Homework_12-solutions

# Homework_12-solutions - toupal(rgt374 Homework 12...

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toupal (rgt374) – Homework 12 – Chiu – (58295) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The reflecting surfaces of two intersecting flat mirrors are at an angle of 54 , as shown in the figure. A light ray strikes the horizon- tal mirror, reflects off the horizontal mirror, impinges on the raised mirror, reflects off the raised mirror, and proceeds in the right-hand direction. 54 φ Figure is not drawn to scale. Calculate the angle φ . Correct answer: 72 . Explanation: Basic Concept: θ incident = θ reflected Solution: θ 1 θ 2 φ θ Figure is to scale. The sum of the angles in a triangle is 180 . In the triangle on the left we have angles θ , 180 - θ 1 2 , and 180 - θ 2 2 , so 180 = θ + 180 - θ 1 2 + 180 - θ 2 2 , or θ 1 + θ 2 = 2 θ . (1) In the triangle on the right we have angles θ 1 , θ 2 , and φ . 180 = θ 1 + θ 2 + φ , so θ 1 + θ 2 = 180 - φ . (2) Combining Eq. 1 and 2, we have φ = 180 - 2 θ = 180 - 2 (54 ) = 72 . As a matter of interest, in the upper-half of the figure the angles (clockwise) in the triangles from left to right are 33 , 33 , and 114 ; 66 , 36 , and 78 ; 102 , 21 , and 57 ; 123 , 21 , and 36 ; and in the lower-half of the figure the angles (counter-clockwise) in the triangles from left to right are 21 , 21 , and 138 ; 42 , 36 , and 102 ; 78 , 33 , and 69 ; 111 , 33 , and 36 . 002 (part 1 of 3) 10.0 points Consider a light ray which enters from air to a liquid, where the index of refraction of the liquid is given by n = 1 . 3. Consider the following three ratios, where each is defined by the specified quantity in the liquid, λ , f , and v , to that in the air, λ, f, and c . n = 1 1 . 3 Air Liquid light ray

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toupal (rgt374) – Homework 12 – Chiu – (58295) 2 What is the ratio of their wavelengths, λ λ ? Correct answer: 0 . 769231. Explanation: The frequency of an electro-magnetic wave is independent of the media in which it is present, that is, f = f . A ray with a frequency f , has a wavelength λ = c f in the vacuum. In a medium with an index of refraction n and a velocity v , and from the definition of the index of refraction, n c v . λ = v f = v parenleftbigg 1 f parenrightbigg = parenleftBig c n parenrightBig parenleftbigg 1 f parenrightbigg = parenleftbigg 1 n parenrightbigg parenleftbigg c f parenrightbigg = λ n . So for n = 1 . 3 λ λ = 1 n = 0 . 769231 , f f = 1 , and v c 1 n = 0 . 769231 . 003 (part 2 of 3) 10.0 points What is the ratio of their frequencies of oscil- lations, f f ? Correct answer: 1. Explanation: See explanation for Part 1. The frequency doesn’t change, so for n = 1 . 3 f f = 1 . 004 (part 3 of 3) 10.0 points What is the ratio of the traveling speeds, v c ? Correct answer: 0 . 769231. Explanation: From the definition of the index of refrac- tion in Part 1, for n = 1 . 3 v c 1 n = 0 . 769231 .
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