I read about the faraday cages and understood that the size of the holes should be very small than the wavelength of the signal for effective shielding.

But I am really confused due to the illustrations of the high and low wavelength signals.

Like, if there is a hole, of a certain size, doesn't matter about the frequency of the signal, the signals (assume a sine wave) with low amplitude , comparable to the size of the hole, will pass through and signals will amplitude higher than the size of the hole, will be blocked. But I know this is wrong. But due to the illustrations , I can think of only the amplitude to be the blocking factor.

Can someone provide an illustration as the why the wavelength of the signal is the limiting factor and not frequency or amplitude.


1 Answer 1


The frequency does matter. But that is because $\lambda = c/f$ - the wavelength and frequency aren't independent.

When you see pictures that are supposed to represent plane electromagnetic waves, like the one below, the amplitude that is shown has nothing to do with the spatial extent of the field. To draw some sinusoidally varying function on a page requires you to draw something going up and down, but this is the size of the electric field at some point in space along the propagation direction, not any indication of where the field is or how it varies perpendicular to the propagation direction. A large amplitude wave does not necessarily have a greater extent in that perpendicular direction. Indeed, if the plane electromagnetic waves are incident normally on the cage material, then the electric field will have the same value at all points on the interface.

electromagnetic wave

If the wavelength of the electromagnetic wave is significantly bigger than the holes, then the interface just acts like a conductive reflector. As the frequency gets higher and the wavelength becomes smaller than the holes, the holes start to act like little waveguides that allow the propagation of the waves. That is not because the incident electric field varies significantly across the hole (it could be the same for normal incidence), but because the fields generated by the oscillating electrons in the conductive mesh have a wavelength shorter than the hole dimensions.

  • $\begingroup$ Thank you for the answer. Any illustration can you provide for the hole incidence scenario? $\endgroup$
    – Newbie
    Jun 27, 2021 at 9:10

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