Polarisation

Malu’s Law

Light from the polarizer is incident on the analyzer

• $I=I_0{\cos }^2{\theta }_i$ , $I_0$ is the initial intensity; ${\theta }_i$ is the angle between the plane of the polarizer and the analyzer.

• For a beam of unpolarized light, it contains light from all angles, so the value of ${\cos }^2\theta$ can be thought of as the average, which is $1/2$

  Hence, $I/I_0=1/2$

Proof of Malu’s Law

• $I=A^2$ , $I$ is the intensity, $A$ is the amplitude
• The electric vector of a light has an amplitude of $E_0$. This can be decomposed into $E_0\cos \theta$ and $E_0\sin \theta$, $\theta$ being the angle between the direction of oscillation of the light wave and the plane of the polarizer
• The polarizer lets in the $E_0\cos \theta$ component and absorbs the $E_0\sin \theta$ component.
• $I_0=(E_0\cos \theta {)}^2$
• $I/I_0=(E_0\cos \theta {)}^2/E_0^2={\cos }^2\theta$