Physics · Wave Optics

JEE Main 2025 — 24 January, Evening Shift — Question 63

In a Young's double slit experiment, three polarizers are kept as shown in the figure. The transmission axes of P1P_{1} and P2P_{2} are orthogonal to each other. The polarizer P3P_{3} covers both the slits with its transmission axis at 45∘45^{\circ} to those of P1\mathrm{P}_{1} and P2P_{2}. An unpolarized light of wavelength λ\lambda and intensity I0\mathrm{I}_{0} is incident on P1\mathrm{P}_{1} and P2\mathrm{P}_{2}. The intensity at a point after P3P_{3} where the path difference between the light waves from s1\mathrm{s}_{1} and s2\mathrm{s}_{2} is λ3\frac{\lambda}{3}, is

Question figure
  1. Option A:

    I02\frac{I_{0}}{2}

  2. Option B:

    I04\frac{I_{0}}{4}

    Correct
  3. Option C:

    I0I_{0}

  4. Option D:

    I03\frac{I_{0}}{3}

Answer: B

Step-by-step solution

After passing through third polerisor, Intensity of both the waves must be I04\frac{\mathrm{I}_{0}}{4}

Now, at a point where path diff is λ3\frac{\lambda}{3}, phase difference

Δϕ=2 K(Δxλ)=2π3\Delta \phi=2 \mathrm{~K}\left(\frac{\Delta \mathrm{x}}{\lambda}\right)=\frac{2 \pi}{3}

∴Ires =(I04)2+(I04)2+2(I04)2cos⁡2π3\therefore \mathrm{I}_{\text {res }}=\sqrt{\left(\frac{\mathrm{I}_{0}}{4}\right)^{2}+\left(\frac{\mathrm{I}_{0}}{4}\right)^{2}+2\left(\frac{\mathrm{I}_{0}}{4}\right)^{2}} \cos \frac{2 \pi}{3}

=I04=\frac{\mathrm{I}_{0}}{4}

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
In a Young's double slit experiment, three polarizers are kept as… | JEE Main 2025 PYQ with Solution · DhiX AI