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Unformatted text preview: mehta (snm425) – Homework 1 – erskine – (58715) 1 This printout should have 21 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 10.0 points For 599 nm light, calculate the critical an gle for the following materials surrounded by water (index of refraction: 1 . 333). diamond. (index of refraction: 2 . 34) Correct answer: 34 . 7264 ◦ . Explanation: Basic Concept: Snell’s law for total internal reflection: n 1 sin θ c = n 2 sin 90 ◦ where θ c is the critical angle for total internal reflection. From Snell’s law, for total internal reflection in diamond, 2 . 34 sin θ 1 = 1 . 333 sin 90 ◦ So, critical angle for for diamond θ diamond = sin − 1 parenleftbigg 1 . 333 2 . 34 parenrightbigg = 34 . 7264 ◦ . 002 (part 2 of 3) 10.0 points flint glass. (index of refraction: 1 . 64) Correct answer: 54 . 3709 ◦ . Explanation: Similarly, critical angle for flint glass θ glass = sin − 1 parenleftbigg 1 . 333 1 . 64 parenrightbigg = 54 . 3709 ◦ . 003 (part 3 of 3) 10.0 points ice. (index of refraction: 1 . 31) 1. sin − 1 parenleftbigg n water n ice parenrightbigg 2. there is no critical angle correct 3. sin − 1 parenleftbigg 1 n ice parenrightbigg 4. sin − 1 parenleftbigg n ice n water parenrightbigg Explanation: For ice, since its refraction index, 1 . 31, is less than that of water, 1 . 333, there is no critical angle! 004 10.0 points Hint: A ray diagram would be helpful. Determine the minimum height of a vertical flat mirror in which a person 71 in . in height can see his or her full image. Correct answer: 35 . 5 in . . Explanation: In the figure, the mirror is labeled AB . A ray from the woman’s foot F strikes the bottom of the mirror at B , with an angle equal to θ and proceeds to the woman’s eye. L θ θ C D A B E T F M The two right triangles EBM and FBM are identical, since they share the common side MB and angle θ . Therefore EM = MF = 1 2 EF which is also the distance BC . Similarly, a ray from the top of the woman’s head T strikes the top of the mirror at A and proceeds to her eye. The same line of reasoning as above leads to the conclusion that DA = 1 2 TE Thus the length AB of the mirror is 1 2 ( TE + EF ), which is one half the woman’s height. Note that the mirror’s bottom edge must be exactly 1 2 EF from the floor for a fullheight image to be possible. Note also that the conclusions reached here are valid regardless of how far she stands from the mirror. mehta (snm425) – Homework 1 – erskine – (58715) 2 005 10.0 points An object is placed at a distance of 1.5 f from a converging lens of focal length f , as shown. 2 f f f 2 f 3 f What type of image is formed and what is the size relative to the object?...
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This note was uploaded on 01/20/2010 for the course PHY 1 taught by Professor Erskine during the Spring '10 term at University of Texas.
 Spring '10
 ERSKINE
 Physics, Work, Light

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