Question

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

• A. 71.6${}^{\circ }$
• B. 90${}^{\circ }$
• C. 18.4${}^{\circ }$
• D. 45${}^{\circ }$

Answer 1

Answer: (C)

Given condition $I=10\%I_0$
$\to I=\frac{10}{100}{I}_0$ $=\frac{I_0}{10}$
We also know form polariser analyses set up that
$I=I_0{\cos }^2\Phi$
Where $\Phi$ is the angle between axes of polariser and analyser.
Refer image.
$\Rightarrow \frac{I_0}{10}=I_0{\cos }^2\Phi \Rightarrow \cos \Phi =\frac{1}{\sqrt{10}}\Rightarrow \Phi =71.6^{\circ }$
Now if we put $\Phi =90^{\circ }$ we get $I=O$
So initially angle being $71.6^{\circ }$ we need to rotate the analyser
through ( 90 $71.6)=18.4^{\circ }$ to get final angle as $90^{\circ }$ and thus $I=0$
$\therefore$ angle to be turned $=18.4^{\circ }$