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 :

Engineering

Physics

Polarisation of Light New

Question

45º

71.6º

18.4º

90º

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I = I₀ cos²θ

$\frac{I_{0}}{10} = I_{0} \cos^{2} θ$

$cosθ = \frac{1}{\sqrt{10}} = 0 .31 < \frac{1}{\sqrt{2}}$ which is 0.77

So θ > 45° and 90 – θ < 45º so only one option is correct

i.e. 18.4º

angle rotated should be = 90° – 71.6° = 18.4°