chapter 33 solutions

# chapter 33 solutions - T HE N ATURE AND P ROPAGATION OF L...

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Unformatted text preview: T HE N ATURE AND P ROPAGATION OF L IGHT 33.1. I DENTIFY : For reflection, r a θ θ = . S ET U P : The desired path of the ray is sketched in Figure 33.1. E XECUTE : 14.0 cm tan 11.5 cm φ = , so 50.6 φ = . 90 39.4 r θ φ = - = and 39.4 r a θ θ = = . E VALUATE : The angle of incidence is measured from the normal to the surface. Figure 33.1 33.2 I DENTIFY : For reflection, r a θ θ = . S ET U P : The angles of incidence and reflection at each reflection are shown in Figure 33.2. For the rays to be perpendicular when they cross, 90 α = . E XECUTE : (a) 90 θ φ + = and 90 β φ + = , so β θ = . 90 2 α β + = and 180 2 α θ = - . (b) 1 1 2 2 (180 ) (180 90 ) 45 θ α = - = - = . E VALUATE : As θ , 180 α . This corresponds to the incident and reflected rays traveling in nearly the same direction. As 90 θ , α . This corresponds to the incident and reflected rays traveling in nearly opposite directions. Figure 33.2 33-1 33 33-2 Chapter 33 33.3. I DENTIFY and S ET U P : Use Eqs.(33.1) and (33.5) to calculate v and . λ E XECUTE : (a) 8 8 2.998 10 m/s so 2.04 10 m/s 1.47 c c n v v n a = = = = (b) 650 nm 442 nm 1.47 n λ λ = = = E VALUATE : Light is slower in the liquid than in vacuum. By , v f λ = when v is smaller, λ is smaller. 33.4. I DENTIFY : In air, c f λ = . In glass, n λ λ = . S ET U P : 8 3.00 10 m/s c = E XECUTE : (a) 8 14 3.00 10 m/s 517 nm 5.80 10 Hz c f λ = = = (b) 517 nm 340 nm 1.52 n λ λ = = = E VALUATE : In glass the light travels slower than in vacuum and the wavelength is smaller. 33.5. I DENTIFY : c n v = . n λ λ = , where λ is the wavelength in vacuum. S ET U P : 8 3.00 10 m/s c = . n for air is only slightly larger than unity. E XECUTE : (a) 8 8 3.00 10 m/s 1.54. 1.94 10 m/s c n v a = = = (b) 7 7 (1.54)(3.55 10 m) 5.47 10 m. n λ λ-- = = = E VALUATE : In quartz the speed is lower and the wavelength is smaller than in air. 33.6. I DENTIFY : n λ λ = . S ET U P : From Table 33.1, water 1.333 n = and benzene 1.501 n = . E XECUTE : (a) water water benzene benzene n n λ λ λ = = . water benzene water benzene 1.333 (438 nm) 389 nm 1.501 n n λ λ = = = . (b) water water (438 nm)(1.333) 584 nm n λ λ = = = E VALUATE : λ is smallest in benzene, since n is largest for benzene. 33.7. I DENTIFY : Apply Eqs.(33.2) and (33.4) to calculate and . r b θ θ The angles in these equations are measured with respect to the normal, not the surface. (a) S ET U P : The incident, reflected and refracted rays are shown in Figure 33.7. E XECUTE : 42.5 r a θ θ = = The reflected ray makes an angle of 90.0 47.5 r θ- = with the surface of the glass. Figure 33.7 (b) sin sin , a a b b n n θ θ = where the angles are measured from the normal to the interface....
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## This note was uploaded on 04/14/2008 for the course PHYS 2303 taught by Professor Hoffman during the Fall '07 term at University of Texas at Dallas, Richardson.

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chapter 33 solutions - T HE N ATURE AND P ROPAGATION OF L...

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