Question

Unpolarized light of intensity $32W/m^2$ passes through three polarizers, such that the transmission axis of last polarizer is crossed with the first. If intensity of emerging light is $3W/m^2$, what is the angle (in degree) between the transmission axis of the first two polarizers?

Answer 1

Answer: 30

Let ${\theta }_{12}$ be the angle between the transmission axes of the first two polarizers $P_1$ and $P_2$ and ${\theta }_{23}$ be the angle between the transmission axes of polarizers $P_2$ and $P_3$.
Then ${\theta }_{12}+{\theta }_{23}=90^{\circ }$

$I_0=32W/m^2$
Let $I_1,I_2$ and $I_3$ be the intensities of light on passing through the first, second and third polarizer, respectively.
Then, $I_1=\frac{1}{2}{I}_0$
$=\frac{1}{2}\times 32$
$=16W/m^2$
From Malus law,
$I_2=I_1{\cos }^2{\theta }_{12}$
$\Rightarrow I_2=16{\cos }^2{\theta }_{12}$
and $I_3=I_2{\cos }^2{\theta }_{23}$
$=\left(16{\cos }^2{\theta }_{12}\right){\cos }^2\left(90^{\circ }-{\theta }_{12}\right)$ ...[From ()i]
$=16{\cos }^2{\theta }_{12}{\sin }^2{\theta }_{12}$
$=4{\sin }^{2}2{\theta }_{12}$
But $I_3=3W{m}^{-2}$
$\Rightarrow 4{\sin }^{2}2{\theta }_{12}=3$
$\Rightarrow \sin 2{\theta }_{12}=\frac{\sqrt{3}}{2}$
$\Rightarrow 2{\theta }_{12}=60^{\circ }$
$\Rightarrow {\theta }_{12}=30^{\circ }$