Physics - Wave Optics Question with Solution | TestHub

Let the intensity of the unpolarized light be $I_0=64{\text{ Wm}}^{-2}$.

When unpolarized light passes through the first polarizer, its intensity becomes $I_1=\frac{I_0}{2}=32{\text{ Wm}}^{-2}$.

Let the angle between the transmission axis of the first and second polarizers be $\theta$. The intensity of light after the second polarizer is $I_2=I_1{\cos }^2\theta =32{\cos }^2\theta$.

Since the third polarizer is crossed with the first, the angle between the transmission axis of the first and third polarizers is $90^{\circ }$. Therefore, the angle between the transmission axis of the second and third polarizers is $90^{\circ }-\theta$.

The intensity of light after the third polarizer is $I_3=I_2{\cos }^2(90^{\circ }-\theta )=I_2{\sin }^2\theta$.

Substituting the value of $I_2$, we get $I_3=32{\cos }^2\theta {\sin }^2\theta$.

Using the identity $\sin 2\theta =2\sin \theta \cos \theta$, we can write $I_3=8(4{\sin }^2\theta {\cos }^2\theta )=8{\sin }^{2}2\theta$.

Given that $I_3=6{\text{ Wm}}^{-2}$, we have $8{\sin }^{2}2\theta =6$.

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

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

$\theta =\frac{1}{2}{\sin }^{-1}\left(\frac{\sqrt{3}}{2}\right)$

Answer: $\frac{1}{2}{\sin }^{-1}\left(\frac{\sqrt{3}}{2}\right)$