Question

A polarizer-analyzer set is adjusted such that the intensity of light coming out of the analyzer is just $10\%$ of the original intensity. Assuming that the polarizer-analyzer set does not absorb any light, the angle by which the analyzer need to be rotated further to reduce the output intensity to be zero is

• A. $45°$
• B. $90°$
• C. $71.6°$
• D. $18.4°$

Answer 1

The correct option is D

$18.4°$

Step 1. Given data

$I=10\%I_0$

$\begin{array}{rcl}I& =& \frac{10}{100}{I}_0\\ & =& \frac{I_0}{10}\end{array}$

Step 2. Finding the angle by which analyzer need to be rotated, $\theta$

Intensity after polarization through Polaroid, $I=I_0{\cos }^2\phi$

Where $\phi$ is the angle between axes of the polarizer and analyzer

$I_0$ is original intensity

$I$ is output intensity

$\Rightarrow$ $\frac{I_0}{10}=I_0{\cos }^2\phi$

$\Rightarrow$$\cos \phi =\frac{1}{\sqrt{10}}$

$\Rightarrow$$\cos \phi =71.6°$

So initially angle being $71.6°$ we need to rotate the analyzer through $(90-71.6)=18.4°$ to get final angle as $90°$ and thus $I=0$

$\therefore$angle to be turned $\mathbf{=}\mathbf{18}\mathbf{.}\mathbf{4}\mathbf{°}$

Hence, option $\left(D\right)$ is correct.