The intensity of transmitted light is maximum when a Polaroid sheet ($P_2$) is rotated between two crossed Polaroids, $P_1$ and $P_3$ when

the angle between $P_1$ and $P_2$ and $P_2$ and $P_3$ will be $\frac{\pi}{4}$

The correct answer is Option (1) → the angle between $P_1$ and $P_2$ and $P_2$ and $P_3$ will be $\frac{\pi}{4}$

For crossed Polaroids $P_1$ and $P_3$, the angle between their transmission axes is $\frac{\pi}{2}$.

If $P_2$ is placed between them, and the angles between $P_1$ and $P_2$, and between $P_2$ and $P_3$, are both $\frac{\pi}{4}$, then:

Using Malus’ law: $I = I_0 \cos^2\theta_1 \cos^2\theta_2$

Substituting $\theta_1 = \theta_2 = \frac{\pi}{4}$,

$I = I_0 \left(\cos^2 \frac{\pi}{4}\right)\left(\cos^2 \frac{\pi}{4}\right) = I_0 \left(\frac{1}{2}\right)\left(\frac{1}{2}\right) = \frac{I_0}{4}$

This configuration gives the maximum transmission through crossed Polaroids with an intermediate sheet.

Answer: The angle between $P_1$ and $P_2$ and $P_2$ and $P_3$ will be $\frac{\pi}{4}$.