Question

Two polaroids are present in the path of unpolarized beam of intensity $I_0$ such that no light is emitted from the second polaroid. If a third polaroid, whose transmission axis makes an angle $\theta$ with the transmission axis of the first polaroid, is placed between these polaroids. Then the intensity of light emerging from the last polaroid will be,

• A. $\left(\frac{I_0}{8}\right){\sin }^{2}2\theta$
• B. $\left(\frac{I_0}{4}\right){\sin }^{2}2\theta$
• C. $\left(\frac{I_0}{4}\right){\cos }^4\theta$
• D. $\left(\frac{I_0}{2}\right){\cos }^4\theta$

Answer 1

The correct option is A $\left(\frac{I_0}{8}\right){\sin }^{2}2\theta$



When unpolarized light of intensity $\left(I_0\right)$, passes through the polarizer, the intensity of transmitted light becomes $I_{\circ }/2$, irrespective of the orientation of the polarizer
So, $I_1=\frac{I_0}{2}$

According to Malus' law
$I_2=I_1{\cos }^2\theta =\frac{I_{\circ }}{2}{\cos }^2\theta I_3=I_2{\cos }^2(90-\theta )=\frac{I_{\circ }}{2}{\cos }^2\theta {\sin }^2\theta$
$=\frac{I_{\circ }}{8}(2\sin \theta \cos \theta {)}^2=\frac{I_{\circ }}{8}(\sin 2\theta {)}^2$

Hence, $\left(A\right)$ is the correct answer.

$\begin{array}{c}\text{Why this question?}\\ \text{To understand the application of Malus law}\end{array}$