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

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1347 Chapter 35 1. The fact that wave W 2 reflects two additional times has no substantive effect on the calculations, since two reflections amount to a 2( /2) = phase difference, which is effectively not a phase difference at all. The substantive difference between W 2 and W 1 is the extra distance 2 L traveled by W 2 . (a) For wave W 2 to be a half-wavelength “behind” wave W 1 , we require 2 L = /2, or L = /4 = (620 nm)/4 =155 nm using the wavelength value given in the problem. (b) Destructive interference will again appear if W 2 is 3 2 “behind” the other wave. In this case, 2 3 2   L , and the difference is 3 620 nm 310nm . 4 4 2 2 L L  2. We consider waves W 2 and W 1 with an initial effective phase difference (in wavelengths) equal to 1 2 , and seek positions of the sliver that cause the wave to constructively interfere (which corresponds to an integer-valued phase difference in wavelengths). Thus, the extra distance 2 L traveled by W 2 must amount to 3 1 2 2 , , and so on. We may write this requirement succinctly as 2 1 where 0,1, 2, . 4 m L m (a) Thus, the smallest value of / L that results in the final waves being exactly in phase is when m = 0, which gives / 1/ 4 0.25 L . (b) The second smallest value of / L that results in the final waves being exactly in phase is when m = 1, which gives / 3/ 4 0.75 L . (c) The third smallest value of / L that results in the final waves being exactly in phase is when m = 2, which gives / 5/ 4 1.25 L . 3. (a) We take the phases of both waves to be zero at the front surfaces of the layers. The phase of the first wave at the back surface of the glass is given by 1 = k 1 L – t , where k 1 (= 2 / 1 ) is the angular wave number and 1 is the wavelength in glass. Similarly, the phase of the second wave at the back surface of the plastic is given by 2 = k 2 L – t , where k 2 (= 2 / 2 ) is the angular wave number and 2 is the wavelength in plastic. The angular frequencies are the same since the waves have the same wavelength in air and the frequency of a wave does not change when the wave enters another medium. The phase difference is

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CHAPTER 35 1348 1 2 k 1 k 2 L 2 1 L . Now, 1 = air / n 1 , where air is the wavelength in air and n 1 is the index of refraction of the glass. Similarly, 2 = air / n 2 , where n 2 is the index of refraction of the plastic. This means that the phase difference is   1 2 1 2 air 2 . n n L The value of L that makes this 5.65 rad is L 1 2 air 2 n 1 n 2 5.65 400 10 9 m 2 1.60 1.50 3.60 10 6 m. (b) 5.65 rad is less than 2
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ch35 - Chapter 35 1. The fact that wave W2 reflects two...

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