a point B, 4 m along x axis. Is re constructive or Consider interference at B if wavelength sound is destructive The path dierence between two sources

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1 214 Spring 99 Problem Set 8 Optional Problems Physics March 16, 1999 Hout Two Slit Interference 1. atwo-slit experiment, slit widths are adjusted so that intensity In light reaching screen from one slit alone is I 0 while intensity light reaching screen from or slit alone is 9I 0. When both are open, what is intensity light at interference minima slits terms I 0? The slits are very narrow. in We can get amplitude eld due to each slit by square root intensity. Ifwe call amplitude eld taking one slit E 0 n amplitude eld from or slit is 3E 0. from minimum, two amplitudes are 180 o out phase so y Atinterference to cancel. At minimum, amplitude will be 3E 0, E 0 =2E 0 tend so intensity will be 4I 0. Double Slits 2. double slit arrangement with slit separation d is illuminated by coherent A wavelength. The top slit is covered by a piece glass thickness light refractive index n. Aninterference pattern is observed on a screen t At what angle,, dowehave principal (m = 0) maximum pattern? (You may assume that is a small angle.) interference To nd phase change caused by insertion slab, consider a segment light's path length t. The glass change experienced by light intraveling this distance is given phase 2t=. Now insert a glass slab thickness t into this space; phase by light traversing slab is now 2t= glass, in which change light in glass is glass = =n. Thus, total phase wavelength between situation with no glass, one with glass shift place, in including this phase shift along with one caused by path Thus, dierence, wehave that, at angle, phase dierence between length this must equal a multiple 2 for an interference maximum. For = 0 (principal maximum) small angles (sin ) wehave that m results has its maxima where minima occured in pattern that (. Given n 0, what is minimum thickness that t 0 can be? part We nowwant to put in a second piece glass to get an phase shift. This means that additional 0, 1) 2(n light from bottom slit from top slit is: 2d sin =, 2t(n, 1)= principal maximum occurs when t(n, 1)=d You now want to put a second piece glass over top slit b) t 0 refractive index n 0 so that new interference pattern thickness 2t 0 (n 0, 1)= = distance D away, where D d. or solving for t 0 : t 0 = Sound Wave Interference 3. 1 S 2 are two coherent point sources sound that are a distance d S apart along y axis: =2t=, 2tn= =,2t(n, 1)= 1

2 a point B, 4 m along x axis. Is re constructive or Consider interference at B if wavelength sound is destructive The path dierence between two sources from a point x x axis is p d 2 + x 2, x. This must be a multiple for maximum on interference, so maxima will occur when p d 2 + x 2, x = constructive for any integer m. Destructive interference occurs when p d 2 + x 2, m, m for any integer m. x = our situation path dierence is p d 2 + x 2,x = p , in Hence Thus 4=1m. for = 1 m, constructive interference (m = 1), but (i) Three Slit Interference 4. Show that intensity for a three-slit interference pattern is I = is phase dierence between waves from two adjacent screen, (= 2d sin = for light wavelength at an angle from slits that slits The phasor for superposition waves from each three slits is amplitude eld from each slit is E, maximum where total amplitude is E 0 = 3E. From diagram, we see that total amplitude is (1 + 2 cos )E 0=3. Squaring this, to get total gives desired result. intensity, di() d Angular widths interference patterns 5. expressions for total angular width central maximum Derive ( a three-slit interference pattern, (b) a four-slit interference pattern, for (c) an N-slit interference pattern. Express your results in terms separation d between adjacent slits. Use phasors to wavelength your results, small-angle approximation to simplify m. derive The phasor diagrams appropriate to light at rst a three-slit, four-slit, N-slit interference pattern are minimum y are a distance d apart). Use phasors to compute your result. S 2 b) Use this result to verify that rst minimum occurs for =2=3. d S 1 x E 0 sources emit sound same frequency, are in phase, are The equal intensity. In analysis below, you may neglect fallo cos φ φ intensity with distance from point sources. E (i) =1m. (ii) =2m. The distance d between sources is 3 m. b) The intensity will be minimum at smallest nonzero for which, 2 sin cos /,sin vanishes. This implies cos =,0:5 or = 2=3. (ii) for = 2 m, destructive interference (m = 0). (4I 0 =9)(1=4 + cos + cos 2 ), where I 0 is intensity at center 2

3 se pictures, phase dierence between light rays coming From adjacent slits must equal 2=N if contributions from all N slits from to cancel completely. This phase dierence is related to angle at are transmitted light is being viewed, so we have which sin 2d 2d Solving for, doubling it to get total total angular width central maximum for N slits. Thus for 3 slits, angular width is for Antenna Array 6. antenna array consists N identical antennas in a straight line one- An a wavelength apart. Each antenna serves as a source electromagnetiquarter waves. The phase dierence between successive antennas in At an angle to array, what is phase dierence between arriving from adjacent antennae? signals This problem is very similar to problem calculating interference pattern from N slits. At some arbitrary angle, phase rst term is phase dierence between each antenna in where second term is due to path dierence. (x is spacing array adjacent antennae.) We are told that x = =4, antenna between so: spacing, Calculate resulting intensity distribution, for in range,90 < b), show that for large N this scheme results in nearly all <90 concentrated in a single direction. What is that direction? Sketch power intensity as a function to show in which direction power goes. Following our derivation in class for multiple slits, we add N phasors with phase angle between m where is given in up c)for N slits, angular width is 2=Nd Note that width decreases as number slits is increased. row is45. dierence between adjacent antennae is: = =4+2xsin = = = =4+(=2) sin = 2=N (1 st minimum): width, we obtain 2=Nd 2=3d (Hints: Use a phasor diagram to calculate intensity. ) b)for 4 slits, angular width is =2d part (. From gure we see 3

4 2R sin(=2) = E 0 sin(n=2) = E 0 E sin(=2) sin 2 (N=2) 0 I 2 I = (=2) sin sin 2 (N(1 + 2 sin )=8) = 2 I 0 2 sin )=8) + ((1 sin intensity peaks in direction for which sin =0,orsin= The =,30.,0:5, At what angles in second order spectrum would you expect to two violet lines wavelengths nm? nd so we can solve for R plug into our expression for E we get or (6.1) E =2Rsin 2 = N E =2Rsin(N=2) We still need to express R in terms known quantities. φ/2 R R E 0 Diraction Grating 7. 1 cm wide diraction grating with 2000 slits is used to measure A wavelengths emitted by hydrogen gas. From gure we see that 4

5 What is resolving power grating in rst second order? b) which order would it be easier to resolve closely spaced ne structure In above, d = 1 From 2000 cm,1 =510,6 m. In second order, interference maximum occurs when 2 = d sin ) = sin,1 ( 2 d resolving power grating is R = Nn, where n is order, b)the slits. In rst order, R = 2000; in second order, R = N=number = so that it is easier to resolve ne structure in second order than smaller, rst order. in on se spectral lines? ). For = 410 nm, =9:44 ; for = 434 nm, =10:0. is The increased resolving power in second order means that 1 R 5