Unpolarized light of intensity 32W m ^ - 2 passes through three...

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

${I}_{1} = \dfrac{1}{2}{I}_{0} = 16W{m}^{- 2}$

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

${I}_{3} = {I}_{2}\cos^{2}\left(90 ° - \theta\right) = 16\cos^{2}\theta\sin^{2}\theta$

$⇒ {I}_{3} = 16\cos^{2}\theta\sin^{2}\theta = 3$

$⇒ 4\cos^{2}\theta\sin^{2}\theta = \dfrac{3}{4}$

$⇒ \sin^{2}\left(2\theta\right) = \dfrac{3}{4}$

$\theta = 30 °$