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Unformatted text preview: Platt, David – Homework 30 – Due: Nov 21 2005, 4:00 am – Inst: Ken Shih 1 This printout should have 9 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Consider two ways that light might hypothet ically get from its starting point S to its final point F by being reflected by a mirror at ei ther point A or point B. Since light travels at a fixed speed in air, the path of the least time will also be the path of the least distance. ⊗ ⊗ S F A B C 44 cm 44 cm 23cm 23cm Show by calculation that the path SBF is shorter than the path SAF and record the difference in the path lengths. Correct answer: 14 . 6585 cm. Explanation: Given : SA = 23 cm and , AB = 44 cm . The path SBF is SBF = p SA 2 + AB 2 + p BC 2 + CF 2 = 2 p SA 2 + AB 2 = 2 q (23 cm) 2 + (44 cm) 2 = 99 . 2975 cm . The path SAF is SAF = SA + p AC 2 + CF 2 = (23 cm) + q (88 cm) 2 + (23 cm) 2 = 113 . 956 cm . The path difference is Δ SF = SAF SBF = (113 . 956 cm) (99 . 2975 cm) 14 . 6585 cm . 002 (part 1 of 2) 10 points Consider the case in which light ray A is in cident on mirror 1 , as shown in the figure. The reflected ray is incident on mirror 2 and subsequently reflected as ray B. Let the an gle of incidence (with respect to the normal) on mirror 1 be θ A = 41 ◦ and the point of incidence be located 20 cm from the edge of contact between the two mirrors. 90 ◦ x A B mirror 1 mirror 2 θ A θ B What is the angle of the reflection of ray B (with respect to the normal) on mirror 2 ? Correct answer: 49 ◦ . Explanation: If the angle of the incident ray A is θ A , the angle of reflection must also be θ A . Since the mirrors are perpendicular to each other, angle θ B is equal to 90 ◦ θ A θ B = 90 ◦ θ A = 90 ◦ 41 ◦ = 49 ◦ ....
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This note was uploaded on 07/26/2009 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas.
 Fall '08
 Turner
 Physics, Work

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