Question

Unpolarized light of intensity $32W{m}^{-2}$ passes through three polarisers such that the transmission axis of the last polariser is crossed with the first, if the intensity of the emerging light is $3W{m}^{-2}$, what is the angle between the transmission axes of the first two polarisers?

• A. ${30}^{\circ }$
• B. ${37}^{\circ }$
• C. ${45}^{\circ }$
• D. ${60}^{\circ }$

Answer 1

The correct option is A ${30}^{\circ }$

From data, we know that the first and the third polarizers are crossed. i.e., $P_1$ and $P_3$ are crossed.
If angle between transmission axes of $P_1$ and $P_2$ is $\theta$, then that between $P_2$ and $P_3$ is ${90}^{\circ }-\theta .$
If $I_0$ is the intensity of unpolarized light falling on $P_1,$ then intensity of light emerging from $P_1$ will be $I_1=\frac{I_0}{2}$
By Malus law, the intensity of light emerging from $P_2$ is given by,
$I_2=I_1{\cos }^2\theta =\frac{I_0}{2}{\cos }^2\theta$
By Malus law, the intensity of light emerging from $P_3$ is given by,
$I_3=I_2{\cos }^2[{90}^{\circ }-\theta ]=\frac{I_0}{2}{\cos }^2\theta {\sin }^2\theta =\frac{I_0{\sin }^2\left(2\theta \right)}{8}\Rightarrow {\sin }^2\left(2\theta \right)=\frac{8I_3}{I_0}=\frac{8\times 3}{32}=\frac{3}{4}\Rightarrow \sin \left(2\theta \right)=\frac{\sqrt{3}}{2}\Rightarrow 2\theta ={60}^{\circ }\Rightarrow \theta ={30}^{\circ }$