Question

A beam of unpolarized 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$^\circ$ relative to that of A. The intensity of the emergent light is:
${\text{A}}{\text{. }}\dfrac{{{I_0}}}{2}{\text{ }} \\\ {\text{B}}{\text{. }}\dfrac{{{I_0}}}{4} \\\ {\text{C}}{\text{. }}\dfrac{{{I_0}}}{8} \\\ {\text{D}}{\text{. }}{I_0} \\\$

Answer 1

The intensity of light reduces to half of its initial value after passing through a polaroid placed perpendicularly in its path. The second polaroid will reduce the intensity of light given by the square of the cosine of the angle at which it is oriented to the first polaroid.
Formula used:
The Brewster’s law for polaroids is given as
$I = {I_0}{\cos ^2}\theta$

Complete answer:
We are given that a beam of unpolarized 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$^\circ$ relative to that of A.
First we need to consider the intensity of light after passing through the first polaroid. The intensity reduces to half after passing through a polaroid placed perpendicular to the direction of propagation of the wave. Therefore, it is given as
$I = \dfrac{{{I_0}}}{2}$
Now for the second polaroid, we need to make use of the Brewster’s law formula for intensity of light. We are given the angle between the transmission axis of the first polaroid and the second polaroid. The value is
$\theta = 45^\circ$
The light with reduced intensity obtained from the first polaroid is now incident on the second. Therefore, the intensity of the emergent light is given as
$I' = I{\cos ^2}\theta$
Inserting the known values, we get
$I' = \dfrac{{{I_0}}}{2}{\cos ^2}45^\circ = \dfrac{{{I_0}}}{2} \times {\left( {\dfrac{1}{{\sqrt 2 }}} \right)^2} = \dfrac{{{I_0}}}{4}$
This is the final intensity of light.

Therefore, the correct answer is option B.

Note:
It should be noted that in polarization, some components of the electric field vector of the light waves get removed after passing through a polaroid. Since the intensity of light is directly related to the magnitude of the electric field vector of the wave, the intensity of light reduces after passing through a polaroid.