A polaroid sheet \( P_2 \) is rotated by an angle \( \theta \) (angle between pass axes of \( P_1 \) and \( P_2 \)) between two crossed polaroids \( P_1 \) and \( P_3 \). The intensity of transmitted light will be maximum for:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \theta=45^\circ \)
By Malus's law a polaroid transmits the fraction \( \cos^{2} \) of the intensity of plane polarised light that falls on it, where the angle is measured between the polarisation of the light and the pass axis. Take unpolarised light of intensity \( I_0 \) falling on \( P_1 \).
After \( P_1 \): \( I_1 = I_0/2 \)
The pass axis of \( P_2 \) makes \( \theta \) with that of \( P_1 \), and \( P_3 \) is crossed with \( P_1 \), so the angle between \( P_2 \) and \( P_3 \) is \( 90^\circ - \theta \).
\( \displaystyle I = \frac{I_0}{2}\cos^{2}\theta\,\cos^{2}(90^\circ - \theta) = \frac{I_0}{2}\cos^{2}\theta\sin^{2}\theta \)
\( \displaystyle I = \frac{I_0}{8}\sin^{2}2\theta \)
This is greatest when \( \sin2\theta = \pm1 \), that is \( \theta = 45^\circ \) (or \( 135^\circ \)). The maximum intensity is \( I_0/8 \).
Shortcut: At \( \theta = 0^\circ,\ 90^\circ \) or \( 180^\circ \) the pass axis of \( P_2 \) coincides with that of \( P_1 \) or \( P_3 \) (or is parallel to it). The crossed pair then still blocks the light and the intensity is zero, so only \( 45^\circ \) is left.
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 – “… Example 10.2 Discuss the intensity of transmitted light when a polaroid sheet is rotated between two crossed polaroids? … Therefore, the transmitted intensity will be maximum when \( \theta = \pi/4 \).”