Malus's Law 

According to malus, when completely plane polarized light is incident on the analyzer, the intensity I of the light transmitted by the analyzer is directly proportional to the square of the cosine of angle between the transmission axes of the analyzer and the polarizer. 

i.e I ∞ cos^{2}θ 



Suppose the angle between the transmission axes of the analyzer and the polarizer is θ. The completely plane polarized light form the polarizer is incident on the analyzer. If E_{0} is the amplitude of the electric vector transmitted by the polarizer, then intensity I_{0} of the light incident on the analyzer is
I ∞ E_{0}^{2}


The electric field vector E_{0} can be resolved into
two rectangular components i.e E_{0} cosθ and E_{0}
sinθ. The analyzer will transmit only the component ( i.e
E_{0} cosθ ) which is parallel to its transmission
axis. However, the component E_{0}sinθ will be absorbed by the analyser.
Therefore, the intensity I of light transmitted
by the analyzer is, 

I ∞ ( E_{0} x cosθ )^{2}


I / I_{0} = ( E_{0} x cosθ )^{2} / E_{0}^{2} = cos^{2}θ 

I = I_{0} x cos^{2}θ


Therefore, I ∞ cos^{2}θ. This proves law of malus.


When θ = 0° ( or 180° ), I = I_{0} cos^{2}0°
= I_{0} That is the intensity of light transmitted by
the analyzer is maximum when the transmission axes of the analyzer
and the polarizer are parallel. 

When θ = 90°, I = I_{0} cos^{2}90° = 0
That is the intensity of light transmitted by the analyzer is minimum when the transmission axes of the analyzer and polarizer are perpendicular to each other.
