Question

A polaroid sheet ‘P’ is placed on another similar polaroid sheet ‘Q’ such that the angle between their axes is \(45^\circ\). - The ratio of the intensities of the light emerged from polaroid ‘Q’ and the unpolarised light incident on polaroid ‘P’ is

• A. \(1:4\)
• B. \(1:2\)
• C. \(1:\sqrt{3}\)
• D. \(1:\sqrt{2}\)

Hint:

Remember that when unpolarized light passes through a polaroid, its intensity is reduced by half. The second polaroid further reduces intensity based on Malus’s Law: \[ I = I_0 \cos^2 \theta \] where \(\theta\) is the angle between the two polaroid axes.

Answer 1

The Correct Option is A

Step 1: Understanding Malus’s Law When unpolarized light passes through a polaroid, its intensity is reduced to half of the initial intensity (\(I_0\)): \[ I_1 = \frac{I_0}{2} \] When the transmitted light passes through a second polaroid at an angle \(\theta\) to the first, the intensity of the emerging light follows Malus’s Law: \[ I_2 = I_1 \cos^2 \theta \] Step 2: Substituting given values Given that \(\theta = 45^\circ\), we substitute into Malus’s Law: \[ I_2 = \frac{I_0}{2} \cos^2 45^\circ \] Since \( \cos 45^\circ = \frac{1}{\sqrt{2}} \), we get: \[ I_2 = \frac{I_0}{2} \times \left(\frac{1}{\sqrt{2}}\right)^2 \] \[ I_2 = \frac{I_0}{4} \] Thus, the ratio of the intensities is: \[ \frac{I_2}{I_0} = \frac{1}{4} \] which corresponds to the answer \( 1:4 \).