Question

Two polarisers $ P_1 $ and $ P_2 $ are placed in such a way that the intensity of the transmitted light will be zero. A third polariser $ P_3 $ is inserted in between $ P_1 $ and $ P_2 $, at the particular angle between $ P_1 $ and $ P_2 $. The transmitted intensity of the light passing the through all three polarisers is maximum. The angle between the polarisers $ P_2 $ and $ P_3 $ is:

• A. \( \frac{\pi}{4} \)
• B. \( \frac{\pi}{6} \)
• C. \( \frac{\pi}{8} \)
• D. \( \frac{\pi}{3} \)

Hint:

To maximize the transmitted intensity through multiple polarizers, the angles between the polarizers should be chosen to align with the conditions of maximum intensity based on the formula for light transmission.

Answer 1

The Correct Option is A

Through polariser \( P_2 \), the intensity \( I_1 \) of the transmitted light is given by: \[ I_1 = I_0 \cos^2 \theta \] where \( \theta \) is the angle between the light incident on \( P_2 \) and the polariser axis. Next, through \( P_3 \), the intensity \( I_{\text{net}} \) becomes: \[ I_{\text{net}} = I_0 \cos \theta \sin \theta \] To maximize the transmitted intensity, we set the angle \( \theta \) such that the product \( \sin(2\theta) \) is maximized. This occurs when: \[ \sin(2\theta) = 1 \quad \text{for} \quad \theta = 45^\circ \]
Thus, the angle between \( P_2 \) and \( P_3 \) is \( \frac{\pi}{4} \).