Why are electromagnetic waves not able to pass through a hole with a diameter smaller than the wavelength? 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!
 A: 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.
A: 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. 
