Physics - Wave Optics Question with Solution | TestHub

Since unpolarized light is passing through the first polarizer, hence the intensity of light after crossing the first polarizer will be

$I_1=\frac{1}{2}{I}_0=16 \mathrm{W} {\mathrm{m}}^{-2}$

Let us assume that the angle between the transmission axis of the first and second polarizer is $\theta$, then from Malus law we can find out the intensity of light after it crosses the second polarizer.

$I_2=I_1{\cos }^2\theta =16{\cos }^2\theta$

Similarly, the intensity of light after crossing the third polarizer is

$I_3=I_2{\cos }^2\left(90°-\theta \right)=16{\cos }^2\theta {\sin }^2\theta$

$\Rightarrow I_3=16{\cos }^2\theta {\sin }^2\theta =3$

$\Rightarrow 4{\cos }^2\theta {\sin }^2\theta =\frac{3}{4}$

$\Rightarrow {\sin }^2\left(2\theta \right)=\frac{3}{4}$

$\theta =30°$