JEE Mains · Physics · STD 12 - Wave optics
- JEE Mains 2025
- Medium
- MCQ
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 :
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(\frac{\pi}{4}\)
Step-by-step Solution
Detailed explanation
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)

\(I_1=I_0 \cos ^2 \theta\)
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)
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