Question

Three polaroid sheets are co-axially placed as indicated in the diagram. Pass axes of the polaroids 2 and 3 make 30^o and 90^o with pass axis of polaroid sheet 1. If I₀ is the intensity of the incident unpolarised light entering sheet 1, the intensity of the emergent light through sheet 3 is

• A. zero
• B. \(\frac{3I_0}{32}\)
• C. \(\frac{3I_0}{8}\)
• D. \(\frac{3I_0}{16}\)

Answer 1

The Correct Option is B

The intensity of unpolarized light entering the first polaroid sheet is \( I_0 \). The first sheet polarizes the light, and the intensity of light passing through it will be half of the incident intensity:

\[ I_1 = \frac{I_0}{2} \]

The second sheet has its pass axis at 30° to the first sheet. The intensity passing through the second sheet is given by Malus' Law:

\[ I_2 = I_1 \cos^2(30^\circ) = \frac{I_0}{2} \cos^2(30^\circ) \]

The third sheet is at a 90° angle to the second sheet. Using Malus' Law again:

\[ I_3 = I_2 \cos^2(60^\circ) = \left( \frac{I_0}{2} \cos^2(30^\circ) \right) \cos^2(60^\circ) \]

Calculating the values: 
\( \cos(30^\circ) = \frac{\sqrt{3}}{2} \) and \( \cos(60^\circ) = \frac{1}{2} \).

\[ I_3 = \left( \frac{I_0}{2} \times \left( \frac{\sqrt{3}}{2} \right)^2 \right) \times \left( \frac{1}{2} \right)^2 \]

\[ I_3 = \frac{I_0}{2} \times \frac{3}{4} \times \frac{1}{4} = \frac{3 I_0}{32} \]

Therefore, the intensity of the emergent light through sheet 3 is \( \frac{3I_0}{32} \).

Answer: (B) \( \frac{3I_0}{32} \) 

Answer 2

In this problem, the three polaroid sheets are aligned such that their axes of polarization are at different angles relative to each other. The intensity of light passing through a polaroid is governed by Malus' law, which is given by: \[ I = I_0 \cos^2 \theta \] where \(I_0\) is the initial intensity of light incident on the polaroid, and \(\theta\) is the angle between the light's polarization direction and the polaroid's pass axis.
First Polaroid (Sheet 1): The light entering the first polaroid is unpolarized. When unpolarized light passes through a polaroid, the intensity of transmitted light is reduced by half: \[ I_1 = \frac{I_0}{2} \]
Second Polaroid (Sheet 2): The pass axis of polaroid 2 makes a 30° angle with the pass axis of polaroid 1. Therefore, the intensity of the light after passing through polaroid 2 can be calculated using Malus' law: \[ I_2 = I_1 \cos^2 30^\circ = \frac{I_0}{2} \times \left( \frac{\sqrt{3}}{2} \right)^2 = \frac{I_0}{2} \times \frac{3}{4} = \frac{3I_0}{8} \] Third Polaroid (Sheet 3): The pass axis of polaroid 3 makes a 90° angle with the pass axis of polaroid 1. Therefore, the intensity of the light after passing through polaroid 3 is: \[ I_3 = I_2 \cos^2 60^\circ = \frac{3I_0}{8} \times \left( \frac{1}{2} \right)^2 = \frac{3I_0}{8} \times \frac{1}{4} = \frac{3I_0}{32} \] Thus, the intensity of the emergent light through sheet 3 is \(\frac{3I_0}{32}\). Therefore, the correct answer is (B).