift for either reflection. n1<n2, n2>n3 ! Rays r3 and r4 have the same path length in air, and after the refraction from 1 to 2 until reaching the 2|3 interface. However, ray r4 travels an additional length ~2L (~0o) in medium 2. ! December 4, 2009 Ch. 35: Interference - Part C 8 Transmission through Thin Layers air 1.33 air n1<n2, n2>n3 Reflection phase shifts Path length difference Thin film index r4 2x0 2L n2 r3 0 Maxima: Minima: 2L n 2 2L = integer or or 2 L = integer n2 n 2 = integer+0.5 odd number 2L = 2 n2 9 December 4, 2009 Ch. 35: Interference - Part C Transmission through Thin Layers 1.59 1.37 1.46 Reflection phase shifts Path length difference Thin film index Maxima: 2 L = m r4 2x0.5 wavelength 2L n2 r3 0 n1>n2, n2<n3 n2 for m=0,1,2,... for m=0,1,2,... 10 2m+1 Minima: 2 L = 2 n2 December 4, 2009 Ch. 35: Interference - Part C Transmission through Thin Layers 1.59 1.37 1.46 If = 522 nm, what is the third least film thickness to see a bright fringe in the transmitted light? Maxima: 2 L = m n2 for m=0,1,2,... n1>n2, n2<n3 m L= 2 n2 3 L(m = 3) = 2 n2 3 522 L(m = 3) = = 572 nm 2 1.37 December 4, 2009 Ch. 35: Interference - Part C 11 HITT Question ! ! The two light rays in air, n~1, and initially in phase, go through different paths by reflecting from various glass surfaces with n=1.5. In what way will the two waves interfere if air = 500 nm, and d = 1.0 m? (a) fully constructively; (b) fully destructively; (c) close to constructively; (d) close to destructively; (e) insufficient information. December 4, 2009 Ch. 35: Interference - Part C 12 HITT Question Reflection phase shifts Path length Index r1 6 x /2 7d nair r2 1 x /2 2d nair Phase in difference terms of wavelength: = = 5d 7d 2d air + ( 6 1) 0.5 nair 5 1000 + 2.5 = + 2.5 = 10 + 2.5 = 12.5 500 (b) fully destructively If n=1.38, then =16.3, (d) close to destructively. 13 eff = 0.5 December 4, 2009 Ch. 35: Interference - Part C Michelsons Interferometer ! An interferometer is a device that can be used to accurately measure distances, relying on light interference. Originally devised by Albert A. Michelson in 1881, 1907 Nobel Prize in Physics. ! Light originates at point P. ! Mirror M (beam splitter) reflects half of the light beam, and refracts the other half. ! The reflected beam is directed to mirror M2 and back, traveling a distance 2d2. ! The refracted beam is directed to mirror M1 and back, traveling a distance 2d1. ! The two beams are then refracted or reflected from M and directed to the telescope T. Ch. 35: Interference - Part C 14 December 4, 2009 Michelsons Interferometer ! ! A pattern of interference fringes is seen at T. The path length difference is: 2 d 2 2 d1 ! ! If mirror M2 is moved by distance /2, the path length difference is changed by and the fringe pattern is shifted by one fringe. If a transparent material with an index of refraction n and thickness L is inserted into one of the optical paths, the phase shift is: Nmat Nair = 2L n 2L nair = 2L ( n 1) http://www.youtube.com/watch?v=8QUhgYaxWao&feature=related http://www.youtube.com/watch?v=h9jjebbOkxo December 4, 2009 Ch. 35: Interference - Part C 15 Diffraction ! ! A diffraction experiment showed for the first time that light is a wave. If a wave encounters a barrier with an opening that has dimensions similar to its wavelength, the wave that passes through the opening will spread out (flare) or diffract into the region beyond the barrier. ! ! Diffraction is produced by all types of waves, not just light. The flaring is consistent with the spreading of the wavelets in the Huygens construction. diffraction of a water wave December 4, 2009 Ch. 35: Interference - Part C 16 Diffraction ! An incident plane wave of wavelength , diffracted through a slit that has width 6.0 (a), 3.0 (b), and 1.5 (c). The screens extend into and out of the page. ! The narrower the slit, the greater the diffraction. Ch. 35: Interference - Part C 17 December 4, 2009 Youngs Interference Experiment ! Thomas Young (1773-1829), an English polymath, proved in 1801 that light is a wave. http://vsg.quasihome.com/interf.htm December 4, 2009 Ch. 35: Interference - Part C 18 Youngs Interference Experiment ! ! ! ! Light from a monochromatic light source illuminates slit S0 in screen A. The light is spread by diffraction to illuminate slits S1 and S2 in screen B. Diffraction of light by the two slits sends overlapping circular waves into the region beyond B, where the waves interfere. The two waves are coherent (constant phase difference). The interference pattern is captured in screen C. 19 December 4, 2009 Ch. 35: Interference - Part C Youngs Interference Experiment ! ! ! ! ! ! The interference pattern is determined by the path length difference L of the rays reaching a point P on the screen. d is the distance between the two slits. Angle specifies the location of the bright and dark fringes on the screen. Need to relate , L, d and . L is the distance S1 to b. Assume that the screen distance D is much greater than d, thus r1 and r2 are parallel. L sin = d December 4, 2009 L = d sin 20 Ch. 35: Interference - Part C Youngs Interference Experiment L = d sin integer (maxima) integer+0.5 (minima) ! Maxima, or bright fringes (bands): d sin = m ! for m=0,1,2,... Minima, or dark fringes (bands): d sin = ( m + 0.5 ) for m=0,1,2,... ! December 4, 2009 =0, m=0, central maximum. Ch. 35: Interference - Part C 21 Youngs Interference Experiment ! Calculate the distance between the second order bright fringes if the screen is 3.3 m away from the slits, = 515 nm (blue-green), and the slit separation is 0.50 mm. d sin = m ! ! for m=0,1,2,... =0, m=0, central maximum. 2nd order maximum, m=2. 2 = sin d sin = 2 d 2 515 109 = sin1 = 0.11803 3 0.50 10 1 December 4, 2009 Ch. 35: Interference - Part C 22 Youngs Interference Experiment ! The distance between the central maximum and the 2nd maximum is: y 0,2 = D tan ! The interference pattern is symmetric with respect to the central maximum, therefore: y -2,2 = 2 y 0,2 = 2 D tan y -2,2 = 2 3.3 tan0.11803 y -2,2 = 0.013596 m= 1.4 cm December 4, 2009 Ch. 35: Interference - Part C 23 Intensity in Double-Slit Interference ! If the two waves interfering at point P have Es: E1 = E0 sin t ! and E2 = E0 sin( t + ) 2d The two waves combine at point P with intensity: 1 I = 4 Io cos 2 2 where = sin December 4, 2009 Ch. 35: Interference - Part C 24 Intensity in Double-Slit Interference ! If the two waves interfering at point P have Es: E1 = E0 sin t E2 = E0 sin( t + ) 1 E = 2 E0 cos where = 2 E = 2 E0 cos 2 I E2 I = 4 Io cos December 4, 2009 2 2 I E2 =2 I o Eo Ch. 35: Interference - Part C 25
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University of Florida - PHY 2048 - PHY
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