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

1 Answer
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*cos^2(theta)#

Sounds difficult but look at the picture:
enter image source here

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 #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 #1/2#!!!)

As a mathematical explanation you have:
If #E0# 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 #E0^2#.

The electric field vector #E0# of the incoming wave can be resolved into two rectangular components projected upon the polarizer axis:
#E0*cos(theta)# and #E0*sin(theta)#
The analyzer will transmit only the component ( i.e #E0*cos(theta)#) which is parallel to its transmission axis.
enter image source here
But intensity is proportional to the square of the electric vector you have:
#IpropE^2# and:
#Iprop(E0*cos(theta))^2# but:
#I/I_0=(E0*cos(theta))^2/(E0^2)#

and finally:
#I=I_0*cos^2(theta)#