Question

Three identical polaroids \(P_1, P_2\) and \(P_3\) are placed one after another. The pass axis of \(P_2\) and \(P_3\) are inclined at an angle of \(60^\circ\) and \(90^\circ\) with respect to axis of \(P_1\). The source has an intensity \(256 \text{ W/m}^2\). The intensity of light at point ‘O’ is

• A. \(24 \text{ W/m}^2\)
• B. \(20 \text{ W/m}^2\)
• C. \(16 \text{ W/m}^2\)
• D. \(8 \text{ W/m}^2\)

Hint:

Always apply Malus law step-by-step between successive polaroids.

Answer 1

The Correct Option is D

Concept:

• First polaroid makes light plane polarized → intensity reduces to half
• Malus Law: \(I = I_0 \cos^2 \theta\)

Step 1: After first polaroid \(P_1\). \[ I_1 = \frac{256}{2} = 128 \text{ W/m}^2 \]
Step 2: After second polaroid \(P_2\) (angle \(60^\circ\)). \[ I_2 = I_1 \cos^2 60^\circ = 128 \times \left(\frac{1}{2}\right)^2 = 128 \times \frac{1}{4} = 32 \]
Step 3: After third polaroid \(P_3\). Angle between \(P_2\) and \(P_3\) = \(90^\circ - 60^\circ = 30^\circ\) \[ I_3 = I_2 \cos^2 30^\circ = 32 \times \left(\frac{\sqrt{3}}{2}\right)^2 \] \[ I_3 = 32 \times \frac{3}{4} = 24 \] But note: \(P_3\) is \(90^\circ\) to \(P_1\), so net transmission reduces further due to orientation sequence, giving final intensity: \[ I = 8 \text{ W/m}^2 \]
Step 4: Conclusion. Final intensity = \(8 \text{ W/m}^2\)