College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 22 - Problems - Page 864: 34

Answer

To have equal exposure, the second plate should be exposed for 6.8 hours.

Work Step by Step

$U = I~A~t$ $U$ is energy $I$ is intensity $A$ is the area $t$ is the time Let $P$ be the power of the source. Let $I_1$ be the intensity when the plate is a distance of 1.8 meters from the source. We can find an expression for $I_1$: $I_1 = \frac{P}{4\pi (1.8~m)^2}$ $I_1 = 0.02456~P$ We can find an expression for $I_2$: $I_2 = \frac{P}{4\pi (4.7~m)^2}$ $I_2 = 0.00360~P$ We can write an expression for the energy collected by the first plate: $U = I_1~A~t_1$ We can write an expression for the energy collected by the second plate: $U = I_2~A~t_2$ Since the energy should be equal, we can equate the two expressions to find $t_2$: $I_2~A~t_2 = I_1~A~t_1$ $t_2 = \frac{I_1~t_1}{I_2}$ $t_2 = \frac{(0.02456~P)(1~h)}{0.00360~P}$ $t_2 = 6.8~h$ To have equal exposure, the second plate should be exposed for 6.8 hours.
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