Question

Unpolarized light of intensity $32{\text{W }}{{\text{m}}^{ - 2}}$ passes through three polarizers such that the transmission axis of the last polarizer is crossed with the first. If the intensity of the emerging light is $2{\text{W }}{{\text{m}}^{ - 2}}$, what is the angle between the transmission axes of the first two polarizers?
$\left( a \right){\text{ 3}}{{\text{0}}^ \circ }$
$\left( b \right){\text{ 4}}{{\text{5}}^ \circ }$
$\left( c \right){\text{ 22}}{\text{.}}{{\text{5}}^ \circ }$
$\left( d \right){\text{ 6}}{{\text{0}}^ \circ }$

Answer 1

Hint So first of all it is given that the transmission axis of the last polarizer is crossed so the angle made between them will be${90^ \circ }$. And also we know that the initial intensity of polarized light will represent the final intensity. Therefore by using this equation, we will get the angle between them.
Formula used:
The intensity of polarized light,
${I_1} = \dfrac{1}{2}{I_0}$
Here,
${I_1}$, will be the intensity of first polarized light.
${I_0}$, will be the initial intensity of polarized light.
The intensity $I$ after being passed from the polarizer will be given by
Malus’ law,
$I = {I_0}{\cos ^2}\theta$

Complete Step By Step Solution First of all let us assume that$\theta$, will be the angle between the transmission axes ${P_1}{\text{ and }}{{\text{P}}_2}$
And similarly$\phi$, will be the angle between the transmission axes ${P_2}{\text{ and }}{{\text{P}}_3}$
Since it is given in the question that the transmission axis of the last polarizer is crossed so the angle made between them will be${90^ \circ }$.
Therefore, we can write it as
$\Rightarrow \theta + \phi = {90^ \circ }$
And from here
$\Rightarrow \phi = {90^ \circ } - \theta$, named it an equation$1$.
Also, we have from the question,
${I_0} = 32W{\text{ }}{{\text{m}}^{ - 2}}$
And from the formula, we have
${I_1} = \dfrac{1}{2}{I_0}$
Now on substituting the values, we get
$\Rightarrow {I_1} = \dfrac{1}{2} \times 32W{\text{ }}{{\text{m}}^{ - 2}}$
On solving it, we get
$\Rightarrow {I_1} = 16W{\text{ }}{{\text{m}}^{ - 2}}$
Therefore, by using the malus’ law
We get
$\Rightarrow {I_2} = {I_1}{\cos ^2}\theta$, and
$\Rightarrow {I_3} = {I_2}{\cos ^2}\phi$
Therefore by using the above equation, we can write it as
$\Rightarrow {I_3} = {I_1}{\cos ^2}\theta {\cos ^2}\phi$
And also it can be written as
$\Rightarrow {I_3} = {I_1}{\cos ^2}\theta {\cos ^2}\left( {{{90}^ \circ } - \theta } \right)$
Hence by putting the values,
$\Rightarrow {I_3} = 16{\cos ^2}\theta {\sin ^2}\theta$
Also, the value of${I_3}$is given to us
Therefore,
$\Rightarrow 2 = 4{\left( {\sin 2\theta } \right)^2}$
And from here, on solving
$\Rightarrow \sin 2\theta = \dfrac{1}{{\sqrt 2 }}$
And it will be equal to the
$\Rightarrow \sin 2\theta = \sin {45^ \circ }$
And the value of the angle will be,
$\Rightarrow \theta = {22.5^ \circ }$

Therefore, the option $\left( c \right)$ is correct.

Note The light wave doesn’t have to be in that exact orientation to come through the lens. That one orientation is where it passes through the lens. As the wave rotates farther away from that specific orientation, light still passes through but diminishes in intensity, more so at greater angles until you reach the normal. However, once through, their orientations become turned to the favored orientation of the lens.