How can I compute the intensity of a polarized wave going through a Polaroid?

Dec 21, 2014

You could use Malus' Law.
Malus' Law tells us that if you have a polarized wave (of intensity ${I}_{0}$) passing through a polarizer the emerging intensity ( $I$ ) will be proportional to the cosine squared of the angle between the polarizing direction of the incoming wave and the axis of the polarizer.

Or: $I = {I}_{0} \cdot {\cos}^{2} \left(\theta\right)$

Sounds difficult but look at the picture:

Knowing the angle $\theta$ and the incoming intensity you´ll be able to evaluate the output intensity.

Special case :
When you have unpolarized light falling upon the polarizer the transmitted intensity will be $\frac{1}{2}$ of the incoming intensity, i.e.:
$I = {I}_{0} / 2$
(to understand this, look at the mathematical explanation that follows and remember that the average value of ${\cos}^{2}$ is $\frac{1}{2}$!!!)

As a mathematical explanation you have:
If $E 0$ is the amplitude of the electric vector of the incoming wave, then the intensity ${I}_{0}$ of the wave incident on the polarizer is proportional to $E {0}^{2}$.

The electric field vector $E 0$ of the incoming wave can be resolved into two rectangular components projected upon the polarizer axis:
$E 0 \cdot \cos \left(\theta\right)$ and $E 0 \cdot \sin \left(\theta\right)$
The analyzer will transmit only the component ( i.e $E 0 \cdot \cos \left(\theta\right)$) which is parallel to its transmission axis.

But intensity is proportional to the square of the electric vector you have:
$I \propto {E}^{2}$ and:
$I \propto {\left(E 0 \cdot \cos \left(\theta\right)\right)}^{2}$ but:
$\frac{I}{I} _ 0 = {\left(E 0 \cdot \cos \left(\theta\right)\right)}^{2} / \left(E {0}^{2}\right)$

and finally:
$I = {I}_{0} \cdot {\cos}^{2} \left(\theta\right)$