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Unformatted text preview: terry (ect328) homework 35 Turner (59130) 1 This print-out should have 12 questions. Multiple-choice 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 . 26. 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 a = 26 with respect to the vertical. air water flashlight ray w a At what angle (with respect to the vertical) is the flashlight being held under transparent liquid? Correct answer: 20 . 3598 . Explanation: By Snells Law n a sin a = n w sin w , where n a and n w are the indices of refraction for each substance and a and 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 . 26. The inci- dent angle in the air is given to be a = 26 . Hence sin w sin a = n a n w sin w sin 26 = 1 1 . 26 sin w = . 438371 1 . 26 w = arcsin(0 . 347914) w = 20 . 3598 . 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: 52 . 528 . Explanation: This is solved in the same fashion as Part 1. When the light ceases to be visible outside the transparent liquid, then a 90 . The sin 90 = 1. Hence (from above), sin w = n a n w w = arcsin parenleftbigg 1 1 . 26 parenrightbigg w = 52 . 528 ....
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- Fall '09