# What is the effect of polarization on diffraction by a narrow slit?

Consider the well known demonstration of diffraction by a narrowing slit. (See for example the demonstration at the 30 minute mark of this lecture at MIT by Walter Lewin)

It is my (possibly mistaken) understanding that the light emerging after the slit becomes substantially slimmer than one wavelength is polarized.
This would seem to imply that light of perpendicular polarization would not be transmitted, thus implying a fairly substantial and dramatic difference in the results of the experiment with parallel and perpendicularly polarized light. That is, instead of spreading out, the light polarized in the wrong direction would essentially just shut down as the slit narrows below one wavelength. Is this true?

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## 2 Answers

I tried to make this a comment, but I couldn't do it. I don't know why.

@Colin K Thank you for your answer, although I’m not sure I understand what you are saying. The filter of parallel conducting wires was indeed what I was thinking about. Let’s consider the case where the wires are spaced one tenth (or less) of a wavelength apart. My expectation is that this filter passes the perpendicularly polarized light, and reflects the parallel polarized light.

Now consider a slit cut into a conducting material the same size and shape as the space between two wires in the above filter. My expectation is that this slit will again reflect parallel polarized light, and pass perpendicularly polarized light. In addition the passed light will “suffer diffraction” and be spread out significantly. Do you agree?
If not please point out exactly where you disagree.
Thank you. Jim Graber

Incidentally, my first guess is that the filter with many wires will have a much smaller, maybe even negligible diffraction angle, an angle based on the width of the entire filter, rather than based on the width of a single slit.

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If you edit your question, I can update my answer. Right now my only option if to answer in a comment to this "answer" – Colin K Feb 27 '11 at 4:03

First, I just want to correct a minor misunderstanding. If anything, it would be the light with its electrical field oscillating in the plane parallel to the slit which would have more difficulty propagating, and even then, only in specific circumstances. The best way to explain this is to ignore, for a moment, the experiment you've described, and consider a simple polarizing filter.

The easiest type of polarizing filter to talk about is simple an array of closely space parallel conductive wires. Such a filter will block light polarized parallel to the wires. Why? Because the E field from this light is parallel to the wires, it can induce currents, which absorb energy from the traveling EM radiation. Perpendicularly polarized light doesn't induce such currents, and is therefore able to pass more easily.

Now, going back to your question. If the slit were cut in a conductive material, and the slit width was on the order of the wavelength, then there may indeed be diminished transmission for light polarized parallel to the long axis of the slit because, to a rough approximation, the edges of the slit would act like the conductive wires I described before. However, while the light transmitted would possibly be dimmer, the form of the resulting diffraction pattern should be unchanged.

Of course, this is all moot if the slits are cut into a non-conductive material. In any event, the only change would be in the intensity of the pattern. It's up to you to decide if you think this difference qualifies as a "substantial and dramatic difference in the results of the experiment."

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The fence polarizer also works with superconducting wires. Once can explain this with the wires being excited oscillators with a 90° phase shift emitting waves both directions. In the forward direction the incident and the excited waves cancel out, where in the reflected direction one ends up with the familiar standing waves. This is the whole essence of reflection/transmission theory: There's some energy transfer yielding emittance of phase shifted waves. – datenwolf Feb 25 '11 at 17:36
@datenwolf: I hadn't thought of that example with superconducting wires, but I do of course understand how to derive reflection/transmission phenomenon. I didn't include it in this answer because I thought it was a bit beyond the scope of the question. – Colin K Feb 25 '11 at 17:45
I believe this is more or less correct; the two situations basically trade E for M, but I have to suspect that there are some near field (or wide angle) effects that distinguish polarities. So I would think there might be some differences for a nearby target and that they would correspond to a change in the overall shape of the interferences (but not the positions or angles). I'll think about this some more. – Carl Brannen Feb 26 '11 at 0:26