Consider the polaroids \( P_1 \) and \( P_2 \) kept with axis parallel to the other. Now the polaroid \( P_2 \) is rotated from 0 to \( 2\pi \) with respect to \( P_1 \). The variation of intensity of transmitted light through \( P_2 \) will be shown as
Answer & explanation
Correct answer: option 1
The correct answer is Option (1)
The light leaving \( P_1 \) is plane polarised. By Malus' law, the intensity transmitted by \( P_2 \) when its axis makes an angle \( \theta \) with that of \( P_1 \) is
\( \displaystyle I_2 = I_1\cos^{2}\theta \)
where \( I_1 \) is the intensity of the light leaving \( P_1 \). As \( \theta \) goes from 0 to \( 2\pi \):
\( \theta = 0:\ I_2 = I_1\ (\text{maximum}) \)
\( \theta = \pi/2:\ I_2 = 0 \)
\( \theta = \pi:\ I_2 = I_1 \)
\( \theta = 3\pi/2:\ I_2 = 0 \)
\( \theta = 2\pi:\ I_2 = I_1 \)
Since \( \cos^{2}\theta \) is a square, the intensity is never negative. The graph starts at its maximum, falls to zero at \( \pi/2 \) and \( 3\pi/2 \), and has maxima at 0, \( \pi \) and \( 2\pi \). Only Option (1) shows this.
Why the other options are incorrect
Option (2) — Incorrect. This is the \( \sin^{2}\theta \) curve: it is zero at \( \theta = 0 \), where parallel axes give the maximum.
Option (3) — Incorrect. This \( \cos\theta \) curve goes negative, and an intensity cannot be negative.
Option (4) — Incorrect. This \( \sin\theta \) curve starts at zero and goes negative.
NCERT (2026-27):
Polarisation – “… If the pass-axis of P2 makes an angle θ with the pass-axis of P1, then when the polarised beam passes through the polaroid P2, the component E cos θ (along the pass-axis of P2) will pass through P2. Thus, as we rotate the polaroid P1 (or P2), the intensity will vary as:
\( \displaystyle I = I_0 \cos^2\theta \qquad (10.18) \)
where I0 is the intensity of the polarized light after passing through P1. This is known as Malus’ law.”