JEE MAIN - Physics (2025 - 4th April Evening Shift - No. 15)
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 $P_2$, at particular angle between $P_2$ and $P_3$. The transmitted intensity of the light passing the through all three polarisers is maximum. The angle between the polarisers $P_2$ and $P_3$ is :
$\pi / 6$
$\frac{\pi}{3}$
$\frac{\pi}{4}$
$\pi / 8$
Explanation
Through $P_2 I_1=I_0 \sin ^2\left(\frac{\pi}{2}-\theta\right)$
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$$I_1=I_0 \cos ^2 \theta$$
Through $P_3 I_{\text {net }}=\left(I_0 \cos ^2 \theta\right) \sin ^2 \theta$
$$I_{\text {net }}=\frac{I_0}{4}[\sin (2 \theta)]^2 \text { for max } I_{\text {net }} \theta=45^{\circ}$$
So angle between $\mathrm{P}_2$ and $\mathrm{P}_3=\frac{\pi}{4}$
Correct Ans. (1)
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