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HW 35 - benavides(jjb2356 homework 35 Turner(59130 This...

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benavides (jjb2356) – 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 . 51. 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 = 28 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: 18 . 114 . 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 . 51. The inci- dent angle in the air is given to be θ a = 28 . Hence sin θ w sin θ a = n a n w sin θ w sin 28 = 1 1 . 51 sin θ w = 0 . 469472 1 . 51 θ w = arcsin(0 . 310908) θ w = 18 . 114 . 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: 41 . 4718 . 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 . 51 parenrightbigg θ w = 41 . 4718 . 003 10.0 points A flashlight on the bottom of a 3.87 m deep swimming pool sends a ray upward and at an angle so that the ray strikes the surface of the water 1.17 m from the point directly above the flashlight.
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