Question

An unpolarised light is passed through 3 polarizers. If the second polariser is at an angle ${30^\circ }$ with the first and the third polariser is at an angle ${60^\circ }$ with the second. Find the final intensity of the light passed through this combination if the initial intensity was $I$.

Answer 1

Hint: Light waves that vibrate in a single plane are known as polarised waves. Plane polarisation light is made up of waves in the same pattern of vibration for both of them. A plane polarised light vibrates on just one plane, as seen in the picture above.

Formula used:
$I = {I_0}{\cos ^2}\theta $
Here $I$ = Intensity of Polarised light, ${I_0}$=Intensity of Unpolarised light and $\theta $= Angle of deviation.

Complete step by step answer:
A polarizer, also known as a polariser, is an optical filter that allows light waves of one polarisation to pass through while blocking light waves of another. It will convert an unknown or mixed polarised light beam into polarised light by filtering it. According to Malus' theorem, the amplitude of plane-polarized light passing through an analyzer varies as the square of the cosine of the angle between the polarizer's plane and the analyzer's propagation axes.

The law allows one to check the nature of polarised light quantitatively. Let's look at how Malus' rule is expressed.
-Regardless of how the polarising axis is focused, when unpolarized light is incident on an optimal polarizer, the intensity of the reflected light is precisely half that of the incident unpolarized light.
-An optimal polarising filter transmits 100% of incident unpolarized light that is polarised in the filter's (Polarizer) Polarizing axis.

Now according to the question, after passing through first polarizer ${\mathbf{I}} = \dfrac{{{{\mathbf{I}}_0}}}{{\mathbf{2}}}$
Now after second polariser ${\text{I}} = \dfrac{{{{\text{I}}_0}}}{2}{\cos ^2}\left( {{\mathbf{3}}{{\mathbf{0}}^\circ }} \right) = \dfrac{3}{8}{{\mathbf{I}}_0}$
Now after third polariser ${\mathbf{I}} = \dfrac{{{\mathbf{3}}{{\mathbf{I}}_0}}}{8}{\cos ^2}\left( {{\mathbf{6}}{{\mathbf{0}}^\circ }} \right) = \dfrac{{{\mathbf{3}}{{\mathbf{I}}_0}}}{{{\mathbf{32}}}}$

Hence final intensity = $\dfrac{{3{{\text{I}}_0}}}{{32}}$.

Note: In physics, radiant energy intensity is defined as the amount of energy transmitted per unit area measured on a plane perpendicular to the energy's propagation path. Luminous intensity is the amount of visible light released per unit solid angle in unit time. The lumen is the unit for the amount of light streaming from a source in one second (luminous power, or luminous flux). The lumen is assessed using visual sensation as a criterion.