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 $I =10 \% I _0$
$\rightarrow 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 $=18.4^{\circ}$