# Transmission of light, sub-wavelength apertures, and cut-off frequencies

I was hoping someone could please explain how the transmission of light through a sub-wavelength aperture in a metal film, at a particular wavelength, changes when the aperture is:

i) above cut-off width and ii) below cut-off width?

From what I've been reading, I gather that above cut-off width there are propagating modes and below cut-off width then tunnelling occurs. If this is correct, could someone please explain the differences between the two processes in this situation?

Also I'm not sure how the metal film thickness affects tunnelling/propagation (i.e. does thickness < skin depth imply propagating modes and thickness > skin depth imply tunnelling? Or vice versa?)

Any help would be appreciated,

Thanks!

• Even just a link to some sources or books explaining this would be great :)
– Sean
Aug 6, 2015 at 21:22

If the aperture width is above cutoff width, then as you say, there are propagating modes. This means that the aperture acts as a waveguide, with a propagation constant $\beta$ with a real part much larger than its imaginary part. (The imaginary part of $\beta$ quantifies damping.) A good reference for waveguides is Snyder & Love, Optical Waveguide Theory.

A propagating waveguide mode has a certain spatial extent. If the aperture width is below cutoff width, then the mode literally does not fit inside the aperture, and so the aperture acts as a very bad waveguide, with a propagation constant $\beta$ which is purely imaginary; so there is no propagation, only damping.

However, electric fields can't just stop discontinuously at the entrance of a waveguide. Even a waveguide mode with purely imaginary $\beta$ will have the field reach into it a little bit. The field will drop off proprortionally to $e^{-i\beta d}$. If the waveguide is short enough that there is any electric field left by the time it gets to the end, then it will turn back into free radiation on the other side. This is called tunneling. An illustration might help to make this intuitive:

The vertical axis is field amplitude (very hand-wavy because I also drew a sine wave to show a propagating mode, but the amplitude certainly doesn't vary like that) and the horizontal axis is distance.

There is no direct relationship between the metal's skin depth and the cutoff width. Rather, the skin depth influences the value of $\beta$, and $\beta$ determines how thin the metal film must be for there to be any appreciable tunneling.

Note that in the case of a perfectly conducting metal film, which is a decent approximation if the film is gold, silver, or aluminum, and the light is of an appropriate wavelength, then there is no skin depth. In this case, it is only the waveguide aperture size and wavelength of the light that decide whether a mode is propagating or not. In the realistic case, this is still the biggest factor.

• Wow thanks for the answer ptomato! I found that really useful, I don't suppose there is any on-line (free) resources for waveguide theory you would particularly recommend? It is an area I really need to get clued up with.
– Sean
Aug 10, 2015 at 14:45
• Also I have an additional question if you don't mind? I was just wondering how, for the case of an EM wave tunnelling through an aperture below cut-off width, the propagation constant is different from an EM wave tunnelling through an optically thin metal sheet? In the case of an aperture being present, does this result in an effective propagation constant being used which is dependent on the metal and the medium filling the aperture?
– Sean
Aug 10, 2015 at 14:58
• I don't know of any free waveguide resource, but Wikipedia is always a good place to start. If you have another question then it's best to open a new question so that others can discover it as well. However, the short answer is that the propagation constant depends on the metal, the aperture medium, and the geometry of the aperture. Aug 10, 2015 at 16:30