Physics - Wave Optics Question with Solution | TestHub

Let the intensity of unpolarized light be $I_0$.

Intensity after passing through the first polaroid is $I_1^{\prime }=\frac{I_0}{2}$.

Let $\theta$ be the angle between the transmission axes of the first and second polaroids. The intensity after the second polaroid is $I_2^{\prime }=\frac{I_0}{2}{\cos }^2\theta$.

Since the third polaroid is crossed with the first, the angle between the second and third polaroids is $90^{\circ }-\theta$. The intensity after the third polaroid is $I_1=I_2^{\prime }{\cos }^2(90^{\circ }-\theta )=\frac{I_0}{2}{\cos }^2\theta {\sin }^2\theta$.

Using the identity $\sin 2\theta =2\sin \theta \cos \theta$, we get $I_1=\frac{I_0}{8}{\sin }^{2}2\theta$.

Given $I_1=\frac{3}{32}{I}_0$, we have:

$\frac{I_0}{8}{\sin }^{2}2\theta =\frac{3}{32}{I}_0$

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

$\sin 2\theta =\frac{\sqrt{3}}{2}$

$\theta =\frac{1}{2}{\sin }^{-1}\frac{\sqrt{3}}{2}$

For maximum transmitted intensity, ${\sin }^{2}2\theta$ must be maximum, which is $1$.

${\sin }^{2}2\theta =1\Rightarrow 2\theta =90^{\circ }\Rightarrow \theta =45^{\circ }$.

Answer: $\frac{1}{2}{\sin }^{-1}\frac{\sqrt{3}}{2}$ and $45^{\circ }$