JEE Physics 2025 Question with Solution

Given: Unpolarized light of intensity $I_0$ is incident on polarizers $P_1$ and $P_2$. Their transmission axes are mutually orthogonal. Polarizer $P_3$ has its transmission axis at $45^\circ$ to both. The path difference between waves from $S_1$ and $S_2$ is $\frac{\lambda}{3}$.

Find: The intensity after $P_3$.

When unpolarized light passes through a polarizer, its intensity becomes half. Therefore, after $P_1$ and $P_2$:

Now each beam passes through $P_3$, whose axis is at $45^\circ$ to the polarization directions of both beams. By Malus's law:

Hence the total intensity after $P_3$ is:

The solution states the correct option is C. Therefore, the intensity is $\frac{I_0}{2}$, so the correct option is C.

Given: Three polarizers are used in a Young's double slit arrangement. $P_1$ and $P_2$ are orthogonal, and $P_3$ is inclined at $45^\circ$ to both. Incident light is unpolarized with intensity $I_0$.

Find: The intensity at the observation point after $P_3$.

For unpolarized incident light, the intensity after a polarizer is reduced to half:

Since the transmission axis of $P_3$ makes an angle $45^\circ$ with each of these polarized beams, each beam is again reduced by a factor of $\cos^2 45^\circ$:

Substitute $\cos^2 45^\circ = \frac{1}{2}$:

Therefore, the final intensity is $\frac{I_0}{2}$ and the correct option is C.