Question

A polarised light of intensity ${I_0}$ is passed through another polarizer whose pass axis makes an angle of ${60^ \circ }$ with the pass axis of the former. What is the intensity of emergent polarized light from the second polarizer?
(A) $I = {I_0}$
(B) $I = \dfrac{{{I_0}}}{6}$
(C) $I = \dfrac{{{I_0}}}{5}$
(D) $I = \dfrac{{{I_0}}}{4}$

Answer 1

This problem is related to the topic, polarisation of the light. The angle of incidence is given in the question. By using Malus’s law, the intensity of emergent polarised light from the second polarizer can be determined.
Malus’s law,
$I = {I_0}{\cos ^2}\theta$
Where, $I$ is the intensity of the light before passing through polarizer, ${I_0}$ is the intensity of the light after passing through polarizer and $\theta$ is the angle of incidence of the light.

Complete step-by-step solution :Given that,
The angle of light incidence, $\theta = {60^ \circ }$
According to the Malus’s law, when completely plane polarised light is incident on the polarizer, the intensity of the light transmitted by the polarizer is directly proportional to the square of the cosine of the angle between the transmission axes of the polarizer.
$I = {I_0}{\cos ^2}\theta \,...............\left( 1 \right)$
By substituting the angle of incidence of the light in the equation (1), then,
$I = {I_0}{\cos ^2}\left( {{{60}^ \circ }} \right)$
From trigonometry, the power of cosine is also written as,
$I = {I_0} \times {\left( {\cos {{60}^ \circ }} \right)^2}$
The value of the $\cos {60^ \circ } = \dfrac{1}{2}$, this value is taken from trigonometry, and substitute this value in the above equation, then the above equation is written as,
$I = {I_0} \times {\left( {\dfrac{1}{2}} \right)^2}$
By squaring the term inside the bracket, then the above equation is written as,
$I = {I_0} \times \left( {\dfrac{1}{4}} \right)$
On multiplying the above equation, then the above equation is written as,
$I = \dfrac{{{I_0}}}{4}$
Thus, the above equation shows the intensity of the emergent polarized light from the second polarizer.

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Hence, the option (D) is the correct answer.
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Note:- In this solution, we have to give more concentration in the step, where the power of cosine is changed to the whole power of the term. And then by substituting the cosine value, the intensity of the light can be determined.