HW30Sol - Platt, David Homework 30 Due: Nov 21 2005, 4:00...

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Unformatted text preview: Platt, David Homework 30 Due: Nov 21 2005, 4:00 am Inst: Ken Shih 1 This print-out should have 9 questions. Multiple-choice 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 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 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 A 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 A , the angle of reflection must also be A . Since the mirrors are perpendicular to each other, angle B is equal to 90 - A B = 90 - A = 90 - 41 = 49 ....
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HW30Sol - Platt, David Homework 30 Due: Nov 21 2005, 4:00...

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