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Two polaroids are placed in the path of unpolarised light beam of intensity $I_0$ such that no light is emitted from the second polaroid. If a third polaroid whose polarisation axis makes an angle $\theta$ with that of the first polaroid is placed between the polaroids, then intensity of light emerging from the last polaroid is
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Verified Answer
The correct answer is:
$\left(\frac{I_0}{8}\right) \sin ^2 2 \theta$
When unpolarised light passes through first polariser, it becomes polarised and its intensity becomes half.
$\therefore$ After first polariser, intensity $I_1=I_0 / 2$
After second polariser, intensity $I_2=\frac{I_0}{2} \cos ^2 \theta$
(Malus law)
After third polariser, intensity
$$
I_3=\frac{I_0}{2} \cos ^2 \theta \cos ^2\left(90^{\circ}-\theta\right)
$$
(because this is at $90^{\circ}$ angle from first polariser)
$$
\Rightarrow I_3=\frac{I_0}{2} \cos ^2 \theta \sin ^2 \theta=\frac{I_0}{8}(2 \sin \theta \cos \theta)^2=\frac{I_0}{8} \sin ^2 2 \theta
$$
$\therefore$ After first polariser, intensity $I_1=I_0 / 2$
After second polariser, intensity $I_2=\frac{I_0}{2} \cos ^2 \theta$
(Malus law)
After third polariser, intensity
$$
I_3=\frac{I_0}{2} \cos ^2 \theta \cos ^2\left(90^{\circ}-\theta\right)
$$
(because this is at $90^{\circ}$ angle from first polariser)
$$
\Rightarrow I_3=\frac{I_0}{2} \cos ^2 \theta \sin ^2 \theta=\frac{I_0}{8}(2 \sin \theta \cos \theta)^2=\frac{I_0}{8} \sin ^2 2 \theta
$$
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