ch35 - Chapter 35 1. The fact that wave W2 reflects two...

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1347Chapter 351. The fact that waveW2reflects two additional times has no substantive effect on thecalculations, since two reflections amount to a 2(/2) =phase difference, which iseffectively not a phase difference at all. The substantive difference betweenW2andW1isthe extra distance 2Ltraveled byW2.(a) For waveW2to be a half-wavelength “behind” waveW1, we require 2L=/2, orL=/4 = (620 nm)/4 =155 nm using the wavelength value given in the problem.(b) Destructive interference will again appear ifW2is32“behind” the other wave. Inthis case,232 L, and the difference is3620 nm310nm .4422LL2. We consider wavesW2andW1with an initial effective phase difference (inwavelengths) equal to12, and seek positions of the sliver that cause the wave toconstructively interfere (which corresponds to an integer-valued phase difference inwavelengths). Thus, the extra distance 2Ltraveled byW2must amount to3122,, and soon. We may write this requirement succinctly as21where0,1, 2,.4mLm(a) Thus, the smallest value of/Lthat results in the final waves being exactly in phaseis whenm= 0, which gives/1/ 40.25L.(b) The second smallest value of/Lthat results in the final waves being exactly inphase is whenm= 1, which gives/3/ 40.75L.(c) The third smallest value of/Lthat results in the final waves being exactly in phaseis whenm= 2, which gives/5/ 41.25L.3. (a) We take the phases of both waves to be zero at the front surfaces of the layers. Thephase of the first wave at the back surface of the glass is given by1=k1L –t, wherek1(= 2/1) is the angular wave number and1is the wavelength in glass. Similarly, thephase of the second wave at the back surface of the plastic is given by2=k2L –t,wherek2(= 2/2) is the angular wave number and2is the wavelength in plastic. Theangular frequencies are the same since the waves have the same wavelength in air and thefrequency of a wave does not change when the wave enters another medium. The phasedifference is
CHAPTER 35134812k1k2L21L.Now,1=air/n1, whereairis the wavelength in air andn1is the index of refraction ofthe glass. Similarly,2=air/n2, wheren2is the index of refraction of the plastic. Thismeans that the phase difference is1212air2.nnLThe value ofLthat makes this 5.65 rad isL12air2n1n25.65 400109m21.601.503.60106m.

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Term
Spring
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Tags
Pythagorean Theorem, Wavelength, NM, phase difference

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