Verified Solution • 2025–26 Session moderate

QJEEM25Ph2173: JEE Main 2025 Physics - Class 12 JEEM25Ph2173 Solution: In a Young's double slit experiment, three polarizers are kept…

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 P1P_1 and P2P_2 . An unpolarized light of wavelength λ\lambda and intensity I0I_0 is incident on P1P_1 and P2P_2 . The intensity at a point after P3P_3 where the path difference between the light waves from s1s_1 and s2s_2 is λ3\frac{\lambda}{3} , is

JEE Main 2025 (Online) 24th January Evening Shift Physics - Wave Optics Question 2 English

Multiple Choice Options

  • A I0\mathrm{I}_0
  • B I03\frac{\mathrm{I}_0}{3}
  • C I04\frac{\mathrm{I}_0}{4}
  • D I02\mathrm{\frac{I_0}{2}}

Step-by-Step Verified Solution

JEE Main 2025 (Online) 24th January Evening Shift Physics - Wave Optics Question 2 English Explanation

after passing through third poleriser, Intensity of both the waves must be I04\frac{I_0}{4}

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

amp;Δϕ=2 K(Δxλ)=2π3amp;∴Ires =(I04)2+(I04)2+2(I04)2cos⁡2π3amp;=I04\begin{aligned}& \Delta \phi=2 \mathrm{~K}\left(\frac{\Delta \mathrm{x}}{\lambda}\right)=\frac{2 \pi}{3} \\& \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} \\& =\frac{\mathrm{I}_0}{4}\end{aligned}

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