A friend of mine is interested in the answer to the following question:
I told him that I’m pretty sure the RF can’t get through the holes in the building if they are smaller than the wavelength, which I’m pretty sure is correct. I don’t know anything about the attenuation issues, though. Help from Smart Physics People is appreciated.
“RF can’t get through the holes in the building if they are smaller than the wavelength”
This is how I understand it, from the point of view of building enclosures to keep electromagnetic interference in or out. Note that the dimension of the hole for this purpose is its longest dimension.
Have your friend get a copy of the book Noise Reduction Techniques in Electronic Systems by Henry W. Ott, John Wiley and Sons.
He has an excellent dissertation on shielding. To quote from the book: As a practical matter, the intrinsic shielding is of less concern that the leakage through seams, joints and holes. … The amount of leakage from a shield discontinuity depends mainly upon 3 things; 1) The maximum linear dimension, 2) The wave impedance, 3) the frequency.
He goes on to say A rectangular [opening] forms a slot antenna. Such and entenna even if very narrow can cause considerable leakage it is longer than 1/100 wavelength.
He then gives a formula for attenuation of a shield having holes in it of maximum dimension l at wavelength L.
S in db = 20log(L/2l)
So it looks like the length of the metal studs is more critical than the spacing.
By the way. That formula was for one hole and the shielding effectiveness (attenuation) is reduced by approximately the square root of the number of holes.
I don’t believe metal studs will attenuate reasonably high frequencies very much.