Explain Malus's law and state its mathematical expression.

(N/A) Malus's law states that when a beam of plane-polarized light is incident on an analyzer,the intensity of the light transmitted through the analyzer is directly proportional to the square of the cosine of the angle between the transmission axes of the polarizer and the analyzer.
Let the pass-axis of the analyzer $(P_{2})$ make an angle $\theta$ with the pass-axis of the polarizer $(P_{1})$. The electric field vector $\vec{E}_{0}$ of the plane-polarized light emerging from $P_{1}$ makes an angle $\theta$ with the pass-axis of the analyzer $P_{2}$.
We can resolve the vector $\vec{E}_{0}$ into two rectangular components:
$(1)$ $E_{0} \cos \theta$,which is parallel to the pass-axis of the analyzer $P_{2}$.
$(2)$ $E_{0} \sin \theta$,which is perpendicular to the pass-axis of the analyzer $P_{2}$.
The component $E_{0} \cos \theta$ passes through the analyzer,while the component $E_{0} \sin \theta$ is blocked (absorbed) by the analyzer.
Since the intensity of light $(I)$ is proportional to the square of the amplitude $(E^{2})$,the intensity of light incident on the analyzer is $I_{0} \propto E_{0}^{2}$,and the intensity of light emerging from the analyzer is $I \propto (E_{0} \cos \theta)^{2}$.
Taking the ratio:
$\frac{I}{I_{0}} = \frac{(E_{0} \cos \theta)^{2}}{E_{0}^{2}} = \cos^{2} \theta$
Therefore,$I = I_{0} \cos^{2} \theta$.
This is known as Malus's law.