larizer:

$$ I_1 = \frac{I_0}{2} = \frac{64}{2} = 32 \ \text{Wm}^{-2} $$

Let the angle between the transmission axes of the first and second polarizers be $\theta$.

Using Malus’ law, intensity after the second polarizer is

$$ I_2 = I_1 \cos^2 \theta $$

$$ I_2 = 32 \cos^2 \theta $$

The last polarizer is crossed with the first polarizer.

So the angle between the first and third polarizers is $90^\circ$.

Hence, the angle between the second and third polarizers is

$$ 90^\circ - \theta $$

Again using Malus’ law, intensity after the third polarizer is

$$ I_3 = I_2 \cos^2(90^\circ - \theta) $$

Since

$$ \cos(90^\circ - \theta) = \sin \theta $$

we get

$$ I_3 = 32 \cos^2 \theta \sin^2 \theta $$

Using

$$ \sin 2\theta = 2\sin\theta \cos\theta $$

$$ \sin^2\theta \cos^2\theta = \frac{1}{4}\sin^2 2\theta $$

Thus,

$$ I_3 = 32 \times \frac{1}{4}\sin^2 2\theta $$

$$ I_3 = 8\sin^2 2\theta $$

Given emerging intensity is $6 \ \text{Wm}^{-2}$,

$$ 6 = 8\sin^2 2\theta $$

$$ \sin^2 2\theta = \frac{3}{4} $$

$$ \sin 2\theta = \frac{\sqrt{3}}{2} $$

So,

$$ 2\theta = \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) $$

$$ \theta = \frac{1}{2}\sin^{-1}\left(\frac{\sqrt{3}}{2}\right) $$

Hence, the correct option is

$$ \boxed{\text{Option A}} $$

Wave Optics

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