In the school I was impressed with the following experiment: there are two polarization filters with orthogonal polarization directions placed at a distance from each other. They do not allow any light through.

Then the experimenter inserts a third filter oriented at 45 degrees, the whole system starts to conduct light, with strength of about 1/8 of the initial source.

Now, my question is: does the same happen with color?

For instance, initially one filter is purple while the other one is green, of opposite colors and do not conduct light. Then a medium color filter is inserted in the middle, will the system conduct more light?

What if instead of RGB space we use wavelength, initial filter for blue light, the other for red, and the middle filter for green?

If not, why it works with polarization but not with wavelength?

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    $\begingroup$ So why not try the experiment? $\endgroup$
    – Farcher
    Jun 18 at 11:41

2 Answers 2


If you have a $|\uparrow\rangle$ polarizer and a $|\to\rangle$ polarizer. A third polarizer of $45$ degrees in the middle creates superposition of $\frac{1}{\sqrt{2}}|\uparrow\rangle + \frac{1}{\sqrt{2}}|\to\rangle$ states. Therefore, with it in the middle, there is a nonzero probability that light will leave the last, say $|\to\rangle$, polarizer.

You can't do the same with wavelength. Wavelengths of light corresponds to energy states. You can't present an energy state as a superposition of other energy states. And therefore you cant reproduce this phenomenon with wavelength filters.

edit: My answer assumes an experiment with three different wavelength filters. Nevertheless, there might be a different type of intermediate filter (as mentioned by @WillO in the comment) which does create a superposition of energy states such that an effect such as the one with the three polarizer might be observed.

  • $\begingroup$ But if the intermediate filter detects some observable whose eigenstates are nontrivial superpositions of energy eigenstates, then it seems like this should work. $\endgroup$
    – WillO
    Jun 18 at 14:01
  • $\begingroup$ That's a good point. I wonder which observable whose operator does not commute with the hamiltonian could be practically measured $\endgroup$
    – Ben
    Jun 18 at 14:17


Light is a combination of oscillating electric and magnetic fields. Polarization is an orientation of light. Think of it as a picket fence. Light gets through if the E field is parallel to the pickets. It is blocked if perpendicular.

If light is at an angle, it partially gets through, and what does get through comes out is parallel to the pickets. This means you can go through another fence at a new angle, and more after that.

Color filters don't work that way. Color is determine by wavelength. Red is longer wavelengths. Blue is shorter. Purple is a mix of red and blue, so a mix of long and short wavelengths.

A red filter removes or reduces the amount of short wavelengths, leaving only the red longer waves. A blue filter removes the longer waves, leaving only the blue shorter waves. Going through both leaves nothing.

  • $\begingroup$ I'm concerned that this answer ignores quantum mechanics --- but if you ignore QM then the polarizer trick shouldn't work either, so this seems like the wrong level to address this question. $\endgroup$
    – WillO
    Jun 18 at 19:12
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    $\begingroup$ @WillO Why do you need QM for the polarizer trick? Wouldn't be separation into directions and passing of the e-vector component parallel to the polarizer be sufficient to explain the trick? $\endgroup$ Jun 18 at 23:39
  • $\begingroup$ Your answer is correct and I upvoted it. So long as the blue and red filters have disjoint transmission bandpasses, no incident light can pass through both filters and adding more filters anywhere is useless. But if the blue and red filters have a non-negligible overlap in their transmission bandpasses, then a little light can get get through. Then there is a cheat possibility: put a uranium glass (green) fluorescent filter between the blue (first) and red (last) filters. So the blue light excites green fluorescence in the uranium glass filter and some of that gets through the red filter. $\endgroup$
    – Ed V
    Jun 19 at 0:22
  • $\begingroup$ I agree with @WillO that I have simplified the answer as much as I could, to the point that it ignores essential physics. I did not explain why polarizers or filters work. I did not explain the spectral range of colors. I just explained the outcome of using the two kinds of filter and how the outcomes are different. That seemed to be the heart of the question. $\endgroup$
    – mmesser314
    Jun 19 at 0:35
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    $\begingroup$ @WillO no need for QM in classical maxwell eqs physics. The only thing that might bother me and is typically used as a way to explain polarizers is the picket fence and the direction. While that is an analogy, it is always omitted that in reality, wiregrid (conducting) polarizers act in the opposite way: wiregrid parallel to polarization means reflection, wiregrid perpendicular means transmission. Because of this omission, even many optics scientists think the way wiregrid polarizers work is the picket fence way. $\endgroup$ Jun 19 at 13:33

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