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

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

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.

Page1 / 2

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online