## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 22 - Problems - Page 864: 42

#### Answer

We can rank the intensities of light transmitted through the second polarizer, from greatest to smallest: $a = e \gt b \gt d \gt c$

#### Work Step by Step

Since the light is randomly polarized initially, the intensity of the light after passing through the first polarizer is $\frac{I_0}{2}$ We can use the law of Malus to determine an expression for the intensity after passing through the second polarizer. $I = \frac{I_0}{2}~cos^2(\vert \theta_2-\theta_1\vert)$ For each situation, we can find an expression for the intensity of the light after passing through the two polarizers. (a) $I = \frac{I_0}{2}~cos^2(30^{\circ}-0^{\circ}) = \frac{3}{8}\times I_0$ (b) $I = \frac{I_0}{2}~cos^2(30^{\circ}-30^{\circ}) = \frac{1}{2}\times I_0$ (c) $I = \frac{I_0}{2}~cos^2(90^{\circ}-0^{\circ}) = 0\times I_0$ (d) $I = \frac{I_0}{2}~cos^2(60^{\circ}-0^{\circ}) = \frac{1}{8}\times I_0$ (e) $I = \frac{I_0}{2}~cos^2(60^{\circ}-30^{\circ}) = \frac{3}{8}\times I_0$ We can rank the intensities of light transmitted through the second polarizer, from greatest to smallest: $a = e \gt b \gt d \gt c$

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