Chapter 17 Interference(Ans)

Chapter 17 Interference21-37(a) We choose a horizontal x axis with its origin at the left edge of the plastic ， the path length difference is221211=+1()(3.50 m)1.60(4.00 m)1.40+(4.00 m 3.50m)=0.50m n L L L n L μμμμμ∆⨯--=--Thus, combining these phase differences with λ = 0.600 μm, we have0.50m =0.833600.0n mn μλ∆==(b) Since the answer in part (a) is closer to an integer than to a half-integer, the interference is more nearly constructive than destructive.21-40Consider the two waves, one from each slit, that produce the seventh bright fringe in the absence of the mica. They are in phase at the slits and travel different distances to the seventh bright fringe, where they have a path difference of 7λ. Now a piece of mica with thickness x is placed in front of one of the slits, and an additional phase difference between the waves develops.()17n x λ∆=-=x n =-=⨯-=⨯--7175501015816641096 m m.c h..21-41The incident light is in a low index medium, the thin film of oil has somewhat higher n = n 2, and the last layer (the glass plate) has the highest refractive index, 123n n n <<, thus the reflection path difference has no half- wavelength loss. To see very little or no reflection, according to the condition212 0,1,2,2n L m m ⎛⎫=+λ= ⎪⎝⎭K ， must hold. With λ = 500 nm and n 2 = 1.30,2112 0,1,2,2n L m m 1⎛⎫=+λ= ⎪⎝⎭K ， And, with λ = 700 nm and the same value of n 2,2212 0,1,2,2n L m m 2⎛⎫=+λ= ⎪⎝⎭K ， For m 1 and m 2 must be integers , we obtain 12=3=2smallest possible choices)m m ，（1217500()6732222(1.30)nm L m nm n 1λ⎛⎫=+== ⎪⎝⎭ 21-42(a)In the sense that the incident light is in a low index medium, the thin film has somewhat higher n = n 2, and the last layer has the highest refractive index, 123n n n <<, thus the reflection path difference has no half- wavelength loss. To see the brightest reflection, the condition22where 0,1,2,...n L m m =λ=must hold. We disregard the m = 0 value (corresponding to L = 0) since there is some oil on the water. Thus, for 1,2,...,m = maximum reflection occurs for wavelengths()()221.20460nm 21104nm , 552nm,368nm...n L m m===l We note that only the 552 nm wavelength falls within the visible light range.(b) The maximum transmission into the water occurs for wavelengths given by21242122L m n n L m =+F HG I K J ⇒=+●● which yields λ = 2208 nm, 736 nm, 442 nm … for the different values of m . We note that only the 442 nm wavelength (blue) is in the visible range, though we might expect some red contribution since the 736 nm is very close to the visible range.21-44Consider the interference of waves reflected from the top and bottom surfaces of the air film. The wave reflected from the upper surface does not change phase on reflection but the wave reflected from the bottom surface changes phase by π rad. So the reflection path difference has half- wavelength loss. To see the brightest reflection, the condition2where 1,2,3 (2)nd m m λλ+== At a place where the thickness of the air film is D, where λ (= 683 nm) is the wavelength and m is an integer.()()max 92482111412268310m m D m μλ-=+=+=⨯There are, therefore, 141 bright fringes in all.21-45Assume the wedge-shaped film is in air, so the wave reflected from one surface undergoes a phase change of π rad while the wave reflected from the other surface does not. At a place where thefilm thickness is L , the condition for fully constructive interference is 212nL m =+b g● where n is the index of refraction of the film, λ is the wavelength in vacuum, and m is an integer. The ends of the film are bright. Suppose the end where the film is narrow has thickness L 1 and the bright fringe there corresponds to m = m 1. Suppose the end where the film is thick has thickness L 2 and the bright fringe there corresponds to m = m 2. Since there are ten bright fringes, m 2 = m 1 + 9. Subtract 21112nL m =+b g ● from 292112nL m =++b g● to obtain 2n ∆L = 9λ, where ∆L = L 2 – L 1 is the change in the film thickness over its length. Thus,∆L n ==⨯=⨯--996301021501891096● m m.c hb g ..21-46Consider the interference pattern formed by waves reflected from the upper and lower surfaces of the air wedge. The wave reflected from the lower surface undergoes a π rad phase change while the wave reflected from the upper surface does not. At a place where the thickness of the wedge isd , the condition for a maximum in intensity is 21d m =+b g● where λ is the wavelength in air and m is an integer. Therefore,d = (2m + 1)λ/4.As the geometry shows,d R R r =--22, where R is the radius of curvature of the lens and r is the radius of a Newton’s ring. T hus, 2122m R R r +=--b g ● . First, we rearrange theterms so the equation becomes()21.4m R +=-lNext, we square both sides, rearrange to solve for r 2, then take the square root. We getr m R m =+-+21221162b g b g ●●.If R is much larger than a wavelength, the first term dominates the second andrm R=+212b g.m=0,1,2……Orr=m=1,2, 3……21-48where L is the length of the tube. The factor 2 arises because the light traverses the tube twice, once on the way to a mirror and once after reflection from the mirror. Each shift by one fringe corresponds to a change in phase of 2π rad, so if the interference pattern shifts by N fringes as the tube is evacuated,nNL=+=+⨯⨯=--12160500102501010003092λmmc hc h...2(1)n L Nλ-=。