An unpolarized light of intensity $I_0$ is passed through a polariser and is incident on an analyser making an angle of ($π/6$) rad with that of the polariser. The intensity of the light transmitted from the analyzer is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $0.375\, I_0$
Given:
Initial unpolarized light intensity: $I_0$
Angle between polariser and analyzer: $\theta = \pi/6$
After the first polariser: $I_1 = \frac{I_0}{2}$
Using Malus's law for analyzer: $I = I_1 \cos^2 \theta = \frac{I_0}{2} \cos^2 (\pi/6)$
$\cos(\pi/6) = \frac{\sqrt{3}}{2} \Rightarrow \cos^2(\pi/6) = \frac{3}{4}$
$I = \frac{I_0}{2} \cdot \frac{3}{4} = \frac{3 I_0}{8}$
∴ Intensity of transmitted light = 3 I₀ / 8