Question

Unpolarized light of intensity $32\,{\text{W}} \cdot {{\text{m}}^{ - 2}}$ passes through three polarizers such that the transmission axis of the first is crossed with third. If intensity of emerging light is $2\,{\text{W}} \cdot {{\text{m}}^{ - 2}}$, what is the angle of transmission axis between first two polarizers?
A. $30^\circ$
B. $45^\circ$
C. $22.5^\circ$
D. $60^\circ$

Answer 1

Use the expression for Malus’ law. This expression gives the relation between the intensity of the incident light on polarizer, intensity of light emerging from polarizer and angle between the transmission axis of polarizer and direction of polarization of light. Using this law, determine the derivation for intensity of the light from the third polarizer and solve it for the angle of transmission axis between the first two polarizers.

Formula used:
The expression for Malus’ law is
$I = {I_0}{\cos ^2}\theta$ …… (1)
Here, $I$ is the intensity of the light emerging from the polarizer, ${I_0}$ is the intensity of the incident light and $\theta$ is the angle between the transmission axis of polarizer and direction of polarization of light.

Complete step by step answer:
We have given that the intensity of the unpolarized light is $32\,{\text{W}} \cdot {{\text{m}}^{ - 2}}$ and the intensity of the light emerged light from third polarizer is $2\,{\text{W}} \cdot {{\text{m}}^{ - 2}}$.
${I_0} = 32\,{\text{W}} \cdot {{\text{m}}^{ - 2}}$
${I_3} = 2\,{\text{W}} \cdot {{\text{m}}^{ - 2}}$
Let $\theta$ be the angle between the transmission axes of the first and second polarizer and $\theta '$ be the angle between the transmission axes of the second and third polarizer.
Since the transmission axes of the first and third polarizer are crossed. The sum of the angles $\theta$ and $\theta '$ must be $90^\circ$.
$\theta + \theta ' = 90^\circ$
$\Rightarrow \theta ' = 90^\circ - \theta$
The intensity ${I_1}$ of the light emergent from the first polarizer is always half of the incident light.
$\Rightarrow {I_1} = \dfrac{{{I_0}}}{2}$
Apply Malus’ law to determine the intensity ${I_2}$ of the light emerging from the second polarizer.
${I_2} = {I_1}{\cos ^2}\theta$
Substitute $\dfrac{{{I_0}}}{2}$ for ${I_1}$ in the above equation.
$\Rightarrow {I_2} = \dfrac{{{I_0}}}{2}{\cos ^2}\theta$

Apply Malus’ law to determine the intensity ${I_3}$ of the light emerging from the third polarizer.
${I_3} = {I_2}{\cos ^2}\theta '$
Substitute $\dfrac{{{I_0}}}{2}{\cos ^2}\theta$ for ${I_2}$ and $90^\circ - \theta$ for $\theta '$ in the above equation.
${I_3} = \dfrac{{{I_0}}}{2}{\cos ^2}\theta {\cos ^2}\left( {90^\circ - \theta } \right)$
$\Rightarrow {I_3} = \dfrac{{{I_0}}}{2}{\cos ^2}\theta \left[ {1 - {{\sin }^2}\left( {90^\circ - \theta } \right)} \right]$
$\Rightarrow {I_3} = \dfrac{{{I_0}}}{2}{\cos ^2}\theta \left[ {1 - {{\sin }^2}90^\circ + {{\sin }^2}\theta } \right]$
$\Rightarrow {I_3} = \dfrac{{{I_0}}}{2}{\cos ^2}\theta \left[ {1 - 1 + {{\sin }^2}\theta } \right]$
$\Rightarrow {I_3} = \dfrac{{{I_0}}}{2}{\cos ^2}\theta {\sin ^2}\theta$
Substitute $32\,{\text{W}} \cdot {{\text{m}}^{ - 2}}$ for ${I_0}$ and $2\,{\text{W}} \cdot {{\text{m}}^{ - 2}}$ for ${I_3}$ in the above equation and solve it for $\theta$.
$\Rightarrow 2\,{\text{W}} \cdot {{\text{m}}^{ - 2}} = \dfrac{{32\,{\text{W}} \cdot {{\text{m}}^{ - 2}}}}{2}{\cos ^2}\theta {\sin ^2}\theta$
$\Rightarrow 1 = 8{\cos ^2}\theta {\sin ^2}\theta$
$\Rightarrow 1 = 2{\sin ^2}2\theta$
$\Rightarrow {\sin ^2}2\theta = \dfrac{1}{2}$
$\Rightarrow \sin 2\theta = \dfrac{1}{{\sqrt 2 }}$
$\Rightarrow 2\theta = {\sin ^{ - 1}}\left( {\dfrac{1}{{\sqrt 2 }}} \right)$
$\Rightarrow 2\theta = 45^\circ$
$\therefore \theta = 22.5^\circ$

Therefore, the angle between the transmission axis of the first two polarizers is $22.5^\circ$.Hence, the correct option is C.

Note: The students may get confused about how the intensity of light from the first polarizer is half of its initial value. For the unpolarized light, the direction of polarization for half of the light is along the transmission axis of the first polarizer and for the remaining half light the angle is perpendicular to the transmission axis of the first polarizer. Hence, the resultant intensity is half of the incident intensity of light.