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Unformatted text preview: Nguyen, Thanh Homework 20 Due: Apr 18 2008, 7:00 pm Inst: Weathers 1 This printout should have 13 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 A 1 . 73 m long pole stands vertically in a lake having a depth of 0 . 81 m. When the Sun is 35 . 7 above the horizon, find the length of the poles shadow on the bottom of the lake. The index of refraction for water is 1 . 33. Correct answer: 1 . 90482 m. Explanation: 2 s 2 pole s 1 l 1 d Given : = 1 . 73 m , d = 0 . 81 m , n = 1 . 33 , and = 35 . 7 . Light just passing the top of the pole has an angle of incidence of 1 = 90  = 90  35 . 7 = 54 . 3 . This light falls on the waters surface at dis tance s 1 = (  d ) tan 1 = (1 . 73 m . 81 m) tan54 . 3 = 1 . 28031 m from the pole. As the light enters the water, Snells Law becomes sin 1 = n sin 2 2 = arcsin sin 1 n = arcsin sin54 . 3 1 . 33 = 37 . 6321 . Then distance between the extremity of the poles shadow on the waters surface and the extremity of the poles shadow on the bottom of the lake is s 2 = d tan 2 = (0 . 81 m) tan37 . 6321 = 0 . 624507 m . The whole shadows length is s = s 1 + s 2 = 1 . 28031 m + 0 . 624507 m = 1 . 90482 m . Note: The shadows length without the water would be s = tan 1 so its length is shortened by the water by s = s  s = (1 . 73 m) tan54 . 3  1 . 90482 m = 0 . 502726 m . keywords: 002 (part 1 of 4) 10 points Light of wavelength 396 nm is incident on the face of a silica prism at an angle of 1 = 56 . 2 (with respect to the normal to the surface). The apex angle of the prism is = 72 . 6 . Given: The value of the index of refraction for silica is n = 1 . 455. 2 3 1 4 Find the angle of refraction at this first surface. Correct answer: 34 . 8286 . Nguyen, Thanh Homework 20 Due: Apr 18 2008, 7:00 pm Inst: Weathers 2 Explanation: Basic Concept: Snells Law n 1 sin 1 = n 2 sin 2 . Solution: Applying Snells law, the refrac tion angle at the first surface 2 = sin 1 sin 1 n = sin 1 sin 56 . 2 1 . 455 = 34 . 8286 . 003 (part 2 of 4) 10 points Find the angle of incidence at the second sur face. Correct answer: 37 . 7714 . Explanation: The incident angle at the second surface is 3 =  2 = (72 . 6 ) (34 . 8286 ) = 37 . 7714 , where is the apex angle of the prism. 004 (part 3 of 4) 10 points Find the angle of refraction at the second surface. Correct answer: 63 . 0252 ....
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This note was uploaded on 10/24/2011 for the course PHYS 2220 taught by Professor Littler during the Spring '00 term at North Texas.
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