Physics · Wave Optics

JEE Main 2025 — 4 April, Evening Shift — Question 51

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 P2P_{2}, at particular angle between P2P_{2} and P3P_{3}. The transmitted intensity of the light passing the through all three polarisers is maximum. The angle between the polarisers P2P_{2} and P3P_{3} is

  1. Option A:

    π3\frac{\pi}{3}

  2. Option B:

    π8\frac{\pi}{8}

  3. Option C:

    π6\frac{\pi}{6}

  4. Option D:

    π4\frac{\pi}{4}

    Correct

Answer: D

Step-by-step solution

I=locos⁡2θI=l o \cos ^{2} \theta

Angle between P1P_{1} and P2P_{2} is 90∘90^{\circ}

I=Iocos⁡2θ⋅cos⁡2(90−θ)I=I_{o} \cos ^{2} \theta \cdot \cos ^{2}(90-\theta)

I=I0cos⁡2θ⋅sin⁡2θI=I_{0} \cos ^{2} \theta \cdot \sin ^{2} \theta

II will be maximum at θ=45∘\theta=45^{\circ}

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Subject
Physics
Chapter
Wave Optics
Topic
Polarization of Light Waves
Two polarisers P 1 and P 2 are placed in such a way that the… | JEE Main 2025 PYQ with Solution · DhiX AI