 # A polarizer - analyser set is adjusted such that the intensityof light coming `
Question:

A polarizer - analyser set is adjusted such that the intensityof 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:

1. $71.6^{\circ}$

2. $18.4^{\circ}$

3. $90^{\circ}$

4. $45^{\circ}$

Correct Option: , 2

Solution:

(2) According to question, the intensity of light coming out of the analyser is just $10 \%$ of the original intensity

$\left(I_{0}\right)$

Using, $I=I_{0} \cos ^{2} \theta$

$\Rightarrow \frac{I_{0}}{10}=I_{0} \cos ^{2} \theta \Rightarrow \frac{1}{10}=\cos ^{2} \theta$

$\Rightarrow \cos \theta=\frac{1}{\sqrt{10}}=0.316 \Rightarrow \theta \approx 71.6^{\circ}$

Therefore, the angle by which the analyser need to be rotated further to reduced the output intensity to be zero

$\phi=90^{\circ}-\theta=90^{\circ}-71.6^{\circ}=18.4^{\circ}$