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Unformatted text preview: Rodriguez, Brandon – Homework 21 – Due: Nov 2 2006, 11:00 am – Inst: Andrew J Rader 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. Please notice that for your homework to be considered towards your grade, it needs to be submitted one hour before the corresponding recitation starts. Work submitted after this time, but before the DUE DATE on top of this page, will be accepted but not graded. PLEASE REMEMBER THAT YOU MUST CARRY OUT YOUR CALCULA- TIONS TO AT LEAST THREE SIGNIFI- CANT FIGURES. YOUR ANSWER MUST BE WITHIN ONE PERCENT OF THE CORRECT RESULT TO BE MARKED AS CORRECT BY THE SERVER. 001 (part 1 of 2) 5 points Given: The index of refraction of transparent liquid (similar to water but with a different index of refraction) is 1 . 49. A flashlight held under the transparent liq- uid shines out of the transparent liquid in a swimming pool. This beam of light exiting the surface of the transparent liquid makes an angle of θ a = 26 ◦ with respect to the vertical. θ θ air water flashlight ray w a At what angle θ w (with respect to the ver- tical) is the flashlight being held under trans- parent liquid? Correct answer: 17 . 1101 ◦ . Explanation: Basic Concepts: 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. Solution: This is a straightforward appli- cation of Snell’s law. We assume that the surface of the transparent liquid is a level hor- izontal plane, thus each angle with respect to the vertical represents the incident angle in...
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This note was uploaded on 10/18/2010 for the course PHYSICS 251 taught by Professor Pedon during the Spring '08 term at IUPUI.
- Spring '08