JEE Main 2025 — Wave Optics Question with 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 $\begin{aligned} & \Delta \phi=2 \mathrm{~K}\left(\frac{\Delta \mathrm{x}}{\lambda}\right)=\frac{2 \pi}{3} \\ & \therefore \mathrm{I}_{\mathrm{res}}=\sqrt{\left(\frac{\mathrm{I}_0}{4}\right)^2+\left(\frac{\mathrm{I}_0}{4}\right)^2+2\left(\frac{\mathrm{I}_0}{4}\right)^2} \cos \frac{2 \pi}{3} \\ & =\frac{\mathrm{I}_0}{4} \end{aligned}$