Question

Unpolarized light of intensity $32W.{{m}^{-2}}$ passes through three polarizers such that the transmission axis of the last polarizer is crossed with that of the first. The intensity of the final emerging light is $3W.{{m}^{-2}}$. The intensity of light transmitted by the first polarizer will be:
A. $32W.{{m}^{-2}}$
B. $16W.{{m}^{-2}}$
C. $8W.{{m}^{-2}}$
D. $4W.{{m}^{-2}}$

Answer 1

Hint: This problem can be solved by making use of the fact that when unpolarized light passes through a polarizer, its intensity becomes half. This comes from Malus' law which gives the relation between the intensity of transmitted light and incident light when passing through a polarizer based upon the alignment of the transmission axis of the polarizer polarizer and the polarization direction of light.

Formula used:
The intensity of emergent light $I$ after passing through a polarizer is given by,
$I={{I}_{0}}{{\cos }^{2}}\theta $ (Malus' law)
where ${{I}_{0}}$ is the intensity of the light before entering the polarizer and $\theta $ is the angle made by the transmission axis of the polarizer with the direction of polarization of the light.

Complete step by step answer:
When unpolarized or polarized light passes through a polarizer kept suitably, its intensity decreases, as the polarizer only allows a certain component of the light to pass through which has the same direction of polarization as its transmission axis.
The relation between the incident intensity and the emergent intensity of the light is given by Mallus’ law.
The intensity of emergent light $I$ after passing through a polarizer is given by,
$I={{I}_{0}}{{\cos }^{2}}\theta $ (Malus' law) --(1)
where ${{I}_{0}}$ is the intensity of the light before entering the polarizer and $\theta $ is the angle made by the transmission axis of the polarizer with the direction of polarization of the light.
Now, let us analyze the question.
Let the initial intensity of the unpolarized light be ${{I}_{0}}=32W.{{m}^{-2}}$.

Now, since the light is unpolarized, it can be said that it has no direction of polarization. There is a component of the light in all directions. Hence, it can be said that half of the light has some component along the transmission axis $\left( \theta =0,\cos \theta =1.{{\cos }^{2}}\theta =1 \right)$ while the other half has no component along the transmission axis (or in essence, it only has component perpendicular to the transmission axis $\left( \theta ={{90}^{0}},\cos \theta =0,{{\cos }^{2}}\theta =0 \right)$.
Hence, from this explanation and using (1), we get the intensity of light $\left( I \right)$ coming out from the polarizer as
$I=\dfrac{1}{2}{{I}_{0}}{{\cos }^{2}}{{0}^{0}}+\dfrac{1}{2}{{I}_{0}}{{\cos }^{2}}{{90}^{0}}$
$\therefore I=\dfrac{1}{2}{{I}_{0}}{{\left( 1 \right)}^{2}}+\dfrac{1}{2}{{I}_{0}}{{\left( 0 \right)}^{2}}=\dfrac{1}{2}{{I}_{0}}+0=\dfrac{1}{2}{{I}_{0}}$ [$\left( \theta =0,\cos \theta =1.{{\cos }^{2}}\theta =1 \right)$,$\left( \theta ={{90}^{0}},\cos \theta =0,{{\cos }^{2}}\theta =0 \right)$]
$\therefore I=\dfrac{{{I}_{0}}}{2}=\dfrac{32}{2}=16W.{{m}^{-2}}$
Hence, the intensity of the light coming out from the first polarizer will be $16W.{{m}^{-2}}$.
Hence, the correct option is B) $16W.{{m}^{-2}}$.

Note: This problem could also have been solved by making use of the fact that when unpolarized light passes through a polarizer, its intensity becomes halved. We arrived at this specific result by using and explaining Malus' Law. This is the more general approach and can be used for finding out the intensity of light coming out from any polarizer. Since the question required the intensity of the light coming out from the first polarizer only, we could have solved using the idea of unpolarized light intensity becoming halved after going through a polarizer.