
In a Young's double slit experiment, three polarizers are kept as shown in the figure. The transmission axes…

- $\frac{I_0}{2}$
- $\frac{I_0}{4}$
- $\frac{\mathrm{I}_0}{3}$
- $I_0$
Solution

after passing through third poleriser, Intensity of both the waves must be $\frac{\mathrm{I}_0}{4}$
now, at a point where path diff is $\frac{\lambda}{3}$, phase difference
$\begin{aligned}
& \Delta \phi=2 \mathrm{~K}\left(\frac{\Delta \mathrm{x}}{\lambda}\right)=\frac{2 \pi}{3} \\ & \therefore \mathrm{I}_{\mathrm{res}}=\sqrt{\left(\frac{\mathrm{I}_0}{4}\right)^2+\left(\frac{\mathrm{I}_0}{4}\right)^2+2\left(\frac{\mathrm{I}_0}{4}\right)^2} \cos \frac{2 \pi}{3} \\ & =\frac{\mathrm{I}_0}{4}
\end{aligned}$
Asked in: JEE Main 2025 (24 Jan Shift 2)