Question

Unpolarized light of intensity $32{\text{Wm}}^{-2}$ passes through three polarizers arranged in a way, such that the transmission axis of the last polarizer is crossed with that of the first. If the intensity of emerging light is $3{\text{Wm}}^{-2}$, what is the angle between the axis of first two polarizers?

• A. ${45}^{\circ }$
• B. ${60}^{\circ }$
• C. ${30}^{\circ }$
• D. ${90}^{\circ }$

Answer 1

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


Now, angle between $P_1$ and $P_3$ is ${90}^{\circ }$

Let,
${\theta }_1\to \text{angle between}{P}_1\&P_2$
And, ${\theta }_2\to \text{angle between}{P}_2\&P_3$
$\Rightarrow {\theta }_1+{\theta }_2={90}^{\circ }$

After passing through $1^{st}$ polarizer, the intensity becomes half, i.e. $I_1=\frac{I_0}{2}$

Now, intensity coming out of the $2^{nd}$ polarizer is,
$I_2=I_1{\cos }^2{\theta }_1$

Intensity coming out of the $3^{rd}$ polarizer is,
$I_3=I_2{\cos }^2{\theta }_1$

Here, $I_3=3$

$I_3=I_1{\cos }^2{\theta }_1{\cos }^2{\theta }_2$

$\Rightarrow 3=16{\cos }^2{\theta }_1{\cos }^2(90-{\theta }_1)$

$\Rightarrow 3=16{\cos }^2{\theta }_1{\sin }^2{\theta }_1$

$\Rightarrow 3=4\times 4{\cos }^2\theta ,{\sin }^2{\theta }_1$

$\Rightarrow 3=4(\sin 2{\theta }_1{)}^2$

$\Rightarrow {\sin }^{2}2{\theta }_1=\frac{3}{4}$

$\Rightarrow \sin 2{\theta }_1=\frac{\sqrt{3}}{2}$

$\Rightarrow 2{\theta }_1=60$

$\Rightarrow {\theta }_1={30}^{\circ }$

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