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JEE Main 20254 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
4 April
Shift
Evening Shift
Subject
Physics
Chapter
Wave Optics

JEE Main 2025 · 4 April · Evening Shift · Physics · Wave Optics

Two polarisers P_1 and P_2 are placed in such a way that the...

Two polarisers P1P_1 and P2P_2 are placed in such a way that the intensity of the transmitted light will be zero. A third polariser P3P_3 is inserted in between P1P_1 and P2\mathrm{P}_2, at the particular angle between P2\mathrm{P}_2 and P3\mathrm{P}_3. The transmitted intensity of the light passing the through all three polarisers is maximum. The angle between the polarisers P2\mathrm{P}_2 and P3\mathrm{P}_3 is :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

Options

  1. Aπ4\frac{\pi}{4}
  2. Bπ6\frac{\pi}{6}
  3. Cπ8\frac{\pi}{8}
  4. Dπ3\frac{\pi}{3}

Correct answer

A. π4\frac{\pi}{4}

Step-by-step solution

Through P2I1=I0sin⁡2(π2−θ)P_2 I_1=I_0 \sin ^2\left(\frac{\pi}{2}-\theta\right)
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
I1=I0cos⁡2θI_1=I_0 \cos ^2 \theta
Through P3Inet =(I0cos⁡2θ)sin⁡2θP_3 I_{\text{net }}=\left(I_0 \cos ^2 \theta\right) \sin ^2 \theta
Inct=I04[sin⁡(2θ)]2 for max Inet θ=45∘I_{n c t}=\frac{I_0}{4}[\sin (2 \theta)]^2 \text{ for max } I_{\text{net }} \theta=45^{\circ}
So angle between P2P_2 and P3=π4P_3=\frac{\pi}{4}
Correct Ans. (1)
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