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Question: Unpolarised light of intensity I is incident on a system of two polarizers, A followed by B. The int...

Unpolarised light of intensity I is incident on a system of two polarizers, A followed by B. The intensity of emergent light is I/2. If a third polarizer C is placed between A and B, the intensity of emergent light is reduced to I/3. The angle between the polarizers A and C is θ. Then
A. cosθ=(23)14\theta = {\left( {\dfrac{2}{3}} \right)^{\dfrac{1}{4}}}
B. cosθ=(13)14\theta = {\left( {\dfrac{1}{3}} \right)^{\dfrac{1}{4}}}
C. cosθ=(13)12\theta = {\left( {\dfrac{1}{3}} \right)^{\dfrac{1}{2}}}
D. cosθ=(23)12\theta = {\left( {\dfrac{2}{3}} \right)^{\dfrac{1}{2}}}

Explanation

Solution

According to Malus, when completely plane polarized light is incident on the analyser, the intensity I of the light transmitted by the analyser is directly proportional to the square of the cosine of angle between the transmission axes of the analyser and the polarizer I= I0cos2θ{I_0}{\cos ^2}\theta .
Where I0, is the intensity of light incident on the system.

Complete step by step answer:
Given that the initial intensity of light is I.
We know that intensity of a beam after passing through a polarizer is given as according to Malus law,
I =I0cos2θ{I_0}{\cos ^2}\theta .
On passing through the first polarizer A, the intensity is given as I1I_1=I/2
The intensity of light after passing through polarizer B is I2I_2= I1I_1 cos2θ^2\theta
Given in the question that the intensity of light after passing through polarizer B is I2I_2=I/2.
I2=I2cos2θ\dfrac{I}{2} = \dfrac{I}{2}{\cos ^2}\theta
cos2θ=1 cosθ=1 θ=cos11 θ=0  \Rightarrow {\cos ^2}\theta = 1 \\\ \Rightarrow \cos \theta = 1 \\\ \Rightarrow \theta = {\cos ^{ - 1}}1 \\\ \therefore \theta = 0^\circ \\\
Therefore the angle between A and B is zero.
Now polarizer C is introduced between A and B.
Intensity of light after passing through C is given by IC=I1I_1 cos2θ^2\theta
Given the question that intensity of light after passing through C is I3\dfrac{I}{3}.
Substituting the value of IC and I1I_1 we get
I3=I2cos2θ     23=cos2θ     cosθ=(23)12  \dfrac{I}{3} = \dfrac{I}{2}{\cos ^2}\theta \\\ \implies \dfrac{2}{3} = {\cos ^2}\theta \\\ \implies \cos \theta = {\left( {\dfrac{2}{3}} \right)^{\dfrac{1}{2}}} \\\

So, the correct answer is “Option A”.

Additional Information:
Polarized light waves are light waves in which the vibrations occur in a single plane. The process of transforming unpolarised light into polarized light is known as polarization. The law stating that the intensity of a beam of plane-polarized light after passing through a rotatable polarizer varies as the square of the cosine of the angle through which the polarizer is rotated from the position that gives maximum intensity is called Malus law.

Note:
Some of the important laws should be learnt by the student.
In this question Malus law of polarisation of light is used.
The process of transforming unpolarised light into polarized light is known as polarization.