Question

Unpolarized light of intensity $32{\text{W/m}}^2$ passes through three polarizers such that the transmission axis of the last polarizer is crossed with the first. If the intensity of emerging light is $3{\text{W/m}}^2$, the angle $\theta$ between the transmission axes of the first two polarizers and the angle $\varphi$ at which transmitted intensity will be maximum are

• A. $\theta ={30}^{\circ },\varphi ={60}^{\circ }$
• B. $\theta ={45}^{\circ },\varphi ={45}^{\circ }$
• C. $\theta ={30}^{\circ },\varphi ={45}^{\circ }$
  
• D. $\theta ={45}^{\circ },\varphi ={30}^{\circ }$

Answer 1

The correct option is C $\theta ={30}^{\circ },\varphi ={45}^{\circ }$


Given:
Intensity of unpolarized light$I_0=32{\text{W/m}}^2$


The intensity of light after passing through $P_3$,

$I=\frac{I_0}{2}{\cos }^2\theta {\sin }^2\theta$

But it is given that $I=3{\text{W/m}}^2$

$\Rightarrow 3=\frac{32}{2}{\cos }^2\theta {\sin }^2\theta$

$\Rightarrow 3=4\times 4{\cos }^2\theta {\sin }^2\theta$

$\Rightarrow \frac{3}{4}=(2\sin \theta \cos \theta {)}^2$

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

$\Rightarrow \sin 2\theta =\frac{\sqrt{3}}{2}=\sin {60}^{\circ }$

$\Rightarrow 2\theta ={60}^{\circ }$

$\therefore \theta ={30}^{\circ }$

Also, $I$ will be maximum when $\frac{dI}{d\theta }=0$

$\Rightarrow \frac{d}{d\theta }$$\left\{\frac{I_0\times 4}{2\times 4}{\cos }^2\theta {\sin }^2\theta \right\}=0$

$\Rightarrow \frac{I_0}{8}\frac{d}{d\theta }\left\{{\sin }^{2}2\theta \right\}=0$

$\therefore 2\sin 2\theta \cos 2\theta \times 2=0$

$\Rightarrow \sin 4\theta =0$

$\Rightarrow 4\theta =0^{\circ },{180}^{\circ }...$

$\therefore {\theta }_{max}={45}^{\circ }$

Angle at which intensity is maximum$\varphi ={\theta }_{max}={45}^{\circ }$

Hence, option $\left(\text{C}\right)$ is correct.