JEE Physics 2025 Question with Solution

Given: Two polarisers $P_1$ and $P_2$ are arranged so that no light is transmitted, so their transmission axes are perpendicular.

Find: The angle between $P_2$ and $P_3$ for maximum transmitted intensity when a third polariser $P_3$ is inserted between them.

If the angle between $P_1$ and $P_3$ is $\theta_1$ and the angle between $P_3$ and $P_2$ is $\theta_2$, then

$$\theta_1 + \theta_2 = \frac{\pi}{2}$$

because $P_1$ and $P_2$ are crossed polarisers.

By Malus's law, after the first polariser the intensity becomes

$$\frac{I_0}{2}$$

After passing through $P_3$ and then $P_2$, the transmitted intensity is

$$I = \left(\frac{I_0}{2}\right) \cos^2 \theta_1 \cos^2 \theta_2$$

For maximum intensity, take

$$\theta_1 = \theta_2$$

Using

$$\theta_1 + \theta_2 = \frac{\pi}{2}$$

we get

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

Therefore, the angle between the polarisers $P_2$ and $P_3$ is $\frac{\pi}{4}$. The correct option is A.

Given: $P_1$ and $P_2$ are crossed, so the angle between them is $\frac{\pi}{2}$.

Find: The angle between $P_2$ and $P_3$ for maximum transmission.

Since the third polariser is inserted between two crossed polarisers, maximum transmission occurs when the total angular gap is shared equally. Hence,

$$\theta = \frac{1}{2}\left(\frac{\pi}{2}\right) = \frac{\pi}{4}$$

So the angle between $P_2$ and $P_3$ is $\frac{\pi}{4}$. This works because equal splitting of the angular change gives the largest product of the cosine factors in Malus's law.