Question

A polarizer and an analyser are oriented so that the maximum amount of lights is transmitted. Diffraction of its maximum value is the intensity of the transmitted light reduced when the analyser is rotated through: (intensity of incident light =\[{I_0}\]​)
A) $30^\circ $
B) $45^\circ $
C) $60^\circ $

Answer 1

Hint: There is no difference between polariser and analyser; they are given different names due to their different roles in optics. These are filters which allow certain directions of light waves to pass through the filters. The polariser allows only that component of light which is perpendicular to the plane of specimen whereas the analyser allows the component which is parallel to the plane of specimen.

Formula Used:
When any polarizer is rotated, the transmitted intensity is always reduced. This can be mathematically represented by the equation given below:
$I = \dfrac{{{I_0}}}{{2{{\cos }^2}\theta }}$
In the above mathematical expression, $\theta $is known as the angle of rotation.

Complete step by step answer:
It has been given that the intensity of the incident light is given as ${I_0}$.
When this intensity of incident light is passed through a polarizer, the intensity becomes halved. Thus the intensity of incident light ${I_0}$becomes$\dfrac{{{I_0}}}{2}$.
Now in the first case that is (A) the angle of rotation is given equal to $30^\circ $. Thus, when we put this in the mathematical formula given above, we get:
$I = \dfrac{{{I_0}}}{2}{\cos ^2}30^\circ $
Putting the value of $\cos 30^\circ $in the above expression and solving further, we get:
$I = 0.375{I_0}$
This is the correct answer for the orientation (A) is equal to $I = 0.375{I_0}$.

In the second case that is (B) the angle of orientation is given equal to $45^\circ $. Thus, when we put this in the mathematical formula given above, we get:
$I = \dfrac{{{I_0}}}{2}{\cos ^2}45^\circ $
Using the value of $\cos 45^\circ $in the above expression and solving further, we get:
$I = 0.25{I_0}$

Similarly in the case (C) we find that the angle of orientation is ${60^ \circ }$. Now, putting this in the mathematical formula, we get:
$I = \dfrac{{{I_0}}}{2}{\cos ^2}60^\circ $
Solving this, we get:
$I = 0.125{I_0}$

Note: A polarizer and an analyser are two necessary mechanisms for a polarized light microscope. The main dissimilarity between polarizer and analyser is that polarizer produces plane polarized light whereas analyser is used to check whether the light has been polarized or not.