I am having some difficulty accepting the implications of the equation governing the intensity of light passing through polarization filters, $$ I = I_0 \space\cos^2\theta $$ with $\theta$ being the angular difference between the two filters.

Here's the difficulty. If I put two filters at an angle of $\frac{\pi}{2}$, then no light makes it through to the other side. But if I then put another filter in between the original two, then we apply the above equation twice successively, neither time getting a result of zero.

That is, if you have two filters that don't allow any light through, you can force them to allow light through by placing a filter in between. It seems to me that, in general, a filter blocks light, so the result is counterintuitive.

What happens to the light when the second filter is placed to allow some light to pass through the three-filter system?

  • $\begingroup$ Think of polarsation as a 2D vector pointing in a certain direction and the filters as projections along a certain direction on the 2D plane. The same phenomenon happens with spin $\endgroup$ – Phoenix87 Oct 23 '16 at 22:43
  • 23
    $\begingroup$ "This seems to imply that if you have two filters that do not allow any light through, you can force them to allow light through by placing a filter in between." Yes it does. And this is a standard class-room demonstration. It is startling enough that I sometimes get gasps out of my otherwise quite jaded classes. $\endgroup$ – dmckee --- ex-moderator kitten Oct 23 '16 at 22:58
  • 3
    $\begingroup$ Check out the end of this Youtube clip. $\endgroup$ – Daniel R Oct 24 '16 at 20:56

Indeed, it can be counterintuitive that adding a polarizing filter can increase the transmitted intensity, when each filter only 'removes' light.

Here's a slightly more intuitive way of thinking about it. A filter doesn't strictly remove light -- what it really does is add a light wave which destructively interferes with part of the incoming light. For example, a horizontal polarizing filter removes the horizontal part by transmitting an additional horizontally polarized wave out of phase by $180^\circ$. (This is true on a microscopic level, too. In the polarizing filter, electrons are driven by the incoming wave and, since they accelerate, they emit radiation of their own.)

Now let's say we add a diagonal polarizing filter in between horizontal and vertical filters. Thinking about the horizontal and vertical filters as "destroying light" in the usual way, you are correct in that none of the original light wave will make it through, regardless of what you put in between.

But the new diagonal filter adds a new diagonally polarized wave! It's part of this wave that makes it out.

  • 7
    $\begingroup$ This is the most important answer, because it mentions that electromagnetic waves induce new electromagnetic waves. For similar reasons, radar doesn't "bounce" waves off of metallic objects, it induces currents that cause the emission of waves in the opposite direction. $\endgroup$ – bright-star Oct 24 '16 at 4:53
  • 2
    $\begingroup$ I'm having a hard time to apply your microscopic picture to the case of non-absorptive thin film polarizers. Can you elaborate how added light is still a good picture there? $\endgroup$ – Emil Oct 24 '16 at 9:07
  • $\begingroup$ @Emil If the initial and final polarizations are $A$ and $B$, all my answer is saying is that you can think of $B$ as $A + (B-A)$. There's strictly speaking no physics there, it's purely to aid intuition. $\endgroup$ – knzhou Oct 24 '16 at 16:25
  • 1
    $\begingroup$ However, "adding waves" is still, to me, the more natural picture, even for a non-absorptive polarizer which appears to split a wave instead. As an even more extreme case, consider a mirror. It feels much more natural to say that it "bounces" a wave off, but there's no term anywhere in Maxwell's equations that describes this. If you compute what the electrons are doing, they're simply spawning a canceling wave in the forward direction and a new wave in the backward direction. We can look at the result and call it reflection, or beam splitting, but this is a secondary interpretation. $\endgroup$ – knzhou Oct 24 '16 at 16:28

There are two effect on a beam of light from passing through a polarizing filter.

  • First the intensity is reduced as you write above. (For unpolarized light incident on a polarizer you average over all possible polarization and get $I = I_{up}/2$.)

  • Second, the light that passes now has the polarization of the filter it just passed.

The second effect is what makes the trick work.


The same experiment can be demonstrated with a swinging rope. Place two slits behind each other (let space between them), the slits both in the same way oriented, take a rope through the slit, fix it at one end and swing the rope, of course in the direction of the slits. Now rotate the second slit. The rope no more will swing behind the second slit. But now if you place a third slit between the other two slits and this under 45°. The rope will swing all other his length and by this the swinging direction will be rotated to 90°.


This is a known effect. It can be explained classically by the vector decomposition of the transverse electric field vector of the light waves. If you have a vertically polarized wave after the first filter propagating in x-direction with electric field amplitude in z-direction $\vec E_0$and this wave encounters a filter that is oriented at $\theta=\pi/4$ to the vertical, then the electric field amplitude component passing this filter is $E_0 cos(\pi/4)=E_0/\sqrt(2)$. When this wave with electrical field amplitude $E_0/\sqrt(2)$ at an angle of $\pi/4$ encounters the polarizer oriented at $\theta=\pi/2$ it has an electric amplitude component in the direction of this last polarizer $\frac {E_0}{\sqrt(2)\sqrt(2)}=\frac {E_0}{2}$ and its intensity is $I=\frac {I_0}{4}$. Thus, indeed, a substantial light intensity passes through when the middle filter is inserted.


"It seems to me that, in general, a filter blocks light, so the result is counterintuitive."

If all a polarization filter does is blocking light then your result would indeed be impossible. However a polarization filter also forces light to take a particular polarization; In the instant case the intermediate filter rotates the polarization of the light, with the consequence that it's no longer normal to the last filter.

  • 1
    $\begingroup$ One can demonstrate this in a classroom easily with three polarized sunglasses lenses. $\endgroup$ – ttw Oct 24 '16 at 15:10

This actually isn't unique to optics. If you take a rope and three solid (e.g. metallic) sheets with slits (one horizontal, one vertical and one diagonal), placed after one another as you place your optical polarizers, then you'll have much the same result.

A wave going through first, let's say horizontal, slit will be polarized horizontally, i.e. the part of the rope on the other side will only move in horizontal direction. Then coming to the diagonal slit, the motion will be forced to be not quite horizontal: the wave will reflect much like if you threw a ball into a tunnel under right angle, like this (the tunnel is analogous to our slit):

how to throw the ball into the tunnel

As you can understand, average motion of the ball shown above will get both horizontal and vertical velocity components.

Now the third pass through the (vertical) slit is just a repetition of the second one (the diagonal slit pass), but rotated by $45^\circ$.


Not the answer you're looking for? Browse other questions tagged or ask your own question.