Malus's Law

When perfectly plane polarized light is incident on an 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 ∝cos2θ

Let the angle between the transmission axes of the analyzer and the polarizer be θ.

The completely plane polarized light form the polarizer hits the analyzer at right angles.

If E0 is the amplitude of the electric vector transmitted by the polarizer, then intensity I0 of the light incident on the analyzer is I ∝ E02

Let us resolve electric field vector into two perpendicular components i.e E0 cosθ and E0 sinθ.

The analyzer can transmit only the component which is parallel to its transmission axis ( i.e E0 cosθ ) and the component E0sinθ will be absorbed by the analyser. Therefore, the intensity I of light transmitted by the analyzer is:

I ∝ ( E0 x cosθ )2

I / I0 = ( E0 x cosθ )2 / E02 = cos2θ

so, I = I0 x cos2θ

∴ I ∝ cos2θ.

When θ = 0° ( or 180° ), I = I0 cos20° = I0 the intensity of light transmitted by the analyzer is a maximum so this occurs when the transmission axes of the analyzer and the polarizer are parallel.

When θ = 90°, I = I0 cos290° = 0 so the intensity of light transmitted by the analyzer is a minimum when the transmission axes of the analyzer and polarizer are perpendicular to each other.