The intensity of transmitted light is maximum when a Polaroid sheet (…

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

Each polaroid passes the component of the light along its axis, and the intensity follows Malus' law \( I = I_0\cos^{2}\theta \). Unpolarised light of intensity \( I_0 \) leaves \( P_1 \) with \( I_0/2 \).

If \( P_2 \) makes angle \( \theta \) with \( P_1 \), it makes \( 90^\circ - \theta \) with the crossed \( P_3 \).

    \( \displaystyle I = \frac{I_0}{2}\cos^{2}\theta\,\cos^{2}(90^\circ - \theta) = \frac{I_0}{2}\cos^{2}\theta\,\sin^{2}\theta = \frac{I_0}{8}\sin^{2}2\theta \)

This is maximum when \( \sin 2\theta = 1 \), that is \( \theta = 45^\circ = \pi/4 \), and then \( I = I_0/8 \).

Option (1) the angle between \( P_1 \) and \( P_2 \) and \( P_2 \) and \( P_3 \) will be \( \displaystyle\frac{\pi}{4} \) — Correct. At \( \pi/4 \) from both \( P_1 \) and \( P_3 \), \( \sin^{2}2\theta = 1 \) and the intensity is the largest, \( I_0/8 \).

Option (2) polaroids \( P_1 \) and \( P_2 \) are crossed — Incorrect. If \( P_1 \) and \( P_2 \) are crossed, no light passes \( P_2 \) at all.

Option (3) angle between \( P_1 \) and \( P_2 \) is \( \displaystyle\frac{\pi}{3} \) and between \( P_2 \) and \( P_3 \) is \( \displaystyle\frac{\pi}{6} \) — Incorrect. Here \( 2\theta = 120^\circ \), so \( \sin^{2}2\theta = 3/4 \) and \( I = 3I_0/32 \), less than \( I_0/8 \).

Option (4) the angle between \( P_2 \) and \( P_3 \) does not matter — Incorrect. The angle between \( P_2 \) and \( P_3 \) is \( 90^\circ - \theta \), and the intensity depends on it through Malus' law.

NCERT (2026-27):
Polarisation – …