Question

Find an expression for intensity of transmitted light when a polaroid sheet is rotated between two crossed polaroids. In which position of the polaroid sheet will the transmitted intensity be maximum?

Answer 1

Let $I_0$ be the intensity of unpolarised light , incident on first polaroid , intensity of light transmitted from first polaroid will be ,

$I_1=I_0{\cos }^2\theta$ ,

or $I_1=I_0\times 1/2=I_0/2$ ........................eq1 ,

let $\theta$ be the angle between the transmission axes of the first and second polaroid (which is placed between two crossed polaroids) and $\varphi$ be the angle between second and hird polaroid ,

then $\theta +\varphi =90$ (as first and third polaroids are perpendicular to each other) ,

or $\varphi =90-\theta$ ,

now intensity of light , transmitted from second polaroid ,

$I_2=I_1{\cos }^2\theta =\left(I_0/2\right){\cos }^2\theta$ ,

andintensity of light , transmitted from third polaroid ,

$I_3=I_2{\cos }^2\varphi =\left(I_0/2\right){\cos }^2\theta \times {\cos }^2\varphi$ ,

$I_3=\left(I_0/2\right){\cos }^2\theta \times {\cos }^2(90-\theta )$ ,

or $I_3=\left(I_0/2\right){\cos }^2\theta {\sin }^2\theta$ ,

or $I_3=\left(I_0/2\right){\cos }^2\theta {\sin }^2\theta$ ,

or $I_3=\left(I_0/2\right)\left({\sin }^{2}2\theta /4\right)$

this is the required expression .

Now $I_3$ will be maximum when $\sin 2\theta$ is maximum i.e. $\sin 2\theta =1=\sin 90$ ,

or $2\theta =90$ ,

or $\theta ={45}^o$