Question

A system of three polarizers $P_1,P_2,P_3$ is set up such that the pass axis of $P_3$ is crossed with respect to that of $P_1$. The pass axis of $P_2$ is inclined at ${60}^{\circ }$ to the pass axis of $P_3.$ When a beam of unpolarized light of intensity $I_0$ is incident on $P_1$, the intensity of light emerging from $P_3$ is $I$. The ratio $\left(I_0/I\right)$ equals (nearly),

• A. $5.33$
• B. $16.00$
• C. $10.67$
• D. $1.80$

Answer 1

The correct option is C $10.67$



When unpolarized light passes through polaroid $P_1$, intensity of light coming out of $P_1$ is, $I_1=\frac{I_0}{2}$
where, $I_0=$ intensity of incident light.

Now, this transmitted light is polarized and it passes through the polarizer $P_2.$
So, intensity $I_2$ transmitted is obtained by Malu's law.
$\Rightarrow I_2=I_1{\cos }^2\theta$

As angle of pass axis of $P_1\text{and}{P}_3\text{is}{90}^{\circ }$ and angle of pass axis of $P_2\text{and}{P}_3\text{is}{60}^{\circ },$ so angle between pass axis of $P_1\text{and}{P}_2\text{is}({90}^{\circ }-{60}^{\circ })={30}^{\circ }$

So, $I_2=\frac{I_0}{2}{\cos }^2{30}^{\circ }=\frac{I_0}{2}\times {\left(\frac{\sqrt{3}}{2}\right)}^2=\frac{3}{8}{I}_0$

When this light pass through third polarizer $P_3$, the intensity $I$ transmitted through $P_3$ is again obtained by using Malu's law.
So, $I=I_2{\cos }^2{60}^{\circ }=\left(\frac{3}{8}{I}_0\right){\cos }^2{60}^{\circ }$
$=\frac{3}{8}{I}_0\times {\left(\frac{1}{2}\right)}^2=\frac{3}{32}{I}_0$

So, ratio $=\frac{I_0}{I}=\frac{32}{3}\approx 10.67$

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