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Unformatted text preview: tovar (jdt436) – homework 35 – Turner – (59070) 1 This printout should have 12 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points The index of refraction of a transparent liquid (similar to water but with a different index of refraction) is 1 . 29. A flashlight held under the transparent liquid shines out of the transpar ent liquid in a swimming pool. This beam of light exiting the surface of the transparent liq uid makes an angle of θ a = 29 ◦ with respect to the vertical. θ θ air water flashlight ray w a At what angle (with respect to the vertical) is the flashlight being held under transparent liquid? Correct answer: 22 . 0751 ◦ . Explanation: By Snell’s Law n a sin θ a = n w sin θ w , where n a and n w are the indices of refraction for each substance and θ a and θ w are the inci dent angles to the boundary in each medium, respectively. Assume that the surface of the transparent liquid is a level horizontal plane, thus each angle with respect to the vertical represents the incident angle in each medium. The index of refraction of air is (nearly) n a = 1 . 0 while the index of refraction of trans parent liquid is given as n w = 1 . 29. The inci dent angle in the air is given to be θ a = 29 ◦ . Hence sin θ w sin θ a = n a n w sin θ w sin 29 ◦ = 1 1 . 29 sin θ w = . 48481 1 . 29 θ w = arcsin(0 . 375822) θ w = 22 . 0751 ◦ . 002 (part 2 of 2) 10.0 points The flashlight is slowly turned away from the vertical direction. At what angle will the beam no longer be visible above the surface of the pool? Correct answer: 50 . 8226 ◦ . Explanation: This is solved in the same fashion as Part 1. When the light ceases to be visible outside the transparent liquid, then θ a ≥ 90 ◦ . The sin 90 ◦ = 1. Hence (from above), sin θ w = n a n w θ w = arcsin parenleftbigg 1 1 . 29 parenrightbigg θ w = 50 . 8226 ◦ . 003 10.0 points A flashlight on the bottom of a 3.54 m deep swimming pool sends a ray upward and at an angle so that the ray strikes the surface of the water 1.42 m from the point directly above the flashlight....
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This note was uploaded on 12/11/2009 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
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