## College Physics (4th Edition)

(a) According to Malus' law, the intensity of the transmitted light would be $I_0~cos^2~90^{\circ}$ which is zero. Therefore, we would need to use more than one polarizing sheet. (b) The transmitted intensity is $\frac{I_0}{4}$ (c) The transmitted intensity is $0.53~I_0$
(a) If we tried to use one polarizing sheet to rotate the direction of polarization, we would have to orient the polarizing sheet at a $90^{\circ}$ angle to the initial direction of polarization. But then, according to Malus' law, the intensity of the transmitted light would be $I_0~cos^2~90^{\circ}$ which is zero. Therefore, we would need to use more than one polarizing sheet. (b) We can use Malus' law to determine the intensity of the light after passing through the first polarizing sheet: $I_1 = I_0~cos^2(45.0^{\circ}) = \frac{I_0}{2}$ We can determine the intensity of the light after passing through the second polarizing sheet: $I_2 = \frac{I_0}{2}~cos^2(45.0^{\circ}) = \frac{I_0}{4}$ The transmitted intensity is $\frac{I_0}{4}$ (c) According to Malus' law, the transmitted intensity of the light when passing through each polarizing sheet would be multiplied by a factor of $cos^2~22.5^{\circ}$. We can find the transmitted intensity after passing through four sheets: $I = I_0~(cos^2~22.5^{\circ})^4$ $I = I_0~(cos~22.5^{\circ})^8$ $I = 0.53~I_0$ The transmitted intensity is $0.53~I_0$