Question 163 of 191

Two polarisers $P_1$ and $P_2$ are placed in such a way that the intensity of the transmitted light will be zero. A third polariser $P_3$ is inserted in between $P_1$ and $\mathrm{P}_2$, at the particular angle between $\mathrm{P}_2$ and $\mathrm{P}_3$. The transmitted intensity of the light passing the through all three polarisers is maximum. The angle between the polarisers $\mathrm{P}_2$ and $\mathrm{P}_3$ is :

A$\frac{\pi}{4}$B$\frac{\pi}{6}$C$\frac{\pi}{8}$D$\frac{\pi}{3}$

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Answer: (A) $\frac{\pi}{4}$

Through $P_2 I_1=I_0 \sin ^2\left(\frac{\pi}{2}-\theta\right)$

$I_1=I_0 \cos ^2 \theta$
Through $P_3 I_{\text {net }}=\left(I_0 \cos ^2 \theta\right) \sin ^2 \theta$
$I_{n c t}=\frac{I_0}{4}[\sin (2 \theta)]^2 \text { for max } I_{\text {net }} \theta=45^{\circ}$
So angle between $P_2$ and $P_3=\frac{\pi}{4}$
Correct Ans. (1)