# Passing Polarized light through multiple polarizers in series

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?

I think the simplest way to understand this is that any vector can be described as the sum of two basis vectors. The vector from, say (0,0) to (1,1) is the vector sum of (0,0) to (1,0) and (0,0) to (0,1) . So, take your original polarized light and run it thru a 45-degree polarizer. You've just tossed half the amplitude, and the output light is now at that 45-degree angle. Send that through a polarizer at another 45 degrees (thus 90 degrees from the input polarization angle), and again one of the two basis vectors gets through.

You've attenuated the signal a lot, of course.

From a physical material standpoint, you could view this as the input photons being absorbed and re-emitted at a new polarization angle, with the absorption coefficients and polarization angles defined by the molecular structure of the material.

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 $30º$, 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).

• Is this the general formula for x polarizers after the first? $I_f = I_i * \left(\left(\cos\left(\frac{\pi}{2x}\right)\right)^{x}\right)^{2}$ given that the total angle change is pi/2 – Caridorc Mar 17 at 12:21
• @Caridorc that looks about right. Of course, assuming ideal polarizers; a good exercise is to work out the expression for real polarizers, where there is some leak in the blocking polarisation, and it is not perfectly transparent in the other. – Davidmh Mar 17 at 18:22
• Given an efficiency $e$ so that only an e fraction of amplitude can pass through each polaroid,with (0 < e < 1) I think it is sufficient to add e^x inside so $If=Ii∗((cos(π/2x))^x*e^x)^2$ – Caridorc Mar 17 at 20:33
• @Caridorc that takes care of the non-transparency part, but not the leakage, ie, the blocked polarisation is not completely blocked (if you put two polarisers at 90º you still see some light through). – Davidmh Mar 18 at 16:18