Homework_12-solutions

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

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Unformatted text preview: 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 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 parenrightbiggparenleftbigg 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 ....
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This note was uploaded on 04/26/2009 for the course PHY 58310 taught by Professor Chiu during the Spring '09 term at University of Texas.

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

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