5
$\begingroup$

This is a follow-up to this question: What happens to waves when they hit smaller apertures than their wavelenghts?

Hans Bethe wrote a paper in 1944, "Theory of Diffraction by Small Holes," Phys. Rev. 66, 163. I don't have access to the paper, but from descriptions online it sounds like he proved the following. Suppose a plane wave impinges on an absorbing sheet, and there is a hole in the sheet of diameter $d$, with $d<\lambda$. Let $P_0$ be the power incident on the hole, and $P$ the power diffracted through the hole. Then the transmission is $T=P/P_0=(d/\lambda)^4$.

As a practical application, I think this explains why microwaves don't leak strongly through the metal grille in the front of a microwave oven.

Questions:

  1. Does the thickness of the sheet matter? From references online, it sounds like the hole is treated as a waveguide...? This seems to relate to cutoff frequencies of waveguides, etc.?

  2. It seems to me that Huygens' principle would give $T=1$ for $d\ll\lambda$, since in this limit the wavelets are in phase. Why is Huygens' principle invalid here? Is this related to question #1?

  3. Is there a simple argument for the proportionality to $(d/\lambda)^4$? Or if not, how does one prove this using the gory details of Bessel functions, etc.?

$\endgroup$
  • 1
    $\begingroup$ The thickness of the sheet matters. I believe that for waveguides that are smaller than the wavelength the solution is damped exponentially with length. This is used for precision microwave attenuators because it doesn't require a lossy material, just a narrow tube. I don't believe that Huygen's principle can give correct absolute amplitudes in this case (can it do that in any case?). The original paper is here web.stanford.edu/class/ee349/Handouts/Bethe_PR1944.pdf and I am not going to argue with Bethe on how to derive this... have fun! :-) $\endgroup$ – CuriousOne Oct 16 '14 at 5:51
  • $\begingroup$ related: skeptics.stackexchange.com/q/4483 $\endgroup$ – Ben Crowell Oct 16 '14 at 23:03
  • $\begingroup$ One should probably make a practical comment about why microwave ovens don't leak strongly: because manufacturers are required to make sure that they don't. This, of course, leaves the realm of physics and enters the realm of law, but as a guideline, even a scientist should have a coarse notion of what can happen when they don't comply with the law: ieee.li/pdf/viewgraphs/legal_aspects_regulatory_compliance.pdf. Sorry for the off-topic comment. $\endgroup$ – CuriousOne Oct 17 '14 at 1:58
2
$\begingroup$

It turns out that there's a pretty detailed analysis of this in Jackson, Classical Electrodynamics, sections 3.13, 5.13, and 9.5. Although Jackson does it in excruciating detail using Bessel functions and infinite series, it's actually pretty easy to pull out the basic ideas.

Start by considering a simpler problem. A thin, conducting sheet in the $x-y$ plane has a circular hole in it of radius $a$. Suppose that at large distances above the sheet, the electric field is uniform with magnitude $E_0$ and is in the $z$ direction, but the field is zero below the sheet. Then the field closer to the hole can be broken down into two terms: one for the field you'd have if the hole wasn't there, and another term that exists because the hole is there. At sufficiently large distances, the second term can be approximated as that of an electric dipole $p$, which by symmetry must be in the $z$ direction, and by linearity must be proportional to $E_0$. On dimensional grounds, we must have $p \propto E_0 a^3$. (The unitless constant of proportionality turns out to be $1/3\pi$, but that doesn't really matter for my purposes.)

A similar analysis holds for a magnetic field in the $x$ direction, with a magnetic dipole appearing at the hole. (The constant of proportionality is $2/3\pi$.)

Now consider an electromagnetic wave with wavelength $\lambda$ coming down from above. If $\lambda$ is large compared to $a$, then the electrostatic and magnetostatic results above still hold at any given time. Therefore we have dipole radiation coming from the hole, with amplitude proportional to $a^3$ and power $P$ proportional to $a^6$. The power $P_0$ incident on the hole is proportional to $a^2$, so the fraction of the power transmitted $T=P/P_0$ is proportional to $a^4$. On dimensional grounds, we must therefore have $T\propto (a/\lambda)^4$, where the constant of proportionality is unitless. This also makes sense because the power radiated by a dipole is proportional to $\omega^4$.

I think it's pretty straightforward to see how this would play out for surface waves on water. I would expect a narrow hole of width $h$ to act as a monopole source, which would radiate power proportional to $\omega^2$. Therefore the transmission must go like $\lambda^{-2}$, and on dimensional grounds we must therefore have $T\propto (h/\lambda)^2$.

All of this follows directly from two very simple considerations: (1) dimensional analysis, and (2) the proportionality of the amplitude of $\ell$-pole radiation to $\omega^{2(\ell+1)}$. Huygens' principle is never needed, and in fact would not have been valid in an even number of spatial dimensions (which is what we have in the case of the water waves).

This is all for a thin sheet. As noted by CuriousOne in a comment, when the sheet is thicker, you can treat the hole like a waveguide. I tried to extract the basic ideas from this web page: http://www.cvel.clemson.edu/emc/tutorials/Shielding02/Practical_Shielding.html The basic idea seems to be that the wave falls off exponentially, with a characteristic length $ L \propto a/ \sqrt{1-((2a)/ \lambda)^2}$ I've approximated $\lambda_c \approx 2a$ as the cutoff frequency, and the dependence on the details of the geometry (circular cross-section, rectangular,...) is all built into a unitless constant of proportionality, which I think is just the quantity $T$ previously discussed in this answer. So in the limit of long wavelength, you basically get exponential attenuation through the thickness of the wall, with characteristic length of order $a$.

In the long-wavelength limit, where the field acts like DC, it might also be possible to compare all of this to a Faraday ice pail.

$\endgroup$
0
$\begingroup$

As the question arose from a question about water I give an answer first according to water waves.

In a slit you observe the wave on the boundary between two mediums. As you want to "dive" to a hole you will see the related movement of particles and the compression and decompression of these particles to each over from a wave in this area. What happens in the "depth" is - if not take into account the reflections from the ground - a dissipation process. Now the wave hits a wall, get reflected and lead on the boundary surface to a standing wave. In the depth this is never possible because or you have or reflection from the ground or dissipation in direction to the far away ground. So you have a nearly pure transverse wave only on the boundary surface between two media. In the depth you have more and more longitudinal waves. This is reason behind Hyugens principle about spherical waves.

Around your hole the reflected components of the incoming transverse AND longitudinal waves dissipate the incoming waves. Then smaller the hole then less energy remains. What remains is not a sinusoidal wave. So about resonance in the holes tube I think can not be a talk. And yes the longer the tube then more the interaction with the tubes wall take energy away from the medium compression and decompression and the mediums oszillation faded fast.

How one can see this part of the answer has nothing to do with an answer in the area of electromagnetic waves. And I'm since some time bewildered about how easy Young concluded from the always moving (!) pattern of water waves to the waves natur of light from fringes on a screen. There are more differences than commonalities. But the story ends in happiness because we know today about the oscillation of the electric and the magnetic fields of the photon.

What Bethe pointed out is much more important then Young's conclusions. He wrote in the above mentioned work: "The method is based on the use of fictitious magnetic charges and currents in the diffracting hole which has the advantage of automatically satisfying the boundary conditions on the conducting screen. The charges and currents are adjusted so as to give the correct tangential magnetic, and normal electric, field in the hole." In my understanding he take in attention the interaction between the EM radiation and the holes material. The same we have to do with every EM radiation so also to light.

So Huygens principle is for EM radiation obsolete or only some approximation. If photons "hit" the edge or if they are nearby a hole they have the chance to be reflected or to be assimilated (that causes some emission then) or - if their action radius is smaller the hole's diameter to go in. Inside the hav a lot of difficulties to escape the electric and magnetic fields of the hole. Some fields are from the surface electrons, some are induced from the photon. Only a action of a lot of collimated photons with the same frequency and the same phase could escape the tube and come out on the other side.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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