How does a Faraday Cage work? or, Why can a satellite dish have holes in it?
Asked by: Undisclosed Visitor
In electromagnetic field theory, when one solves for the distribution of a field in and around a
conducting surface, one finds that electromagnetic waves do not penetrate very far through holes
that are less than about a wavelength across. Equivalently, the reflection of electromagnetic
waves from a conducting surface is not much affected by holes (or other irregularities) in the
surface less than about a wavelength across. Therefore, it is possible to make a Faraday Cage or a
satellite dish with 'holes' without materially affecting performance, provided that the holes are
less than about a wavelength across.
In the case of a satellite dish, it is the maintenance of the reflection characteristics of the
dish, even though it may consist of a mesh with substantial holes (to the eye, which, not
accidentally, utilizes wavelengths very much shorter than radio waves), that matters. In the case
of a Faraday Cage, it is the maintenance of adequate attenuation of radio waves, due to the
difficulty of penetrating the holes, that matters.
If the wavelength is reduced to the extent that the holes become substantially greater than a
wavelength across, then the performance characteristics of a Faraday Cage or satellite dish are
materially affected. So, for a given mesh size, a Faraday Cage fails to shield adequately above
some 'cutoff' frequency. Likewise, a satellite dish fails to perform efficiently above some
Generally, the tolerance of one wavelength of the radiation is taken to be the approximate cross
over point between satisfactory and unsatisfactory performance. However, as a practical matter,
holes, or surface variations, must usually be less than a fraction of a wavelength across in
order not to impose an unacceptable performance degradation. This is why, for example, optical
telescope mirrors are figured to within a fraction (a 1/10th, say) of a wavelength of visible
Another familiar consequence of this same principle is that, in imaging, it is not possible to
resolve features that are much less than approximately a wavelength across. So, for example, to
produce finer masks for photoetching of transistors on silicon wafers, consideration is now being
given to using ultraviolet light, rather than light in the visible part of the spectrum, to take
advantage of the shorter wavelength, and better resolution, in the ultraviolet. Likewise, an
optical microscope cannot resolve features much less than a wavelength of visible light across.
Answered by: Warren Davis, Ph.D., President, Davis Associates, Inc., Newton, MA USA
'Where the telescope ends, the microscope begins. Which of the two has the grander view?'