Polarisation question from JEE Main 2024 (27 Jan, Shift 2)

Answer: (c)

1. Middle sheet at $\theta$ to the first, $90^\circ-\theta$ to the last: $I = I_1\cos^2\theta\sin^2\theta = \dfrac{I_1}{4}\sin^2 2\theta$.
2. $I$ is maximum when $\sin 2\theta = 1$, i.e. $2\theta = 90^\circ$.
3. So the answer is (c) $45^\circ$.