Question

An analyzer is inclined to a polarizer at an angle of 30°. The intensity of light emerging from the analyzer is 1nth of that incident on the polarizer. Then $n$ is equal to
(A) $4$
(B) $\dfrac {4}{3}$
(C) $\dfrac {8}{3}$
(D) $\dfrac {1}{4}$

Answer 1

Hint
According to the law of Malus, when a beam of completely plane-polarized light is incident on an analyzer, the resultant intensity of light ($I$) transmitted from the analyzer varies directly as the square of the cosine of the angle ($θ$) between the plane of transmission of analyzer and polarizer, i.e. ${\rm{I}} \propto {\cos ^2}{\rm{\theta }}$. The law will provide the relation between the intensities and results in the value of n.

Complete step by step solution
Given, the angle of inclination between an analyzer and a polarizer is-
${\rm{\theta }} = 30^\circ $
Let the intensity of incident light be ${{\rm{I}}_0}$.
The intensity of light passing through the analyzer is given by,
${\rm{I}} = \dfrac{{{{\rm{I}}_0}}}{{\rm{n}}}$ … (1)
When the light incident on the polarizer, the intensity of light will become half of its original value. According to Malus law,
${\rm{I'}} = {\rm{I}}{\cos ^2}{\rm{\theta }}$
$\dfrac{{{{\rm{I}}_0}}}{{\rm{n}}} = \dfrac{{{{\rm{I}}_0}}}{2}{\cos ^2}30^\circ $
$\dfrac{1}{{\rm{n}}} = \dfrac{1}{2}{\left( {\dfrac{{\sqrt 3 }}{2}} \right)^2}$
$\dfrac{1}{{\rm{n}}} = \dfrac{3}{8}$
or, ${\rm{n}} = \dfrac{8}{3}$
Therefore, (C) $\dfrac{8}{3}$ is the required solution.

Note
Polaroids are the sheets consisting of a long chain of molecules aligned in a particular direction. When an unpolarized light incident on these polaroids, then the electric vectors of the wave along the direction of the aligned molecules get absorbed and only polarised will emerge from it. The point should be remembered that the intensity of light will become half when the incident on the polarizer.