# CH-35 - Chapter 35 1 Comparing the light speeds in sapphire...

This preview shows pages 1–3. Sign up to view the full content.

Chapter 35 1. Comparing the light speeds in sapphire and diamond, we obtain v v v c n n s d s d F H G I K J F H G I K J 1 1 2 998 10 1 177 1 2 42 4 55 10 8 7 . . . . . m s m s c h 2. (a) The frequency of yellow sodium light is f c 2 998 10 589 10 509 10 8 9 14 . . m s m Hz. (b) When traveling through the glass, its wavelength is n n 589 152 388 nm nm. . (c) The light speed when traveling through the glass is   14 9 8 5.09 10 Hz 388 10 m 1.97 10 m s. n v f l 3. The index of refraction is found from Eq. 35-3: n c v 2 998 10 192 10 156 8 8 . . . . m s m s 4. Note that Snell’s Law (the law of refraction) leads to 1 = 2 when n 1 = n 2 . The graph indicates that 2 = 30° (which is what the problem gives as the value of 1 ) occurs at n 2 = 1.5. Thus, n 1 = 1.5, and the speed with which light propagates in that medium is v = = 2.0 × 10 8 m/s. 5. 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 . 1407

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
CHAPTER 35 (a) For wave W 2 to be a half-wavelength “behind” wave W 1 , we require 2 L = /2, or L = /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 L L 3 4 4 2 310 nm . 6. In contrast to the initial conditions of problem 30, we now 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 which 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 1 2 3 2 , , and so on. We may write this requirement succinctly as 2 1 where 0,1, 2, . 4 m L m K (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 . 7. (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
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 44

CH-35 - Chapter 35 1 Comparing the light speeds in sapphire...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online