State and explain Malus' law.
The intensity of the plane polarised light transmitted through the analyser varies as the square of the cosine of the angle between the plane of transmission of the analyser and the plane of the polariser.
Let $E$ be the amplitude of the light transmitted by the polariser and $\theta$ be the angle between the planes of the polariser and the analyser.
Resolve $E$ into two components:
(i) $E \cos \theta$ along $OP$ (i.e. parallel to the plane of transmission of analyser)
(ii) $E \sin \theta$ along $OV$ (i.e. perpendicular to the plane of transmission of analyser).
Only $E \cos \theta$ component is transmitted through the analyser.
We know that, Intensity
$\propto (\text{Amplitude})^2$
i.e.
$I \propto E^2$
Intensity of the transmitted light through the analyser is given by,
$I \propto (E \cos \theta)^2 \quad \text{i.e.} \quad I = k E^2 \cos^2 \theta.$
But
$k E^2 = I_0 \quad \text{i.e., the intensity of the incident polarised light}$
$I = I_0 \cos^2 \theta$
or
$I \propto \cos^2 \theta \quad \text{(It is Malus' Law)}$
Special Cases :
(i) When, $\theta = 0^\circ$ or $180^\circ$ i.e. the planes (or polarising directions) of polariser and analyser are parallel or antiparallel to each other, then
$\cos \theta = \pm 1$
and
$I = I_0 \cos^2 \theta = I_0 (\pm 1)^2 = I_0$
$(\because \cos 0^\circ = 1 \text{ and } \cos 180^\circ = -1)$
Thus, the intensity of the incident plane polarised light does not change, when passes through the analyser.
(ii) When, $\theta = \frac{\pi}{2}$ i.e. the planes of polariser and analyser are perpendicular to each other, then
$\cos \theta = \cos \frac{\pi}{2} = 0.$
$I = I_0$
Hence, the intensity of the incident plane polarised light becomes zero, when passes through the analyser.
The variation of intensity of light passing through analyser with the variation of angle $\theta$ between polariser and analyser is shown in figure.
