Question

A beam of un polarized light of intensity ${I_0}$ is passed through a polaroid $A$ and then through another Polaroid $B$ which is oriented so that its principal plane makes an angle of ${45^0}$ relative to that of $A$. The intensity of the emergent light is
A. ${I_0}$
B. $\dfrac{{{I_0}}}{2}$
C. $\dfrac{{{I_0}}}{4}$
D. $\dfrac{{{I_0}}}{8}$

Answer 1

In order to get the solution for this numerical, we should know the concept of malus law which says that the intensity of light will reduces to half of its initial value after passing through the polaroid placed perpendicular to it. By applying this law we can calculate the intensity of the emergent light.

Complete step by step answer:
From the given data:
$\theta = {45^0}$
When an un polarized light passes through a polarizer its intensity becomes half of the initial intensity and then the intensity of this polarized light becomes ${I_0}{\cos ^2}\theta$ times of the total intensity. Where $\theta$ is the angle of the polarizer.
So here the initial intensity of the light is I. So the intensity of this polarized light becomes $\dfrac{{{I_0}}}{2}$ after passing through the polarizer A.
The polarizer B has an angle${45^0}$.
So, the final intensity will be
${I^1} = I{\cos ^2}\theta$
Substituting the values in it
$\ {I^1} = \dfrac{{{I_0}}}{2}{\cos ^2}{45^0} \\\ \Rightarrow{I^1} = \dfrac{{{I_0}}}{2}{\left( {\dfrac{1}{{\sqrt 2 }}} \right)^2} \\\ \therefore{I^1} = \dfrac{{{I_0}}}{4}$
Hence this is the final intensity of light.

Therefore the correct option is C.

Additional information:
Malus law helps us to verify the nature of polarized light. The expression for Malus’ law explains that when Unpolarized light is incident on polarizer then the intensity of the emitted light is equal to half of the incident light(unpolarized) doesn’t matter how the polarizing axis is oriented.An ideal polarizer filter produces the 100% of incident unpolarized light that is polarized in the direction of the filter's Polarizing axis.

Note: It is important that some components of the electric field vectors produced from the light waves are removed after passing through the Polaroid. Because the magnitude of electric field vectors are directly related to the intensity of light and the intensity of light reduces after passing through the Polaroid.