Unpolarised light of intensity 32 Wm⁻² passes through the combination of three polaroids such that the pass axis of the last polaroid is perpendicular to that of the pass axis of first polaroid. If intensity of emerging light is 3 Wm⁻², then the angle between pass axis of first two polaroids is ______________ °.

Solution

Physics sir · 4 steps

Halve it at the first polaroid, then use Malus's law twice. The two middle angles must add up to 90°.

1. First polaroid. Unpolarised light loses half its intensity at the first polaroid, whatever its orientation. So I₁ = 32/2 = 16 W m⁻².
2. Second and third. Let the angle between the first two pass axes be θ. The third axis is at 90° to the first, so it makes 90° - θ with the second. Malus's law gives I₂ = 16 cos² θ and I₃ = 16 cos² θ cos²(90° - θ) = 16 cos² θ sin² θ.

   I₃ = 16 sin² θ cos² θ = 4 sin² 2θ = 3

3. Solve for the angle. sin² 2θ = 3/4, so sin 2θ = √3/2 and 2θ = 60°. That gives θ = 30°. The other root, 2θ = 120°, gives 60°. That is the same arrangement with the two middle angles swapped, so the expected answer is the smaller one.
4. So. The angle between the pass axes of the first two polaroids is 30°.

The idea to keep. For crossed outer polaroids with one in between, I = I₀/2 cos² θ sin² θ = I₀/8 sin² 2θ. Remember the half at the first sheet for unpolarised light.

Answer from the NTA final answer key. Solution by PrepWiser.