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

Let's analyze the situation with the polarisers and determine the condition for maximizing the transmitted light intensity when a third polariser \( P_3 \) is inserted between \( P_1 \) and \( P_2 \).

1. Initially, when two polarisers \( P_1 \) and \( P_2 \) are placed such that the intensity of the transmitted light is zero, they must be oriented at an angle of \( 90^\circ \) (or \( \frac{\pi}{2} \) radians) to each other. This is because orthogonal polarisers block all light.
2. Now, we insert a third polariser \( P_3 \) between \( P_1 \) and \( P_2 \). To find the angle that maximizes the transmitted light intensity through \( P_1 \), \( P_3 \), and \( P_2 \), we apply Malus's Law, which states that the intensity \( I \) of light passing through a polariser at an angle \( \theta \) is given by: \(I = I_0 \cos^2 \theta\), where \( I_0 \) is the initial intensity of light.
3. The intensity after the first polariser, \( P_1 \), is \( I_0/2 \).
4. If \( \theta_1 \) is the angle between \( P_1 \) and \( P_3 \) and \( \theta_2 \) is the angle between \( P_3 \) and \( P_2 \), the intensity \( I \) after passing \( P_3 \) is: 
   \(I = \left( \frac{I_0}{2} \right) \cos^2 \theta_1 \cos^2 \theta_2\).
5. To maximize the final intensity, which is a function of both \( \theta_1 \) and \( \theta_2 \), one effective choice is when \( \theta_1 = \theta_2 \). Given \( \theta_1 + \theta_2 = \frac{\pi}{2} \): 
   \(\theta_1 = \theta_2 = \frac{\pi}{4}\).

This configuration effectively maximizes the light intensity passing through all three polarisers. Thus, the angle between the polarisers \( P_2 \) and \( P_3 \) is \( \frac{\pi}{4} \).

Answer 2

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 \( \frac{\pi}{4} \).