Question

Two Polaroids $P_1 $ and $P_2$ are placed with their axis perpendicular to each other. Unpolarised light $I_0$ is incident on $P_1$. A third polaroid $P_3$ is kept in between $P_1$ and $P_2$ such that is axis makes an angle 45$^{\circ}$ with that of $P_1$ .The intensity of transmitted light through $P_2$ is : -

• A. $ \frac{I_0}{4}$
• B. $ \frac{I_0}{8}$
• C. $ \frac{I_0}{16}$
• D. $ \frac{I_0}{2}$

Answer 1

The Correct Option is B

To solve the problem, we need to understand the behavior of light through a system of polaroids. When unpolarized light passes through a polaroid, its intensity is reduced to half. Further, when polarized light passes through another polaroid at an angle, its intensity is described by Malus's Law.

1. The initial unpolarized light intensity is I_0. After passing through the first polaroid P_1, the light becomes polarized, and its intensity reduces to half. Therefore, the intensity after P_1 is: \frac{I_0}{2}.
2. The second step involves the light passing through the third polaroid P_3, which is placed at an angle of 45° with respect to the axis of P_1. According to Malus’s Law, the intensity I after passing through a polaroid at an angle \theta with the incoming light is given by: I = I_0 \cdot \cos^2(\theta). Substituting the given values, \theta = 45^\circ and the intensity from step 1: I = \frac{I_0}{2} \cdot \cos^2(45^\circ) = \frac{I_0}{2} \cdot \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{I_0}{2} \cdot \frac{1}{2} = \frac{I_0}{4}.
3. Next, this light with intensity \frac{I_0}{4} passes through the second polaroid P_2, which is perpendicular to P_1. Again using Malus’s Law with \theta = 45^\circ (as P_3 and P_2 are orthogonal through P_2): I = \frac{I_0}{4} \cdot \cos^2(45^\circ) = \frac{I_0}{4} \cdot \frac{1}{2} = \frac{I_0}{8}.

The intensity of the transmitted light through P_2 is \frac{I_0}{8}. Thus, the correct answer is the option \frac{I_0}{8}.