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

Step 1: Understanding the Concept:
When unpolarised light passes through the first polariser, its intensity is reduced by exactly half, and it becomes linearly polarised. When this light passes through a second polariser (analyser), its intensity is governed by Malus's Law.
Step 2: Key Formula or Approach:
Intensity after first polariser: \(I_1 = \frac{I_0}{2}\)
Malus's Law for the analyser: \(I_2 = I_1 \cos^2\theta\)
Step 3: Detailed Explanation:
The final intensity \(I_2\) is given as \(\frac{3I_0}{8}\). Substitute \(I_1\) into Malus's Law: \[ I_2 = \left(\frac{I_0}{2}\right) \cos^2\theta = \frac{3I_0}{8} \] Divide both sides by \(I_0\): \[ \frac{1}{2} \cos^2\theta = \frac{3}{8} \] Multiply by 2 to isolate \(\cos^2\theta\): \[ \cos^2\theta = \frac{6}{8} = \frac{3}{4} \] Take the square root: \[ \cos\theta = \frac{\sqrt{3}}{2} \] The angle whose cosine is \(\frac{\sqrt{3}}{2}\) is \(30^\circ\). \[ \theta = 30^\circ \]
Step 4: Final Answer:
The angle between the polariser and analyser is \(30^\circ\).