I am doing research on Faraday cages for school, and I want to know how it works. Faraday cages can have holes in them, and if the diameter is smaller than the wavelength of waves you want to block, the Faraday cage blocks the waves. I have found a formula that the transmission of electromagnetic waves through a hole with diameter $d (< λ)$ equals $(d/λ)^4$. However, I cannot find anywhere why. It does not make sense in my head, for me it seems like a truck cannot be too long to go through a tunnel, but why is this not the same for waves? My guess is that it has to do with quantum mechanics, can someone please explain to me how this works? Thanks!
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$\begingroup$ A Faraday cage is not supposed to shield EM waves. It is about static electric fields. The charges distribute along the cage so that the E field cancels in the interior. Hole size is not directly important. That EM waves cannot penetrate holes with small diameter is property of wave propagation, and not about quantum mechanics. It is also true for other wave types. But I have no graphic explanation why this is so (Well it comes out of the wave equation..., but this is probably not really an explanation to you). $\endgroup$– Andreas H.Commented Dec 27, 2014 at 11:32
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4$\begingroup$ @AndreasH. Why do you think a Faraday cage is not supposed to shield EM waves? I think the cage on the front door of a microwave oven is supposed to do exactly that. $\endgroup$– ProfRobCommented Dec 27, 2014 at 11:41
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$\begingroup$ @AndreasH. I also think that hole size is important. If the hole becomes big, the cage is ineffective. What I know is that an excited atom placed in a cavity with linear dimension less than the wavelength of the photon to be emitted, doesn't undergo de-excitation because of destructive interference. But of a formula with $(d/ \lambda)^4$ I never heard. $\endgroup$– SofiaCommented Dec 27, 2014 at 11:53
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3$\begingroup$ duplicate of physics.stackexchange.com/q/141556 and physics.stackexchange.com/q/141562 $\endgroup$– user4552Commented Dec 27, 2014 at 15:49
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1$\begingroup$ @RobJeffries: Andreas makes an important point in that a Faraday screen doesn't work because of the holes - it works because it's a conductor and in spite of the holes. $\endgroup$– John RennieCommented Dec 28, 2014 at 6:19
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2 Answers
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The original work on this, and where your formula come from is Bethe (1944) who considered plane waves passing through a zero-thickness conductor, with a hole of radius much smaller than the wavelength of the light. Unbelievably, this paper is behind a paywall, so I can't (won't) read it. From what I can gather, a solution to Maxwell's equations is found such that the scenario can be treated as if there were a small magnetic dipole (current loop) located in the plane of the hole. The incoming light excites oscillations of this dipole, which radiates electromagnetic radiation. The power radiated from a magnetic dipole is proportional to $(d/\lambda)^4$, where $d$ is the diameter of the current loop.
In general, it is wrong to think of some of the light "hitting" the screen and some of it passing through the hole. The light consists of oscillating electric and magnetic fields that induce oscillating currents in the conductor. These currents then themselves emit electromagnetic waves (at the same frequency) that can either constructively or destructively interfere. The wavelength dependence here should really be thought of as a frequency dependence $f = c/\lambda$ and controls the efficiency with which the electrons in the conductor can interact with electromagnetic waves. The same frequency dependence is found from the scattering of electromagnetic radiation by small particles/atoms - Rayleigh scattering.
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$\begingroup$ Thank you so much for this answer! If I understand correctly, this is how it works: When a lightwave approaches a hole with diameter d, the oscillations with a certain frequency push away and attract electrons in the material, such that those electrons themselves will cause electromagnetic waves. The amount depends on (d/λ)^4 and thus some light is blocked and some is not. The stronger the (kind-of virtual ?) dipole the less light comes through. Is this correct? $\endgroup$ Commented Dec 31, 2014 at 12:29
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$\begingroup$ @EHekkelman I think it would be well worth looking at the duplicate flagged by Ben Crowell. The first part of your paraphrasing looks ok, but the virtual magnetic dipole is what is radiating from the hole, so the smaller that is (and the longer the wavelength), the less efficient it becomes. $\endgroup$– ProfRobCommented Dec 31, 2014 at 16:55
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$\begingroup$ A link to Bethe, not behind a paywall web.stanford.edu/class/ee349/Handouts/Bethe_PR1944.pdf $\endgroup$ Commented May 29, 2021 at 5:53
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The diameter-to-the-fourth relationship is easy to understand. If you are looking a hole in a metal sheet, the size of that hole represents the source of the radiation coming through from the other side. If you have two such sources far apart, then you get twice as much power. But if they are close together...closer than the wavelength of the radiation...then they represent two coherent sources whose amplitudes add. So you get FOUR times the power. Diameter-to-the-fourth. (Because it doesn't matter if the two holes are merged into one.)
I don't think this is helpful for explaining the mechanism of the Faraday cage. I think the Faraday cage becomes highly effective when the cage dimension is anywhere close to the wavelength...in other words, I think the attenuation is probably much more than 10,000 for a cage spacing of lambda-by-10. (More precisely, I think if the attenuation is x at lambda-by-y, it is much more than 10000x at lambda-by-10y.)
I'm basing this gut feeling on the analysis I did of the parallel-wire filter here:Particles, waves and parallel wire filters. Transmission formula? The key to this analysis is to look at the reflection/transmission properties of a conducting sheet. Naturally, if the resistance is zero, you get total reflection. The funny thing turns out to be that if you add resistance, there are no solutions whereby
(incoming power) = (transmitted power) + (reflected power)
because all solutions must include some power loss in the resistance of the sheet. So for the faraday cage, assuming no resistance, once the wires are close enough together, where are the losses? So it becomes very difficult for the filter to do anything other than total reflection.
answered Dec 27, 2014 at 21:37
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$\begingroup$ But if they are close together...closer than the wavelength of the radiation...then they represent two coherent sources whose amplitudes add. So you get FOUR times the power. Diameter-to-the-fourth. (Because it doesn't matter if the two holes are merged into one.) Two mistakes here. (1) Two holes side by side do not constitute a hole with twice the diameter. (2) If power goes like the fourth power of the diameter, and you double the diameter, then you get 16 times the power, not four times. For a correct explanation of the exponent, see the questions that this question duplicates. $\endgroup$– user4552Commented Dec 29, 2014 at 3:15
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$\begingroup$ Oh come on. If you merge the two holes into one, the diameter becomes squrt(2). And the power is 4 times: that is, diameter to the fourth. Like I said in the first place. For a correct explanation of the exponent, re-read what I wrote in my answer. $\endgroup$ Commented Dec 29, 2014 at 3:41