# Particles, waves and parallel wire filters. Transmission formula?

If I think of a photon as a particle, I think a parallel wire filter should transmit proportionally to the uncovered area. (and reflect proportionally to the covered area). Obviously polarization adds complexity by introducing a second factor to be considered.

On the other hand, if I consider the photons as waves, I think the transmission should be less sensitive to the area covered. Both a fairly thick and a fairly thin filter should transmit most of the light in one polarization and reflect most of the light in the opposite polarization.

Which is correct?

To be more precise, consider a spectrum of filters with thick and thin wires. (on the same spacing). At one extreme end of the spectrum the thinest wires approach an empty space or vacuum, which transmits everything and reflects nothing. At the other end the thickest wires merge together and form a continuous surface, in effect a mirror which reflects everything and transmits nothing. My question is what happens in between these two extremes? Of course there are two answers, one for parallel polarization and one for perpendicular polarization. What are the two formulas for transmission as a function of wire thickness, or percent of space blocked?

My guess is the answer for the transmitted polarization is close to a step function, only rounded off in the corners. Other possibilities is a linear function decreasing from one hundred percent to zero, or from fifty percent to zero. What are the two correct equations for the two complementary polarizations? Or more specifically, what is the correct formula for the transmitted polarization?

For part credit, just give a reference or a convincing derivation showing whether this function is closer to a step function or closer to a linear approximation. For full credit, provide a reasonably accurate formula with source or derivation showing the full transition from zero to one hundred percent wire coverage. It should be accurate enough to get the round off in the corners right to a percent or two (at least 5%) or prove that there is no significant round off. A good answer will have either a derivation or a reference to a reputable source.

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""On the other hand, if I consider the photons as waves,"" That is not easy to do. I'd recommend to think either of photons or waves. – Georg May 29 '11 at 12:24
Lets see if we can get on the same page on this. Are you talking of a grating, like a diffraction grating?How are these wires constructed as spacing: wavelenth. Also photons and in general electromagnetic waves are transversely polarized except in special constructs, like waveguides: en.wikipedia.org/wiki/Polarization_%28waves%29 – anna v May 29 '11 at 15:19
@Anna He means a wire grid polarizer, the simplest case of an absorptive polarizer, and want to consider how the transmission coefficient varies as a function of wire thickness / wire spacing. There are a lot of subtle issues here of course when the wavelength of the light is comparable to the grid period.. – BjornW May 29 '11 at 18:13
For concreteness say we had an incident unpolarized beam of 1 cm radiation The beam passes through two 10 cm holes at say 2 meters and one meter in front of the filter.the filter is one meter square. the parallel wires are set on a period of 1/10 (one tenth) millimeter. We have a set of 101 filters in which the width of the wires varies from 0 to 100% of the period. The first filter will be an empty space and totally transparent. The last filter will be a mirror like totally filled space and will be totally reflective, i.e. zero percent transparency. what do the intermediate 99 filters do? – Jim Graber May 29 '11 at 23:47
As I tried to explain in my answer, the relative cross-section area of the wires is irrelevant. Only their resistance matters. If you are assuming ideal (perfectly conducting) wires, then the transmission is zero for all 99 filters. – Marty Green May 30 '11 at 3:44

An infinite conducting sheet with a resistance of 377 ohms (per square) is a matched impedance to an em wave, so it reflects a quarter the power, transmits a quarter, and absorbs half. Your parallel wires will do the same when their equivalent resistance is 377 ohms per square. If you're thinking copper, don't forget to use the skin effect when calculating the resistance.

For much lower resistances, you have a simple reflector; as the resistance becomes much higher, it simply ceases to influence the wave at all.

There is very seldom any benefit to be gained by trying to analyze these kinds of things in terms of photons.

EDIT: I see now that I have done something very embarassing, for which my only excuse is that I have never before heard of these egg-slicer type polarizers. I meant to calculate the transmission in the PASSING orientation, so I allowed my egg-slicer to be parallel to the "arrows" of the electric field. Of course, this is the BLOCKING orientation! The passing case is much harder for me to analyze.

Furthermore, I find myself second-guessing even my calculation for the blocking case, so I am going to try to do a better job of justifying it. It was easy to believe in 100% blocking when the electric field "arrows" where slamming into the wires of the egg slicer and just falling down. Now that I see they are lined up so they might pass between the wires, I have to ask again: How do I get 100% blocking? Here is my revised justification.

The analysis is based on estimating the far field based on the current flowing in the strips. It turns out that conservation of energy limits us in surprising ways. Since the strips are perfect conductors, they can have no losses. All power must be accounted in the reflected and transmitted waves.

Let the width of the strips be x (as a fraction of 1) and the current density be I. On the reflected side, the far field is proportional to Ix (the total current per strip), so the reflected power is (Ix)^2. For x=1, the reflected power is 100% (no slits) so the proportionality becomes

P(reflected) = (Ix)^2 = 1 (for unit incident power)

Note that I=1 when x=1: this means that a unit current density generates exactly enough transmitted field to cancel the incident field. We can therefore allow our incident field to have the numerical value of 1. Note that a current density of 1 over the full extent of the sheet gives us a far field of 1.

On the transmitted side, one might think that a current I over an average area x should also generate a far field of Ix, giving a transmitted power of (Ix)^2. But we must remember to add it to the incident field before we square. Since the incident field is 1, and the phase must be such as to cancel the incoming wave immediately behind the strips, the average far field is therfore 1-Ix. So the transmitted power is

P(transmitted) = (1-Ix)^2 (again, for unit incidet power)

Now we can try and add up the reflected and transmitted powers to see if they equal one:

P(reflected) + P(transmitted) = (Ix)^2 + (1-Ix))^2 = 1

There are no solutions for this equation except for 100% reflection or 100% transmission. If any current flows at all in the strips, it must be sufficient to reflect all incoming power. In other words, the current in each strip is a constant, regardless of how thin you cut the strips.

To check my methodology, I attempted to apply the same logic of energy conservation to a conducting sheet of resistance 377 ohms (the matched case). It is a useful case because we can check it against the simple case of the full sheet, with half the power absorbed, a quarter reflected and a quarter transmitted. I will spare everyone the calculation (unless I am asked for it) and simply present the results in simplest form.

Where x is the percentage width of the strips, the powers are:

P(reflected) = (x^2)/k

P(absorbed) = (2x)/k

P(transmitted) = 1/k

where k= (x+1)^2. You can see for x=1 (full sheet) you get the proportions I originally suggested.

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Second try at adding this comment: I really appreciate the added information about finite conductivity of the screen. – Jim Graber May 30 '11 at 15:50
I really appreciate both of the calculations you have added. I believe that the reflecting polarization is a step function from zero to one hundred percent transmission with the step located at the zero strip-width edge. It is interesting to me that the partially conducting case appears to be a quadratic rather than a linear function as the strip width increases. And of course there is a trifurcation rather than a bifurcation of the energy. Also I am not surprised that this is a purely classical computation with no consideration of photons or any other quantum factor. – Jim Graber May 31 '11 at 0:59
I am still very interested in the case of the complementary polarization. Here the only conduction can be what I guess to be a nearly negligible capacitive effect across, rather than along, the strips. Hence little or no current, (big question mark) hence no reflection or absorption, hence near perfect transmission?? Again, the partially conducting case might be much more interesting and difficult. – Jim Graber May 31 '11 at 1:00
Thanks, Jim. Yes, I'd like to do the other polarization but I'm finding it much harder. I'm very tempted to think that the 377-ohm case embodies a kind of complimentarity, where both polarizations will yield similar formulas. But I'm not very close to putting it together. – Marty Green May 31 '11 at 2:36
There are some interesting complications I didn't include in my calculations relating to current distribution across the strips. As the strips get thinner, the current seems to want to bunch up near the edges. This throws my power calculation of a bit; but more importantly, it is suggestive of what happens with charge buildup in the cross-cut case. I wonder if the distributions are the same? – Marty Green May 31 '11 at 2:39

I can't comment on the blocked polarization (it was also covered in another answer) but the transmitted polarization effectively sees a periodical pattern of apertures right (the wires and the spacings)? Thus at least as a first approximation the standard reasoning of diffractive apertures should help. The light field after passing through an aperture is modelled by the Fourier transform of the aperture. The intensity is the square of the field and its spatial integral is the total transmitted intensity. In the case of the variable aspect-ratio apertures (variable width 0-100% at a fixed spacing) the transmitted total intensity will vary linearily from 0 to a maximum.

The form of each aperture is a "top hat" function, i.e. a hat-like shape whose width varies according the aperture ratio. Its Fourier-transform is a sinc function: $$F(k)=a sin(ka/2)/(ka/2)$$ given that a is the aperture width. Since the right part is the normalized sinc-function with a constant integral both at powers of 1 and 2, you see that the total intensity will vary linearily with a (for a single aperture).

An infinitely repeating array of apertures is modelled by a Dirac-comb (an infinite train of Dirac spikes) convolved with the top-hat aperture. When Fourier-transforming this, the convolution turns into a multiplication and the transform of the Dirac-comb is another Dirac-comb. Thus, the resulting field is a repeating array of Dirac spikes whose height (envelope) is modulated by the aperture-shape and its transform. By a handwaving argument, this means the same linear dependency as for the single aperture (you just have the same sinc-shape but "chopped up" and stretched with the Dirac comb).

Furthermore, if you restrict the array to a certain width, this is the same as multiplying the infinite array with another top-hat function. Multiplication turns to convolution in the Fourier transform, and again the top-hat turns into a sinc shape which is convolved with the Dirac spike structure, replacing every spike with the shape of the sinc. Again due to the constant integrals of these sinc's, this doesn't change the linear outcome either.

I'm sorry I didn't have the time to write the tex-equations for the above two convolution and multiplication fourier-pairs, I'll see if I can fix that later.

Thus, making the total polarizer width thinner increases the width of the individual diffraction lines. Changing the aspect ratio of the apertures changes the overall shape of the envelope over the spikes.

Notice that there is a hidden dependency on the light wavelength, the Fourier coordinate k above depends inversely on the wavelength, so for longer wavelengths the intensity distribution widens. When the wavelength approaches the grid period or goes below it, I guess you'll get some special cases here though.

For some nice examples including applets to see the intensity distributions see here and examples of Fourier theory are here.

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When the grid period is longer than the wavelength you get multiple maxima and your analyisis is relevant. For practical applications of this kind of filter I think the period is supposed to be much smaller than the wavelength, so we are only interested in the central maximum. Another point: your intensity is proportional to the open area, but at assumes the light hitting the grid is simply absorbed. For the ideal conducting grid case, the currents flowing in the edges are very significant and lead to different results. – Marty Green May 31 '11 at 11:36
Re my last comment: at any rate, the calculations I've done are based on the assumption that the grid period is much smaller than the wavelength and the total grid diameter is much larger than the wavelength. – Marty Green May 31 '11 at 11:42
I guess we can conclude that it's pretty difficult to estimate either of these behaviours quantitatively? :) Is there some EM solver toolbox program which can simulate stuff like this, if you input the correct parameters for the propagation medium and wires etc? – BjornW May 31 '11 at 20:36
The case I had in mind originally is wavelength much longer than period, but shorter than grid size in both dimensions. I began with the ideal near infinite conductivity case in mind (i.e copper or silver wires), but now I am also interested in the absorbing case as well. – Jim Graber Jun 1 '11 at 1:53
I have tried to find electromagnetic finite element analysis software that could handle this problem and was also free or low cost, but I am not even sure if the professional expensive ones can do it. The problem is simple but requires a lot of elements. I think the use of an infinite Dirac fence is a brilliant idea. I had never heard of it before. I have played with the applets, but they do seem to cover widely separated slits rather than a close spaced grid. – Jim Graber Jun 1 '11 at 1:54

I've posted an answer for the blocking case and now I'm going to post an answer for the transmitting case, since the argument for the linear transition still seems to be holding sway. I'm going to argue for the step function much the way I did in the blocking case. The only sticky points are the transitions, where the wires become infinitely thin (blocking case: how do such thin wires block everything?) and where the slits become infinitely thin (transmitting case: how do such thin slits transmit everything?). I'll come back to those later.

First the ideal blocking case: since the strips are parallel to the electric field, they function as a perfect conductor and block all transmission. Similarly, in the ideal transmitting case, the strips are perpendicular to the electric field, so they block current flow, acting as a perfect insulator. Hence, no reflection and total transmission.

I've already argued that for the blocking case, the strips carry the same amount of current as they are made thinner and thinner, thereby functioning as an perfect reflector. But it's harder to see how the complimentary procedure works in the transmitting case. How does power get through the thin little spaces in the venetian blinds? Answer: the slats of the venetian blinds are perfect conductors, so the current induced on the front surface loops around the back. It makes a complete loop, which it must in order to completely cancel the magnetic field inside the superconductor. And since it makes a very thin loop (because the slats are presumed to be very thin), it does not radiate at all because it has negligible magnetic dipole moment. Since it has no external electromagnetic effect, it might as well not be there, and the power seems to flow right through it. (Although technically you might argue that the power is annihilated on the front face and regenerated on the back face so it does not actually penetrate the superconductor.)

How do these arguments of perfect reflection/transmission break down for the limiting case of very thin wires/gaps? I've already argued that the very thin wires eventually show non-zero resistivity. What about the very thin spaces? By analogy, we can argue that they are not perfect insulators, and as their non-zero conductivity becomes significant, things break down.

These notions of non-zero resitivity/conductivity are of course classical, but the actual mechanisms might actually be quantum-mechanical. When do very thin superconducting wires become resistive? when the magnetic field exceeds its quantum-mechanical threshold. And when do very thin insulating gaps become conductive? When they are small enough for the current to tunnel through.

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The answer depends very much on the wavelength of the photons considered.

When the wire spacing (not thickness) is large compared to the wavelength of the photons we are in the "classical particle" (as opposed to quantum wave) limit; then the amount of absorption is simply the percentage of area blocked. This limit applies when buildings block sunlight. The amount of sunlight blocked is proportional to the size of the buildings. In this limit polarization effects are negligible.

When the wire spacing is small compared to the wavelength of the photons we are in the wave limit. This is the limit of a perfect polarization filter. If we assume that the wires are perfect conductors, then the polarization effect does not decrease as the wire size decreases. In practice, wires of excessively small diameter become resistive so the limit is not achieved.

So I'm convinced that (when you assume perfect conductors) there will be a discontinuous change between an infinitesimal wire and no wire at all. Similarly there will be a discontinuous change between an infinitesimal slot and no slot at all. In both cases, it is the presence or absence of electrical conduction that distinguishes the change.

In between the thin and thick slit limit, I believe there will be no change in transmission and reflection coefficients (in the limit of closely spaced wiring). This is because what is happening is that the electric field is either getting shorted out by wires, or it is not. There is no intermediate condition (unless one takes into account finite wire impedance).

It may seem bizarre that one could have full transmission when the slots are very thin. One can think of the slots as each being at some voltage. As the wires become wider, that voltage is maintained over the width of the wire. This means that the voltage difference between adjacent wires does not depend (much) on the thickness of the wires. (We are in the long wavelength limit.) Consequently, as the separations get narrower, the electric field (i.e. volts/meter) becomes larger. As the separation begins to close, the driving electric field goes to infinity. So it is not unintuitive that the transmission is still 100% when the wires are very fat.

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Carl, Definitely the small wire spacing case. Your argument seems to support the step function answer, but your last statement contradicts that. What is your opinion re the transmitted polarization? – Jim Graber Jun 8 '11 at 2:07
I understand a "perfect polarization filter" to mean perfect reflection of one polarization and perfect transmission of the perpendicular or " opposite" polarization. In the perfect polarization filter limit, both the transmission and the reflection are a step function that "turns on" as soon as the wires are present or significantly conducting. This is what happens at the thin wire end of the sequence. This is counterintuitive for the blocking direction, but we all know it happens. – Jim Graber Jun 8 '11 at 8:17
The question is what do the transmitted waves do as the aperture is more and more blocked? Do they still manage to squeeze through until the bitter end, or are they progressively blocked? Is the transmission only the sum of the indivdual slits, or is there some constructive interference? As soon as the slits appear, or as soon as the conductivity in one direction is reduced to near zero, the transmission of long waves should turn on? This is the counterintuitive conclusion I am questioning. – Jim Graber Jun 8 '11 at 8:18
I'm convinced that the thin end of the limit and the thick end will act the same. That is, it will be a step function. When I wrote "step function" in the above answer, I was assuming that the step function was posited to occur somewhere between the thin and thick limit. I'll edit the answer so that the terminology matches everyone else. – Carl Brannen Jun 8 '11 at 21:58

Here is a 2004 paper.....(Sorry, link died and I cannot find it again)

Try: "Enhanced transmission of transverse electric waves through subwavelength slits in a thin metallic film" Yu Qian Ye, Phys. Rev. E 80, 036606 (2009) [3 pages]

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This paper is definitely relevant, however, it seems to consider primarily one slit or briefly two. I am interested now in many slits. Also their two slit example seems to be very widely separated compared to my target case. Still, a substantial help. – Jim Graber May 29 '11 at 23:43
Plus the paper deals with a slit that has considerable thickness in the direction of propagation, many wavelengths, actually a waveguide. – Bill Slugg May 29 '11 at 23:52