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 :
Solution
I = I<sub>0</sub> cos<sup>2</sup> $\theta$
<br><br>$\Rightarrow$ ${{{I_0}} \over {10}}$ = I<sub>0</sub> cos<sup>2</sup> $\theta$
<br><br>$\Rightarrow$ cos $\theta$ = ${1 \over {\sqrt {10} }}$
<br><br>$\Rightarrow$ $\theta$ = 71.6<sup>o</sup>
<br><br>$\therefore$ $\phi$ = 90 - 71.6 = 18.4<sup>o</sup>
About this question
Subject: Physics · Chapter: Optics · Topic: Reflection and Mirrors
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