Unpolarized light of intensity I passes throu | vedclass

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(B) When unpolarized light of intensity $I$ passes through the first polarizer $A$,the intensity of the transmitted light is $I_A = \frac{I}{2}$.Since

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$45$

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(B) When unpolarized light of intensity $I$ passes through the first polarizer $A$,the intensity of the transmitted light is $I_A = \frac{I}{2}$.Since $A$ and $B$ are parallel,the intensity after $B$ is also $\frac{I}{2}$.When polarizer $C$ is placed between $A$ and $B$ at an angle $	heta$ with respect to $A$,the intensity after $C$ is $I_C = I_A \cos^2 	heta = \frac{I}{2} \cos^2 	heta$.Since $B$ is parallel to $A$,the angle between $C$ and $B$ is $(90^{\circ} - 	heta)$ if we assume $A$ and $B$ were crossed,but here the problem states $A$ and $B$ are parallel. However,the standard formula for this setup with $A$ and $B$ parallel and $C$ at angle $	heta$ to $A$ gives the final intensity as $I_B = I_A \cos^2 	heta \cos^2(0 - 	heta) = \frac{I}{2} \cos^4 	heta$.Given $I_B = \frac{I}{8}$,we have $\frac{I}{2} \cos^4 	heta = \frac{I}{8}$.$\cos^4 	heta = \frac{1}{4} \Rightarrow \cos^2 	heta = \frac{1}{2}$.$\cos 	heta = \frac{1}{\sqrt{2}}$,which gives $	heta = 45^{\circ}$.

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