Question

An unpolarised light of intensity I is passed through two polaroids kept one after the other with their planes parallel to each other. The intensity of light emerging from second polaroid is \(\frac{I}{4}\). The angle between the pass axes of the polaroids is

• A. 0°
• B. 60°
• C. 30°
• D. 45°

Answer 1

The Correct Option is D

Given: 
Initial intensity of unpolarised light = \( I \)
Intensity after second polaroid = \( \frac{I}{4} \)

Step-by-Step Explanation:

Step 1: After passing through the first polaroid

The intensity of unpolarized light after passing through the first polaroid becomes exactly half:

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

Step 2: After passing through the second polaroid (Malus' Law)

According to Malus' law, the intensity \( I_2 \) after passing through the second polaroid is given by:

\[ I_2 = I_1 \cos^2\theta \]

where \( \theta \) is the angle between the pass axes of the two polaroids.

Substitute given values clearly:

\[ \frac{I}{4} = \frac{I}{2}\cos^2\theta \]

Simplify to clearly find the angle \( \theta \):

\[ \cos^2\theta = \frac{1}{2} \]

Thus, clearly:

\[ \cos\theta = \frac{1}{\sqrt{2}} \quad\Rightarrow\quad \theta = 45^\circ \]

Final Conclusion:
The angle between the pass axes of the polaroids is clearly 45°.

Answer 2

When unpolarized light passes through the first polaroid, the intensity of the light is reduced by half. The intensity after passing through the first polaroid is: \[ I_1 = \frac{I}{2} \] When the light passes through the second polaroid, the intensity is further reduced according to Malus's law, which states: \[ I_2 = I_1 \cos^2(\theta) \] where \( \theta \) is the angle between the pass axes of the polaroids. Given that the intensity of the emerging light is \( \frac{I}{4} \), we have: \[ \frac{I}{4} = \frac{I}{2} \cos^2(\theta) \] Simplifying: \[ \cos^2(\theta) = \frac{1}{2} \] \[ \cos(\theta) = \frac{1}{\sqrt{2}} \] \[ \theta = 45^\circ \] Thus, the angle between the pass axes of the polaroids is \( 45^\circ \).