Question

An unpolarized light of intensity $I_0$ passes through three successive Polaroids $P_1,P_2$ and $P_3$. The polarization axes of $P_1$ make an angle ${60}^{\circ }$ with $P_2$ and $P_2$ makes an angle ${30}^{\circ }$ with $P_3$. The intensity of the transmitted light from $P_3$ is

• A. $\frac{3I_0}{32}$
• B. $\frac{5I_0}{32}$
• C. $\frac{I_0}{32}$
• D. $\frac{3I_0}{16}$

Answer 1

The correct option is A $\frac{3I_0}{32}$

Let $I_1,I_2\text{and}{I}_3$ be the intensities of transmitted light from $P_1,P_2\text{and}{P}_3$ respectively.

As unpolarized light passes through $P_1$,

$I_1=\frac{I_0}{2}$

According to Malus law,

$I_2=I_1{\cos }^2\theta$

$\Rightarrow I_2=\frac{I_0}{2}\left({\cos }^2{60}^{\circ }\right)$

$\Rightarrow I_2=\frac{I_0}{2}{\left(\frac{1}{2}\right)}^2=\frac{I_0}{8}$

Applying Malus law for $P_3$

$I_3=I_2{\cos }^2\theta$

$\Rightarrow I_3=\frac{I_0}{8}{\cos }^2{30}^{\circ }$

$\Rightarrow I_3=\frac{I_0}{8}{\left(\frac{\sqrt{3}}{2}\right)}^2=\frac{3I_0}{32}$

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

Why this Question?: Application :- The question involves the change in intensity (decreases) when ordinary light passes through different filters