Question 184 of 191

Unpolarized light of intensity $I$ passes through an ideal polariser $A$. Another identical polariser $B$ is placed behind $A$. The intensity of light beyond B is found to be $\dfrac{I}{2}$. Now another identical polariser $C$ is placed between $A$ and $B$. The intensity beyond B is now found to be $\dfrac{I}{8}$ . The angle between polariser $A$ and $C$ is

A$60^{\circ}$B$0^{\circ}$C$30^{\circ}$D$45^{\circ}$

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Answer: (D) $45^{\circ}$

Transformation axis $A$ and $B$ are parallel. If the transformation axis of $C$ is making angle $\theta$ with $A$, it will make angle $\left(90-\theta \right)$ with $B$. Therefore,

$\Rightarrow \dfrac{I}{2}{\cos }^{2}{\theta cos}^{2}(90-\theta )=\dfrac{I}{8}$$\Rightarrow \dfrac{1}{4}{\sin }^{2}2\theta =\dfrac{1}{4}$
$\Rightarrow 2\theta =\dfrac{\pi }{2}$$\Rightarrow \theta =\dfrac{\pi }{4}$