Two polaroids \( P_1 \) and \( P_3 \) are placed such that their pass-axis are mutually perpendicular. Another polaroid \( P_2 \) is rotated between \( P_1 \) and \( P_3 \). For what angle \( \theta \) between \( P_1 \) and \( P_2 \) the intensity of light emerging from \( P_3 \) will be maximum?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{\pi}{4} \)
Light from \( P_1 \) is plane polarised along its pass-axis. Each later polaroid passes the fraction \( \cos^2 \) of the angle between the light's direction of polarisation and its own pass-axis (Malus's law).
Let the intensity after \( P_1 \) be \( I_1 \). If \( P_2 \) makes angle \( \theta \) with \( P_1 \), it makes \( 90^\circ - \theta \) with \( P_3 \).
\( \displaystyle I_2 = I_1\cos^{2}\theta \)
\( \displaystyle I_3 = I_2\cos^{2}(90^\circ - \theta) = I_1\cos^{2}\theta\,\sin^{2}\theta \)
Using \( 2\sin\theta\cos\theta = \sin 2\theta \):
\( \displaystyle I_3 = \frac{I_1}{4}\sin^{2}2\theta \)
This is largest when \( \sin 2\theta = 1 \), that is \( 2\theta = \pi/2 \), so \( \theta = \pi/4 \).
\( \displaystyle I_{3,\max} = \frac{I_1}{4} \)
Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.
NCERT (2026-27):
Polarisation – …
