Chapter22

Chapter22 - C HAPTER 22 Quick Quizzes 1. (a). In part (a),...

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Unformatted text preview: C HAPTER 22 Quick Quizzes 1. (a). In part (a), you can see clear reflections of the headlights and the lights on the top of the truck. The reflection is specular. In part (b), although bright areas appear on the roadway in front of the headlights, the reflection is not as clear and no separate reflection of the lights from the top of the truck is visible. The reflection in part (b) is mostly diffuse. 2. Beams 2 and 4 are reflected; beams 3 and 5 are refracted. 3. (b). When light goes from one material into one having a higher index of refraction, it refracts toward the normal line of the boundary between the two materials. If, as the light travels through the new material, the index of refraction continues to increase, the light ray will refract more and more toward the normal line. 4. (c). Both the wave speed and the wavelength decrease as the index of refraction increases. The frequency is unchanged. 207 C H A P T E R 2 2 Problem Solutions 22.1 The total distance the light travels is ( 29 8 6 6 8 2 2 3.84 10 6.38 10 1.76 10 m 7.52 10 m center to Earth Moon center d D R R ∆ =-- =λ-…- = Therefore, 8 7.52 10 m 2.51 s d v t ∆ = = = ∆ 8 3 .0 0 1 0 m s &#2; 22.2 If the wheel has 360 teeth, it turns through an angle of 1 720 rev in the time it takes the light to make its round trip. From the definition of angular velocity, we see that the time is ( 29 5 1 720 rev 5.05 10 s 27.5 rev s t θ ϖ- = = = &#0; Hence, the speed of light is ( 29 5 2 7500 m 2 5.05 10 s d c t- = = = 8 2 .9 7 1 0 m s &#2; 22.3 The experiment is most convincing if the wheel turns fast enough to pass outgoing light through one notch and returning light through the next. Then, ( 29 1 2 rev 2 rad rev rad 720 720 π θ π ∆ = = , and ( 29 ( 29 ( 29 8 3 2.998 10 m s 2 rad 2 2 720 2 11.45 10 m c t d c d θ θ θ π ϖ ∆ ∆ ∆ = = = = = ∆ 1 1 4 r a d s 208 C H A P T E R 2 2 22.4 (a) The time for the light to travel to the stationary mirror and back is ( 29 3 4 8 2 35.0 10 m 2 2.33 10 s 3.00 10 m s d t c- ∆ = = = “ At the lowest angular speed, the octagonal mirror will have rotated 1 8 rev in this time, so 4 1 8 rev 2.33 10 s min t θ ϖ- ∆ = = = ∆ 5 3 6 r e v s (b) At the next higher angular speed, the mirror will have rotated 2 8 rev in the elapsed time, or ( 29 2 2 2 536 rev s min ϖ ϖ = = = 3 1 .0 7 1 0 r e v s &#2; 22.5 (a) For the light beam to make it through both slots, the time for the light to travel distance d must equal the time for the disks to rotate through angle θ . Therefore, if c is the speed of light, d t c θ ϖ = = , or d c ϖ θ = (b) If 2.500 m d = ,...
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This homework help was uploaded on 04/21/2008 for the course PHY 213 taught by Professor Cao during the Spring '08 term at Kentucky.

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Chapter22 - C HAPTER 22 Quick Quizzes 1. (a). In part (a),...

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