We have a question on cooking asking about the safety of a microwave with a hole in it. The question isn't totally clear whether the hole has actually penetrated shielding or just the plastic wall of the microwave, but for the sake of argument, let's say you have a microwave with a hole clean through it.

How big does that hole have to be, compared to the wavelength of the microwaves, to leak enough energy to worry about? (Surely not the full wavelength as claimed by the existing answer?)

I've looked through several past questions about this sort of thing, but the best I've found so far is this answer:

However you wish to visualize the principals that govern how a Faraday cage works, it is well established that to block a transmission of a particular frequency, size of the largest hole in the Faraday cage must be AT MOST 1/2 the wavelength of the frequency of the undesired transmission.

Which is a bit less clear than I was hoping for - does that mean that at half the wavelength, nothing gets out? Does that apply even to the thin type of shielding typically used in microwaves, at least on the door? And if it does apply, how does the effectiveness decrease once you pass that cutoff?


The leaking of the microwave radiation through a hole out of the total metallic enclosure of a microwave oven becomes probably significant around a sizable fraction (like 1/4 to 1/2 ) of the wavelength of the radiation, which is 12.2 cm at the typical frequency of 2.45 GHz.The exact leakage depends on the size of the metal chamber and the position and shape of the hole. So when the hole (like the ones in the metallic grid of the door) is much smaller than this there will be negligible leakage. The metal thickness can be much thinner to be effective for shielding.

  • $\begingroup$ Do you by chance have sources or explanation for the 1/4 to 1/2? $\endgroup$ – Cascabel Sep 20 '16 at 6:12
  • $\begingroup$ In essence the problem is comparable to the escape of an electromagnetic wave through an aperture in a thin metal shield (be it light or microwave radiation). The escaping power is proportional to the forth power of the diameter to wavelength ratio. You will find more information on this in the answers given in physics.stackexchange.com/questions/141562/… . There you will also find a link to a paper by Hans Bethe on this problem. $\endgroup$ – freecharly Sep 20 '16 at 7:10
  • $\begingroup$ Yes, I'm aware that microwaves are electromagnetic waves and I asked about them escaping through an aperture :) Could be good information to put in your answer then? I guess the 1/4 to 1/2 is based on you considering 1/256 to 1/16 to be significant? (Also looks like that calculation is for a plane wave, but I guess the scaling is probably still right?) $\endgroup$ – Cascabel Sep 20 '16 at 12:15
  • $\begingroup$ I found this other answer when looking for a source for the explanation. Although is is obviously not a plane wave in the microwave oven, the result should be very similar, because you can decompose the standing wave pattern in the cavity into plane waves and the boundary conditions at the hole are similar. The result should give you an idea of the magnitude of power leakage for a given hole diameter. What one considers significant number is an assumption, it is relative to the power flow in a plane wave with the area of the hole. For the cavity waves you have to make appropriate estimates. $\endgroup$ – freecharly Sep 20 '16 at 15:14

When the hole is smaller than the wavelength of the signal, the transmission of the signal through the hole is proportional to $(r/\lambda)^4$ where $r$ is the radius of the hole and $\lambda$ is the wavelength of the signal.


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