Two polarisers P_1 and P_2 are placed such th | vedclass

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Question

(A) Let the intensity of light incident on $P_1$ be $I_0$. Since $P_1$ and $P_2$ are crossed,the angle between them is $90^{\circ}$ or $\frac{\pi}{2}$

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$\frac{\pi}{4}$

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(A) Let the intensity of light incident on $P_1$ be $I_0$. Since $P_1$ and $P_2$ are crossed,the angle between them is $90^{\circ}$ or $\frac{\pi}{2}$.Let the angle between $P_1$ and $P_3$ be $	heta$. Then the angle between $P_3$ and $P_2$ is $(\frac{\pi}{2} - 	heta)$.According to Malus' Law,the intensity after passing through $P_3$ is $I_1 = I_0 \cos^2 	heta$.The intensity after passing through $P_2$ is $I_{	ext{net}} = I_1 \cos^2(\frac{\pi}{2} - 	heta) = I_0 \cos^2 	heta \sin^2 	heta$.$I_{	ext{net}} = I_0 (\sin 	heta \cos 	heta)^2 = I_0 (\frac{\sin 2	heta}{2})^2 = \frac{I_0}{4} \sin^2(2	heta)$.For maximum intensity,$\sin^2(2	heta)$ must be $1$,so $2	heta = 90^{\circ}$ or $	heta = 45^{\circ}$ or $\frac{\pi}{4}$.The angle between $P_3$ and $P_2$ is $\frac{\pi}{2} - 	heta = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$.

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