Three identical polaroid sheets \( P_1, P_2, P_3 \) are oriented so that pass axis of \( P_2 \) and \( P_3 \) are inclined to \( P_1 \) at \( 60^\circ \) and \( 90^\circ \) respectively. A monochromatic source of intensity \( I_0 \) is used to pass through \( P_1, P_2, P_3 \). Intensities after passing through \( P_1, P_2, P_3 \) will be respectively :
(A) \( I_0, \displaystyle\frac{I_0}{2}, \frac{I_0}{8} \)
(B) \( \displaystyle\frac{I_0}{2}, \frac{I_0}{8}, \frac{I_0}{32} \)
(C) \( \displaystyle\frac{I_0}{2}, \frac{I_0}{8}, \frac{3I_0}{32} \)
(D) \( I_0, \displaystyle\frac{I_0}{2} \), zero
Choose the correct answer from the options given below :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (C)
A polaroid passes only the component of the electric field along its pass axis. Light from the source is unpolarised, so the first polaroid passes half its intensity whatever its orientation. After that, the light is plane polarised and each later polaroid follows Malus' law, \( I = I'\cos^{2}\theta \), where \( \theta \) is the angle between the pass axes of that polaroid and the one just before it.
After \( P_1 \):
\( \displaystyle I_1 = \frac{I_0}{2} \)
After \( P_2 \), at \( 60^\circ \) to \( P_1 \):
\( \displaystyle I_2 = I_1\cos^{2}60^\circ = \frac{I_0}{2}\times\frac{1}{4} = \frac{I_0}{8} \)
After \( P_3 \): its axis is at \( 90^\circ \) to \( P_1 \), so the angle between \( P_2 \) and \( P_3 \) is \( 90^\circ - 60^\circ = 30^\circ \).
\( \displaystyle I_3 = I_2\cos^{2}30^\circ = \frac{I_0}{8}\times\frac{3}{4} = \frac{3I_0}{32} \)
The three intensities are \( I_0/2,\; I_0/8,\; 3I_0/32 \), which is set (C).