## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 22 - Problems - Page 865: 45

#### Answer

At an angle of $45^{\circ}$, the transmitted intensity is a maximum.

#### Work Step by Step

(a) Since the light is polarized in the x-direction initially, the intensity of the light after passing through the first polarizer is $I_0~cos^2~\theta$ After passing through the second polarizer, we can find the intensity of the light that is transmitted: $I = (I_0~cos^2~\theta)~cos^2(90^{\circ}-\theta)$ $I = I_0~cos^2~\theta~sin^2~\theta$ $I = I_0~(sin~\theta~cos~\theta)^2$ $I = I_0~(\frac{1}{2}~sin~2\theta)^2$ $I = \frac{1}{4}~I_0~sin^2~2\theta$ The intensity of the light that is transmitted through the second sheet is $\frac{1}{4}~I_0~sin^2~2\theta$ (Note that $~~sin~2\theta = 2~sin~\theta~cos~\theta$) (b) The intensity of the transmitted light is a maximum when $sin~2\theta = 1$, which is true when $\theta = 45^{\circ}$. At an angle of $45^{\circ}$, the transmitted intensity is a maximum.

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