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 limiting frequency.

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 light.

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



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