Question

Unpolarized light falls first on polarizer $\left(P\right)$ and then on analyzer $\left(A\right)$. If the intensity of the transmitted light from the analyser is $\frac{1}{8}$th of the incident unpolarized light. What will be the angle between optic axes of $P$ and $A$?

• A. ${45}^o$
• B. ${30}^o$
• C. Zero
• D. ${60}^o$

Answer 1

Answer: (D)

${60}^o$

Given,
$I=\frac{I_0}{2}$ ....(i)
$I^{{}^{\prime }}=I{\cos }^2\theta$ $(\because I^{{}^{\prime }}=\frac{I_0}{8})$
$\therefore \frac{I_0}{8}=\frac{I_0}{2}{\cos }^2\theta$
From the equation (i), we have
$\frac{1}{4}={\cos }^2\theta \Rightarrow \cos \theta =1/2$
$\Rightarrow \cos \theta =\cos {60}^o$
$\Rightarrow \theta ={60}^o$