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(A) Since no light is emitted from the second polaroid $(P_2)$,the first polaroid $(P_1)$ and the second polaroid $(P_2)$ must be crossed,meaning thei

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$\left( \frac{I_0}{8} ight) \sin^2 2	heta$

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(A) Since no light is emitted from the second polaroid $(P_2)$,the first polaroid $(P_1)$ and the second polaroid $(P_2)$ must be crossed,meaning their polarization axes are perpendicular to each other.Let the initial intensity of the unpolarized light be $I_0$.The intensity of light after passing through the first polaroid $(P_1)$ is $I_1 = \frac{I_0}{2}$.The third polaroid $(P_3)$ is placed at an angle $	heta$ with respect to $P_1$. According to Malus' Law,the intensity of light emerging from $P_3$ is $I_2 = I_1 \cos^2 	heta = \frac{I_0}{2} \cos^2 	heta$.The angle between the polarization axis of $P_3$ and $P_2$ is $(90^\circ - 	heta)$.Applying Malus' Law again for the light passing through $P_2$,the final intensity $I_f$ is:$I_f = I_2 \cos^2(90^\circ - 	heta) = I_2 \sin^2 	heta$$I_f = \left( \frac{I_0}{2} \cos^2 	heta ight) \sin^2 	heta = \frac{I_0}{2} (\sin 	heta \cos 	heta)^2$Using the identity $\sin 2	heta = 2 \sin 	heta \cos 	heta$,we have $\sin 	heta \cos 	heta = \frac{\sin 2	heta}{2}$.$I_f = \frac{I_0}{2} \left( \frac{\sin 2	heta}{2} ight)^2 = \frac{I_0}{2} \cdot \frac{\sin^2 2	heta}{4} = \frac{I_0}{8} \sin^2 2	heta$.

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