Question

The intensity of transmitted light when a polaroid sheet, placed between two crossed polaroids at \(22.5^\circ\) from the polarization axis of one of the polaroids (\(I_0\) is the intensity of polarised light after passing through the first polaroid):

• A. \( \frac{I_0}{4} \)
• B. \( \frac{I_0}{8} \)
• C. \( \frac{I_0}{16} \)
• D. \( \frac{I_0}{2} \)

Hint:

Remember Malus's Law \(I = I_0 \cos^2 \theta\) and the half-angle formulas for cosine to solve polarization problems involving intermediate polaroids.

Answer 1

The Correct Option is B

Step 1: Intensity after the first polaroid
Let the intensity of light after the first polaroid be I₀. The second polaroid is at an angle θ₁ = 22.5° with the first. The intensity after the second polaroid (I₁) is given by Malus's Law:
I₁ = I₀ cos²(θ₁) = I₀ cos²(22.5°)
Using the half-angle formula, cos²(22.5°) = (1 + cos(45°)) / 2 = (1 + √2 / 2) / 2 = (2 + √2) / 4.
So, I₁ = I₀ (2 + √2) / 4.

Step 2: Intensity after the third polaroid
The third polaroid is crossed with the first, so it is at an angle of 90° with the first. The angle between the second and the third polaroid is θ₂ = 90° - 22.5° = 67.5°. The intensity after the third polaroid (I₂) is:
I₂ = I₁ cos²(θ₂) = I₁ cos²(67.5°)
Using the half-angle formula, cos²(67.5°) = (1 + cos(135°)) / 2 = (1 - √2 / 2) / 2 = (2 - √2) / 4.
Substituting I₁:
I₂ = (I₀ (2 + √2) / 4) × (2 - √2) / 4 = I₀ (2 + √2)(2 - √2) / 16 = I₀ (4 - 2) / 16 = I₀ / 8.