Question

Unpolarised light with intensity \(I_0\) is incident on a polariser. Find angle between axis of polariser and analyser such that intensity of emergent light becomes \( \frac{3I_0}{8} \).

• A. \(60^\circ\)
• B. \(30^\circ\)
• C. \(90^\circ\)
• D. \(0^\circ\)

Answer 1

The Correct Option is B

Concept: When unpolarised light passes through a polariser, its intensity becomes half. \[ I' = \frac{I_0}{2} \] When this polarised light passes through an analyser, intensity follows Malus' law: \[ I = I' \cos^2 \theta \] Step 1: Apply Malus' law.} Given final intensity \[ I = \frac{3I_0}{8} \] Substitute \(I' = \frac{I_0}{2}\) \[ \frac{3I_0}{8} = \frac{I_0}{2}\cos^2\theta \]
Step 2: Solve for \(\theta\).} Cancel \(I_0\): \[ \frac{3}{8} = \frac{1}{2}\cos^2\theta \] \[ \cos^2\theta = \frac{3}{4} \] \[ \cos\theta = \frac{\sqrt{3}}{2} \] \[ \theta = 30^\circ \] Final Result \[ \theta = 30^\circ \]