Question

Two polaroids ${P_1}$and ${P_2}$ are placed with their axis perpendicular to each other. Unpolarized light ${I_0}$is incident on ${P_1}$. A third polaroid ${P_3}$is kept in between ${P_1}$and ${P_2}$ such that its axis makes an angle ${45^ \circ }$with that of ${P_1}$. The intensity of transmitted light through ${P_2}$ is:
(A) $\dfrac{{{I_0}}}{{16}}$
(B) $\dfrac{{{I_0}}}{2}$
(C) $\dfrac{{{I_0}}}{4}$
(D) $\dfrac{{{I_0}}}{8}$

Answer 1

The first polaroid behaves as a polarizer whereas the two other polaroids behave as analyzer. Light coming out of the polaroid ${P_1}$ is incident on the second polaroid and the intensity of the light coming out from it can be determined by using Malus’ law.
Formula used:
$I = {I_0}{\cos ^2}\theta$ where ${I_0}$is the light incident on the analyzer and $\theta$is the angle which the analyzer makes with the polarizer.

Complete step by step answer
Malus’ law is used to obtain a relation which shows how the intensity of light transmitted by the analyzer varies with the angle that its plane of transmission makes with that of the polarizer.
As unpolarized light is incident on a polaroid, it gets resolved into two components- the cos component which behaves as an E-ray and the sin component which behaves as O-ray. The sin component undergoes total internal reflection and is absorbed while the cos component transverses through the polaroid. Since only one-half of the unpolarized light transverses through the polaroid, its intensity is also halved.
So the intensity of the light coming out of polaroid ${P_1}$ is $\dfrac{{{I_0}}}{2}$
By Malus’ law, we know,
$I = {I_0}{\cos ^2}\theta$ where ${I_0}$is the light incident on the analyzer and $\theta$is the angle which the analyzer makes with the polarizer.
Now, the polarizer ${P_3}$ makes an angle of ${45^ \circ }$with respect to ${P_1}$.
${I_1} = \dfrac{{{I_0}}}{2}{\cos ^2}{45^ \circ } = \dfrac{{{I_0}}}{4}$
Again the polaroid ${P_3}$ makes an angle of $\left( {\dfrac{\pi }{2} - 45} \right)$ $= {45^ \circ }$ with ${P_2}$
By Malus’ law, the intensity of light coming out of ${P_2}$ is,
$I = {I_1}{\cos ^2}{45^ \circ } = \dfrac{{{I_0}}}{4} \times \dfrac{1}{2} = \dfrac{{{I_0}}}{8}$
Therefore the intensity of transmitted light through ${P_2}$ is $\dfrac{{{I_0}}}{8}$.

So, the correct option is D.

Note: Similar to polaroids there are certain crystals in which when light is passed through, gets resolved into two components. One of the components is absorbed more strongly than the other such that the emergent beam is linearly polarized. This phenomenon is known as dichroism and such crystals are called dichroic crystals.