Question

Unpolarized light of intensity $32{\text{W m}}^{-2}$ passes through three polarizers such that transmission axis of first is crossed with third. If the intensity of emerging light is $2{\text{W m}}^{-2}$, what is the angle between the transmission axes of the first two polarizers?

• A. ${30}^{\circ }$
• B. ${60}^{\circ }$
• C. ${22.5}^{\circ }$
• D. ${45}^{\circ }$

Answer 1

The correct option is C ${22.5}^{\circ }$

The transmission axis of the first polarizer is crossed with the third.

Let $\theta$ be the angle between the axis of the first two polarizers then the angle between $2^{nd}$ and $3^{rd}$ polarizers will be $({90}^{\circ }-\theta )$.

Let $I_0$ is the intensity of the unpolarized light incident on the first polarizer.
The intensity of light emerging from first polarizer is, $I_1=\frac{I_0}{2}$

According to Malu's law, The intensity of light emerging from second polarizer is,
$I_2=\frac{I_0}{2}{\cos }^2\theta$

And, The intensity of light emerging from third polarizer is,

$I=\left(\frac{I_0}{2}\right){\cos }^2\theta {\cos }^2({90}^{\circ }-\theta )$

$\therefore 2=\frac{32}{2}{\cos }^2\theta {\cos }^2({90}^{\circ }-\theta )$

$2=16{\cos }^2\theta {\sin }^2\theta$

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

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

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

$\therefore \sin 2\theta =\frac{1}{\sqrt{2}}$

And, $2\theta ={45}^{\circ }\Rightarrow \theta ={22.5}^{\circ }$

Hence, $\left(C\right)$ is the correct answer.