JEE Main 2025 Jan 24 Shift 2, Physics Q20: Young’s Double Slit Experiment

Track the intensity through each polarizer:

1. Unpolarized light of intensity $I_0$ falls on $P_1$ (and $P_2$). Passing through a single polarizer halves the intensity, so after $P_1$: $I_0/2$ along $P_1$'s axis. After $P_2$: $I_0/2$ along $P_2$'s axis (perpendicular to $P_1$'s).
2. Both beams then pass through $P_3$ at $45^{\circ }$ to $P_1$ and $P_2$. By Malus's law, each loses a factor ${\cos }^{2}45^{\circ }=1/2$. So after $P_3$: each beam has intensity $\frac{I_0}{2}\cdot \frac{1}{2}=\frac{I_0}{4}$, and both are now polarized along the same axis (that of $P_3$), so they can interfere.
3. The path difference $\Delta x=\lambda /3$ gives a phase difference $\Delta \varphi =\frac{2\pi }{\lambda }\cdot \frac{\lambda }{3}=\frac{2\pi }{3}$.
4. For two coherent waves of equal intensity $I=I_0/4$:

$I_{\text{res}}=I+I+2\sqrt{I\cdot I}\cos (\Delta \varphi )=2\cdot \frac{I_0}{4}+2\cdot \frac{I_0}{4}\cdot \cos \frac{2\pi }{3}$

$=\frac{I_0}{2}+\frac{I_0}{2}\cdot \left(-\frac{1}{2}\right)=\frac{I_0}{2}-\frac{I_0}{4}=\frac{I_0}{4}$