Physics 272 Lecture 27 Interference (Ch ) Diffraction (Ch )

Transcription

1 Physics 272 Lecture 27 Interference (Ch ) Diffraction (Ch )

2 Thin Film Interference 1 2 n 0 =1 (air) t n 1 (thin film) n 2 Get two waves by reflection off of two different interfaces. Ray 2 travels approximately 2t further than ray 1.

3 Reflection + Phase Shifts Incident wave Reflected wave n 1 n 2 Upon reflection from a boundary between two transparent materials, the phase of the reflected light may change. If n 1 > n 2 there is no phase change upon reflection. If n 1 < n 2 there is a phase change of 180º upon reflection. (This is equivalent to the wave shifting by λ/2.)

4 Thin Film Interference Determine δ, number of extra wavelengths for each ray. 1 2 n=1 (air) t n 1 (thin film) n 2 Reflection Distance Ray 1: δ 1 = 0 or Ray 2: δ 2 = 0 or t/ λ film Note this is wavelength in film! If (δ 2 δ 1 ) = 0, 1, 2, 3. m constructive If (δ 2 δ 1 ) = 0.5, 1.5, 2.5. (m+0.5) destructive

5 Thin Film Interference A thin film of gasoline (n gas =1.20) and a thin film of oil (n oil =1.45) are floating on water (n water =1.33). The thickness of the two films is exactly one wavelength. The gasoline film appears to be A. Bright B. Dark Since n air < n gasoline < n water, there is no additional phase difference between the two reflected rays - the only phase difference is due to thickness, which is a multiple of λ. Therefore, this results constructive interference.

6 Thin Film Interference A thin film of gasoline (n gas =1.20) and a thin film of oil (n oil =1.45) are floating on water (n water =1.33). The thickness of the two films is exactly one wavelength. The oil film appears to be A. Bright B. Dark Since n oil > n air & n oil > n water, there is an additional phase difference (180 o ) between the two reflected rays. Therefore, this results destructive interference.

7 Constructive Interference 3 Rays L 1 d d 2 3 All 3 rays are interfering constructively at the point shown. If the intensity from ray 1 is I 0, what is the combined intensity of all 3 rays? 1) I 0 2) 3 I 0 3) 9 I 0 Each slit contributes amplitude E o at screen. E tot = 3 E o. But I α E 2. I tot = (3E 0 ) 2 = 9 E 0 2 = 9 I 0

8 Destructive Interference 3 Rays L 1 d d θ θ 2 dsin θ = λ 3 2 these add to zero this one is still there! When rays 1 and 2 are interfering destructively, is the intensity from the three rays a minimum? 1) Yes 2) No Rays 1 and 2 completely cancel, but ray 3 is still there. Expect intensity I = 1/9 I max

9 9I 0 9 Three slit interference I λ 3 λ 2 2λ 3 dsin θ = λ 3 λ 2 2λ 3 λ

10 Multiple Slit Interference (Diffraction Grating) L 1 d 2 θ d θ d 3 4 Constructive Int: Path length difference 1-2 Path length difference 1-3 Path length difference 1-4 dsin θ = mλ = d sinθ = 2d sinθ = 3d sinθ Constructive interference for all paths when =λ =2λ =3λ

11 Multiple Slit Interference (Diffraction Grating) For many slits, maxima are still at sin θ = m λ d Region between maxima gets suppressed more and more as no. of slits increases. N=2 intensity 0 λ 2λ dsin θ N=10 intensity 0 λ 2λ dsin θ

12 Diffraction

13 Diffraction Diffraction Grating Single Slit Resolving power 4/20/

14 Interference Young s Double Slit Experiment intensity 0 λ 2λ 4/20/

15 Diffraction Huygen s Principle: each point on the wave front is a point source Diffraction: These sources interfere Light can bend around corners (like sound) 4/20/

16 Single Slit Interference?! 4/20/

17 D Diffraction: Interference of light from within just one slit D 2 θ θ θ D D When rays 1 and 1 2 sinθ 2 sinθ = λ 2 will interfere destructively. Rays 2 and 2 also start D/2 apart and have the same path length difference. Condition for every ray originating in top half of slit to interfere destructively with the corresponding ray originating in bottom half. 1 st minimum at sin θ = λ/d 4/20/

18 D Diffraction: Interference of light from within just one slit D 4 θ θ θ D When 4 sinθ = λ 2 rays 1 and 1 D 4 sinθ will interfere destructively. Rays 2 and 2 also start D/4 apart and have the same path length difference. Condition for every ray originating in top half of slit to interfere destructively with the corresponding ray originating in bottom half. 2 nd minimum at sin θ = 2λ/D 4/20/

19 Diffraction Summary Condition for halves of slit to destructively interfere Condition for quarters of slit to destructively interfere Condition for sixths of slit to destructively interfere sin(θ) = λ D sin(θ) = 2 λ D sin(θ) = 3 λ D Dsinθ = mλ THIS FORMULA LOCATES MINIMA!! Narrower slit => broader pattern (m=1, 2, 3, ) 4/20/

20 Diffraction from Circular Aperture 1 st diffraction minimum Central maximum θ Diameter D light Maxima and minima will be a series of bright and dark rings on screen First diffraction minimum is at sin θ = 1.22 λ D 4/20/

21 Intensity from Circular Aperture I λ 1.22 D First diffraction minima sinθ 4/20/

22 These objects are just resolved Two objects are just resolved when the maximum from one is at the minimum of the other. 4/20/

23 Resolving Power Two objects are just resolved when the maximum from one is at the minimum of the other. θ min To see objects distinctly, need θ θ min sin θ min θ min = 1.22 λ D θ This is why cameras with bigger apertures are better 4/20/

24 Resolving Power Question sin θ min θ min = 1.22 λ D How does the maximum resolving power of your eye change when the brightness of the room is decreased. 1) Increases 2) Constant 3) Decreases When the light is low, your pupil dilates (D can increase by factor of 10!) 4/20/

25 Huygen s Construction for a Plane Wave At t = 0, the wave front is indicated by the plane AA The points are representative sources for the wavelets After the wavelets have moved a distance c t, a new plane BB can be drawn tangent to the wavefronts

26 Huygen s Construction for a Spherical Wave The inner arc represents part of the spherical wave The points are representative points where wavelets are propagated The new wavefront is tangent at each point to the wavelet

27 Huygen s Principle and the Law of Reflection The Law of Reflection can be derived from Huygen s Principle AA is a wave front of incident light The reflected wave front is CD

28 Huygen s Principle and the Law of Reflection Triangle ADC is congruent to triangle AA C Angles θ 1 = θ 1 This is the Law of Reflection

29 Huygen s Principle and the Law of Refraction In time t, ray 1 moves from A to B and ray 2 moves from A to C From triangles AA C and ACB, all the ratios in the Law of Refraction can be found n 1 sin θ 1 = n 2 sin θ 2