Question

An analyser is inclined to a polarizer at an angle of ${{30}^{o}}$. The intensity of light emerging from the analyzer is ${{\left( \dfrac{1}{n} \right)}^{th}}$ of that is incident on the polarizer. Then n is equal to
A. $4$
B. $\dfrac{4}{3}$
C. $\dfrac{8}{3}$
D. $\dfrac{1}{4}$

Answer 1

Hint: We are given a situation when a light is passed through a polarizer to an analyser. To find the unknown term in the intensity of light coming from the analyser we need to find the intensity of the light coming out of the analyser. By applying Malus law in this situation we can find the solution.

Formula used:
Malus law $I'=I{{\cos }^{2}}\theta $

Complete answer:
In the question it is said that there is a system that consists of an analyser and a polarizer.
It is said that the analyser is inclined to the polarizer at an angle ${{30}^{o}}$, i.e. the light that passes through the polarizer falls on the analyser at an angle ${{30}^{o}}$.
It is given in the question that the intensity of the light coming out of the analyser is ${{\left( \dfrac{1}{n} \right)}^{th}}$ the intensity of the light incident on polarizer, i.e.
${{I}_{a}}=\left( \dfrac{1}{n} \right){{I}_{0}}$, were ‘${{I}_{a}}$’ is the intensity of light from the analyser and ‘${{I}_{0}}$’ is the intensity of the incident light on the polarizer.
Since ${{I}_{0}}$ is the intensity of the light falling on the polarizer we know that the intensity of light coming out of the polarizer will be $\dfrac{{{I}_{0}}}{2}$.
Now let us consider Malu’s law. According to Malu’s law we now that,
$I'=I{{\cos }^{2}}\theta $, were $I'$ is the intensity of light coming from the analyser, $I$is the intensity of light coming from the polarizer and $'\theta '$ is the angle between plane of transmission and the incident light.
In this case we can write this equation as,
${{I}_{a}}=\dfrac{{{I}_{0}}}{2}{{\cos }^{2}}\left( 30 \right)$, here value of $\theta ={{30}^{o}}$
$\begin{align}
  & \Rightarrow {{I}_{a}}=\dfrac{{{I}_{0}}}{2}\times \dfrac{3}{4} \\
 & \Rightarrow {{I}_{a}}={{I}_{0}}\times \dfrac{3}{8} \\
\end{align}$
$\Rightarrow {{I}_{a}}=\left( \dfrac{{{I}_{0}}}{\left( \dfrac{8}{3} \right)} \right)$
In the question it is given that,
${{I}_{a}}=\left( \dfrac{1}{n} \right){{I}_{0}}$
Comparing these two equations we get $n=\dfrac{8}{3}$

Hence the correct answer is option C.

Note:
According to Malu’s law the intensity of a plane-polarized light that is passing through an analyzer will vary with the square of the cosine of the angle between the plane of the polarizer and the transmission axes of the analyzer.
While solving the Malus law in the given question, don’t confuse the intensity of the light coming from the polarizer and from the analyser. If you get confused and interchange the intensities the whole answer will go wrong.