Chapter 22

Chapter 22 - CHAPTER 22 Quick Quizzes 1. (a). In part (a),...

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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

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CHAPTER 22 Problem Solutions 22.1 The total distance the light travels is Therefore, ( ) 866 2 23.84 10 6.38 1.76 m7 . 52 10 m center to Eart hM o o n center dD R R ⎛⎞ ∆= ⎜⎟ ⎝⎠ = ×−×−× 8 8 7.52 2.51 s d v t ∆× == = 8 3.00 s × 22.2 If the wheel has 360 teeth, it turns through an angle of 1720 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 ( ) 5 1720 rev 5.05 s 27.5 rev s t θ ω = × Hence, the speed of light is ( ) 5 2 7500 m 2 s d c t = × 8 2.97 s × 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, () 12 rev 2 rad rev rad 720 π θπ = , and ( ) ( ) 8 3 2.998 10 m s 2 22 7 2 0 2 11.45 c td c d θθ × ∆∆ = = = × 114 rad s 208
CHAPTER 22 22.4 (a) The time for the light to travel to the stationary mirror and back is ( ) 3 4 8 235. 01 0 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 18 rev in this time, so 4 s min t θ ω == = ∆× 536 rev s (b) At the next higher angular speed, the mirror will have rotated 28 rev in the elapsed time, or ( ) 2 22 5 3 6 r e v s min ωω = 3 1.07 10 rev s × 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 . Therefore, if c is the speed of light, d t c , or d c = (b) If 2. , 500 m d = () ( ) 11 rev 1 rev degree 60 360 degree 60 360 ⎛⎞ ⎜⎟ ⎝⎠ , and 5555 rev s = , ( )( ) 360 2.500 m 5555 s1 c 8 3.000 10 m s × 22.6 Mirror 1 Light beam Mirror 2 40.0° P 1.25 m Mirror 1 i 1 = 40° 40° d 1.25 m Mirror 2 40° 50° 50° 50° i 2 = 50° 209

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CHAPTER 22 (a) From geometry, 1.25 ms i n40.0 d = ° , so d = 1.94 m (b) 50. 0 above horizontal ° , or parallel to the incident ray. 22.7 11 2 si ns i n nn 2 θ = 1 n1 . 333 sin 45.0 1 n( 1.333)(0.707) 0.943 == 1 70. 5 19. 5 above the horizontal ° θ 1 n 1 = 1.00 θ 2 n 2 = 1.333 22.8 The speed of light in water is water c v n = , and in Lucite Lucite . Thus, the total time required to transverse the double layer is 1 Lucit ewa t er eLu ite dd d vv + ∆= The time to travel the same distance in air is 2 , so the additional time required for the double layer is ( ) 12 i te dn tt +− ∆=∆ −∆ )( 22 8 1.0010 m1. 3331 0.50010 591 3.0010 ms −−
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This homework help was uploaded on 03/21/2008 for the course PY 211 / 212 taught by Professor Chilton during the Spring '08 term at N.C. State.

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Chapter 22 - CHAPTER 22 Quick Quizzes 1. (a). In part (a),...

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