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# qz8f08sol - November 14, 2008 PHY2054 Discussion-Fall 08...

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November 14, 2008 PHY2054 Discussion-Fall ‘08 Quiz 8 (Chapter 21.8-21.13, 22.1-22.7) Name: UFID: *1. (5pts) Assume that the intensity of solar radiation incident on Earth is 1 340 W/m 2 . The Earth is 1.50×10 11 m away from the Sun. Calculate the intensity of the solar radiation incident on Venus, which is 1.08×10 11 m away from the Sun. The wave front of solar radiation is a sphere centered at the Sun. Thus the total power of the solar radiation is given by intensity times area of the sphere. P = I E ×4πr E 2 ( = 1340×4×π×(1.50×10 11 ) 2 = 3.79×10 26 W) At the surface of the Venus, the intensity is I V = P/(4πr V 2 ) = I E ×4πr E 2 /(4πr V 2 ) = I E ×(r E /r V ) 2 = 1340×(1.5/1.08) 2 = 2.58×10 3 W/m 2 *2. (5pts) A beam of light both reflects and refracts at the surface between air and glass. If the index of refraction of the glass is 1.55, find the angle of incidence in the air that would result in the reflected ray and the refracted ray being perpendicular to each other. Since the angle of reflection equals the angle of incidence and a straight line

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## This note was uploaded on 12/13/2011 for the course PHY 2054 taught by Professor Avery during the Fall '08 term at University of Florida.

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qz8f08sol - November 14, 2008 PHY2054 Discussion-Fall 08...

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