Lecture 15 Interference Chp. 35

Size: px

Start display at page:

Download "Lecture 15 Interference Chp. 35"

Mae Walker
5 years ago
Views:

1 Lecture 15 Interference Chp. 35 Opening Demo Topics Interference is due to the wave nature of light Huygen s principle, Coherence Change in wavelength and phase change in a medium Interference from thin films Examples Young s Interference Experiment and demo Intensity in double slit experiment Warm-up problem Demos

2 Huygen s Principle, Wavefronts and Coherence E = E m sin(kx "t) E = E m sin( " # f t) k Examples of coherence are: Laser light Small spot on tungsten filament Wavefront Most light is incoherent: Two separate light bulbs Two headlight beams on a car Sun is basically incoherent

3 Interference is the combination of two or more waves to form a composite wave, based on the principle of superposition

4 In order to form an interference pattern, the incident light must satisfy two conditions: (i) The light sources must be coherent. This means that the plane waves from the sources must maintain a constant phase relation. For example, if two waves are completely out of phase with _ =, this phase difference must not change with time. (ii) The light must be monochromatic. This means that the light consists of just one wavelength _ = /k. Light emitted from an incandescent lightbulb is incoherent because the light consists of waves of different wavelengths and they do not maintain a constant phase relationship. Thus, no interference pattern is observed.

5 Once light is in phase there are three ways to get light out of phase 1. Rays go through different material with different index of refraction. Reflection from a medium with greater index of refraction 3. The selected rays travel different distances. Now lets look at examples

6 In Phase Out of Phase by 180 degrees or radians or "/ In between

$Concept of path length difference, phase and index of refraction Vacuum c = f " n=c/v Vacuum v n = f " n =f "/n Ln1 = Ln = L n Path length difference = Phase difference = " n 1 Rays are in phase if$

7 Concept of path length difference, phase and index of refraction Vacuum c = f " n=c/v Vacuum v n = f " n =f "/n Ln1 = Ln = L n Path length difference = Phase difference = " n 1 Rays are in phase if L( n " n ) 1 = m where m=1,, 3.. L ( n ) ( n1 ' 1 $ Rays are out of phase if = where m=1,,3 % m + " & # 1 " is the same as radian (rad), "/ is the same as rad, etc. N N N 1 L = n1 L = " N 1 n = L ( n " n ) ( ) 1

8 n 1 air E n water Wave reflects 180 degrees out of phase when n 1 < n

9 Reflection air 1.0 Thin film Interference Phenomenon: Reflection soap 1.30 air 1.00 E = E m sin(kx "t) 1 eye n = n L First consider phase change upon reflection E 1 = E m sin(kx "t + #) E 1 = E m sin(kx "t) E = E m sin(k(x + L) "t) Suppose the soap thickness L is such that L = 1 $ # &' " % n Show wave demo Now consider the path length differences Constructive Interference 180 deg phase change E = E m sin(kx + kl "t) = E m sin(kx + # $ $ "t) = E = E m sin(kx "t + #) L = m + 1 $ # &' " % n where m = 0, 1,,...

10 Thin film Interference Phenomenon Transmission Transmission Constructive interference air 1.0 soap 1.30 L L = ( m) n = nl m air eye m = 1,,3,4. n =1.30 No phase changes upon reflection

11 39. A disabled tanker leaks kerosene (n = 1.0) into the Persian Gulf, creating a large slick on top of the water (n = 1.30). (a) If you are looking straight down from an airplane while the Sun is overhead at a region of the slick where its thickness is L=460 nm, for which wavelength(s) of visible light is the reflection brightest because of constructive interference? air Path difference between ray 1and ray = L. Phase changes cancel out For constructive interference path difference must = integral number of wavelengths Kerosene 1.0 Water deg phase change L L = m ( ) n = n L m ( )( 460nm) = n L m = 1.0 =1104, 55, 368 nm m for m =1, m =, and m = 3 respectively We note that only the 55 nm wavelength falls within the visible light range.

12 (b) If you are scuba diving directly under this same region of the slick, for which wavelength(s) of visible light is the transmitted intensity strongest? (Hint: use figure (a) with appropriate indices of refraction.) For transmission, ray undergoes 180 deg phase shift upon reflection at the Kerosene-water interface. Therefore, for constructive interference L= integral number of wavelengths in n plus half a wavelength. L = m + 1 $ # &' " % n where m = 0, 1,,... ' n = ' n L = # m + 1 $ & ' where m = 0, 1,,... " % n Solve for ' air 1.0 Kerosene 1.0 Water L Scuba diver

13 L = m + 1 $ # & ' " % n ' = 4n L m +1 Solving for ' = 4n L m +1 = 4(1.)(460) 1 = 08 nm m = 0 = 4n L m +1 = 4(1.)(460) 3 = 4n L m +1 = 4(1.)(460) 5 = 736 nm m =1 = nm m = Visible spectrum is 430 nm nm We note that only the nm wavelength (blue) is in the visible range,

14 7. S 1 and S in Fig are point sources of electromagnetic waves of wavelength 1.00 m. They are in phase and separated by d = 4.00 m, and they emit at the same power. x detector (a) If a detector is moved to the right along the x-axis from source S 1, at what distances from S 1 are the first three interference maxima detected? d + x. The wave from S 1 travels a distance x and the wave from S travels a distance d + x. The path difference is d + x x

15 x d + x For constructive interference we have path difference = d + x x = m" m =1,,3.. The solution for x of this equation is

16 Solve for x d + x x = m" d + x = m" + x Now square both sides d + x = ( m + x) d + x = m + mx + x Now cancel x d = m + mx solve for x x = d m " m" for m =1,, 3,..

17 m=3 m= m=1 x = d " m m. ( ) For m = 3 x = 16 3 ( ) ( 3) For m = x = 16 ( ) ( )( ) =1.17m. = 3.0m. x = 16 m m ( ) ( )( 1) For m =1 x = 16 1 = 7.5m. What about m = 4? This corresponds to x=0. Path difference =4 meters.

18 Where do the minima occur? path difference = d + x x = (m + 1 )" m = 0,1,,3 m=3 m= m=1 m=1 m=0 x = d (m + 1 ) " (m + 1 )". m=0 x=15.75 m m=1 x =4.55 m m= x=1.95 m m=3 x= 0.55 m x = 16 (m + 1 ) (m + 1 ). Although the amplitudes are the same at the sources, the waves travel different distances to get to the points of minimum intensity and each amplitude decreases in inverse proportion to the square of the distance traveled. The intensity is not zero at the minima positions. I 1 = P 0 4x I = P 0 4(d + x ) I 1 I = x d + x = (0.55) 4 + (0.55) ~ 1 64

19 Demo with speakers using sound waves Set oscillator frequency to 137 Hz, Then wavelength of sound is 343/137 =0.5 m " m m Set speakers apart by 1m. Then maxima occur at x =. d x = 1 m (1 /16) m(1 / 4) x = 16 m 8 m m = 4 x = 0 m = 3 x= = 7 4 m 0.33m m = x= = 1 16 m 0.75m m = 1 x= = 15 7 m.0m

Electricity & Optics

Physics 24100 Electricity & Optics Lecture 26 Chapter 33 sec. 1-4 Fall 2017 Semester Professor Koltick Interference of Light Interference phenomena are a consequence of the wave-like nature of light Electric