This is a question that has been troubling me from many days:
Suppose we pass a linearly polarized light through a system of 3 successive polarizers.
The 1st polarizer is offset 30$^{\circ}$ from the plane of polarization. The second is offset 30$^{\circ}$ from the first. The third is offset 30 $^{\circ}$ from the second.
Applying Malus' Law thrice, we get the final intensity as:
$I=I_{o} \cdot \cos^{2}(30^{\circ}) \cdot \cos^{2}(30^{\circ}) \cdot \cos^{2}(30^{\circ}) \neq 0$
But this light is polarized perpendicular to the incident light!
How did the polarizers create a perpendicular field, when the incident wave has no component in that direction?
Best Answer
I also found this surprising when first introduced.
The light has no memory. When it passes through the second polariser, there is no information whatsoever of its previous polarisation. It could have been whatever!
When light goes through a polariser at $30ยบ$, it gets tilted at the cost of some lost intensity. In other words, it gets projected on the polariser axis, that is rotated a bit with respect to the light polarisation vector. This new light then goes on to the next one, and gets tilted again.
You get the picture. You can use this idea to arbitrarily change the polarisation angle of light by putting many of them in series. The more steps you do, the more polarisers you use, and thus, the smaller the steps, more light will come out on the other side (in reality, you always loose some light because they are not completely transparent, so you don't want to use many of them).