This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 1371 Chapter 35 1. The index of refraction is found from Eq. 353: n c v = = × × = 2 998 10 192 10 156 8 8 . . . . m s m s 2. Note that Snell’s Law (the law of refraction) leads to θ 1 = θ 2 when n 1 = n 2 . The graph indicates that θ 2 = 30° (which is what the problem gives as the value of θ 1 ) occurs at n 2 = 1.5. Thus, n 1 = 1.5, and the speed with which light propagates in that medium is 8 8 1 2.998 10 m s 2.0 10 m s. 1.5 c v n × = = = × 3. Comparing the light speeds in sapphire and diamond, we obtain ( 29 8 7 1 1 1 1 2.998 10 m s 4.55 10 m s. 1.77 2.42 s d s d v v v c n n ∆ = = = × = × 4. (a) The frequency of yellow sodium light is f c = = × × = × c 2 998 10 589 10 509 10 8 9 14 . . m s m Hz. (b) When traveling through the glass, its wavelength is c c n n = = = 589 152 388 nm nm. . (c) The light speed when traveling through the glass is ( 29 ( 29 14 9 8 5.09 10 Hz 388 10 m 1.97 10 m s. n v f = = × × = × l 5. (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 φ 1 = k 1 L – ϖ t , where k 1 (= 2 π / λ 1 ) is the angular wave number and λ 1 is the wavelength in glass. Similarly, the phase of the second wave at the back surface of the plastic is given by φ 2 = k 2 L – ϖ t , where k 2 (= 2 π / λ 2 ) is the angular wave number and λ 2 is the wavelength in plastic. The angular frequencies are the same since the waves have the same wavelength in air and the CHAPTER 35 1372 frequency of a wave does not change when the wave enters another medium. The phase difference is ( 29 1 2 1 2 1 2 . k k L L φ φ  = = 1 2 1 p l l Now, λ 1 = λ air / n 1 , where λ air is the wavelength in air and n 1 is the index of refraction of the glass. Similarly, λ 2 = λ air / n 2 , where n 2 is the index of refraction of the plastic. This means that the phase difference is ( 29 1 2 1 2 air 2 . n n L π φ φ λ = The value of L that makes this 5.65 rad is L n n = = × = × φ φ 1 2 1 2 9 6 2 565 400 10 2 360 10 b g b g c h b g c b bfm hFMfm ¡F air m m. . . (b) 5.65 rad is less than 2 π rad = 6.28 rad, the phase difference for completely constructive interference, and greater than π rad (= 3.14 rad), the phase difference for completely destructive interference. The interference is, therefore, intermediate, neither completely constructive nor completely destructive. It is, however, closer to completely constructive than to completely destructive. 6. In contrast to the initial conditions of Problem 355, 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 integervalued phase difference in wavelengths). Thus, the extra distance 2 L traveled by W 2 must amount to 1 2 3 2 λ λ , , and so on. We may write this , and so on....
View
Full
Document
This note was uploaded on 08/09/2010 for the course PHY 202 101A2 taught by Professor Prof.yang during the Summer '10 term at National Taiwan University.
 Summer '10
 Prof.Yang
 Physics

Click to edit the document details