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(B) Let the initial intensity of unpolarized light be $I_0$. After passing through the first polarizer,the intensity becomes $I_1 = \frac{I_0}{2}$. Th

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$\frac{9}{32}$

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(B) Let the initial intensity of unpolarized light be $I_0$. After passing through the first polarizer,the intensity becomes $I_1 = \frac{I_0}{2}$. The angle between the first and second polarizer is $	heta_1 = 60^o - 30^o = 30^o$. Using Malus' Law,the intensity after the second polarizer is $I_2 = I_1 \cos^2(30^o) = \frac{I_0}{2} 	imes (\frac{\sqrt{3}}{2})^2 = \frac{I_0}{2} 	imes \frac{3}{4} = \frac{3I_0}{8}$. The angle between the second and third polarizer is $	heta_2 = 30^o - 60^o = -30^o$. Using Malus' Law,the intensity after the third polarizer is $I_3 = I_2 \cos^2(-30^o) = \frac{3I_0}{8} 	imes (\frac{\sqrt{3}}{2})^2 = \frac{3I_0}{8} 	imes \frac{3}{4} = \frac{9I_0}{32}$. Thus,the fraction of initial intensity transmitted is $\frac{9}{32}$.

An unpolarized light passes through three polarizing sheets whose polarizing directions make an angle of $30^o$,$60^o$,and $30^o$ with the $y$-axis in the same sense. What fraction of the initial intensity is transmitted by the system?

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