Question

A polarizer-analyzer set is adjusted such that the intensity of light coming out of the analyzer is just $10\%$ of the original intensity. Assuming that the polarizer-analyzer set does not absorb any light, the angle by which the analyser need to be rotated further to reduce the output intensity to be zero is

• A. ${45}^o$
• B. ${71.6}^o$
• C. ${90}^o$
• D. ${18.4}^0$

Answer 1

The correct option is D ${18.4}^0$

$$\begin{array}{cc}& \text{Intensity after polarisation through polaroid}=I_{o}co{s}^2\varphi \\ & \text{So,}0.1I_o=I_{o}co{s}^2\varphi \\ & \Rightarrow cos\varphi =\sqrt{0.1}\\ & \Rightarrow cos\varphi =0.316\end{array}$$
Since, $\cos \varphi <\cos {45}^o$ therefore, $\varphi >{45}^o$
If the light is passing at ${90}^o$ from the plane of polaroid, then its intensity will be zero.
Then, $\theta ={90}^o-\varphi$ therefore, $\theta$ will be less than ${45}^o$. So, the only option matching is option d which is ${18.4}^o$