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Unpolarised light passes through a polarised ${P_1}$. When this polarised beam passes through another polarised ${P_2}$ and if the pass axis of ${P_2}$ makes angle $\theta $ with the pass axis of ${P_1}$, then write the expression for the polarised beam passing through ${P_2}$. Draw a plot showing the variation of intensity when $\theta $ varies from $0$ to $2\pi $.

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Last updated date: 19th Jun 2024
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Answer
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Hint: Write unpolarised light definition, through this it could be found and analysed. For polarisation of light by polaroids using melus law is the key to solve these types of doubts.

Formula used:
The polarised beam when passing through the ${P_2}$ is
$ \Rightarrow $ $I = {I_0}{\cos ^2}\theta $
Where,
I is the intensity of polarised light
The angle between ${P_1}_{}$ and ${P_2}$ polarizer
It is also known as malus’s law.

Complete step by step solution:
When the beam of completely plane polarised light is incident on an analysis resultant intensity of light transmitted from analysed varies directly as the square if also become plane of analyse and polarize as square of $\cos \theta $ between the plane of analyse and polarize
$ \Rightarrow $ $I\alpha {\cos ^2}\theta $
$ \Rightarrow $ $I = {I_0}{\cos ^2}\theta $
When the unpolarised light passes through a polarised ${P_1}$ and it again this polarised light will pass through the polariser ${P_2}$. When the pass axis ${P_2}$ makes an angle $\theta $ with the pass axis of ${P_1}$ it will give an expression for polarised beam.
When this unpolarised light passes through ${P_1}$
The average value of ${\cos ^2}\theta $ (where $\theta$ lies between $0$ to $2\pi $) = $\dfrac{1}{2}$
Hence, $I = {I_0}{\cos ^2}\theta $
$I = {I_0} \times \dfrac{1}{2}$
$I = \dfrac{{{I_0}}}{2}$
Here the intensity reduces by $\dfrac{1}{2}$
When this unpolarised light passes through polarises ${P_1}$ the intensity drops to half is $\dfrac{{{I_0}}}{2}$ and when this polarised light passed through polarises ${P_2}$ the intensity of emitted light is,
Now when the polarised light from ${P_1}$, passes through ${P_2}$
$ \Rightarrow $ $I = {I_0}{\cos ^2}\theta $
$ \Rightarrow $ $I = \dfrac{{{I_0}}}{2}{\cos ^2}\theta $
$I = \dfrac{{{I_0}}}{2}{\cos ^2}\theta $
$\theta $ is the angle between ${P_1}$ and ${P_2}$

Note: In electromagnetic waves the filament is the superposition of wave trains and each has its own polarisation direction. Natural light is partial because the multiple scattering and the reflection. In polarised light circumstances of spatial orientation is defined.