http://www.physicsforums.com/showthread.php?t=286513
2.43 GHZ equals a wavelegth of 12.3 cm, clearly this is much larger than the hole size of 0.5 cm found on my microwave's mesh screen, 24x larger in fact. So even if we were corrent on the 'half' wavelength theory, one can still see my microwave's holes are much smaller than they need to be. This makes me wonder if the mesh screen gives added protection around our 2.43 GHZ operating frequency, maybe 2.43 GHZ is the peak of the microwave's emission spectra and less intense radiation is emmited on either end of this 2.43 GHZ amplitudal peak? Maybe we are being given added protection since 0.5 cm corresponds to 60 GHZ.

http://answers.yahoo.com/question/index?qid=20110715153735AAz66ce
Look at a single hole as a circular waveguide of length = thickness of the oven shield.

A circular waveguide will not conduct waves with frequencies below the so-called "cutoff frequency". The lowest possible cutoff freq. for a circular waveguide of radius r is for the so-called TM01 mode, for which the cutoff freq. fc = p01/2pi*r*sqrt(mu0*e0) where

p01 is the first root of the zero-order Bessel function J0(x)=0 so x = 2.405 (you can look up Bessel functions in your CRC math handbook);

mu0 = permeability of free space = 1.26e-6 H
e0 = permittivity of free space = 8.86e-12 F/m.

I have looked at my oven & gauge that r ~ 3mm, not larger than that.

So fc = 2.405/[2pi*3e-4m*sqrt(mu0*e0)] =2.405/(6.28*3e-4*1.4e-9) = 9.1e11 Hz = 911 GHz.
Since f of your uwave oven is only about 2.3 GHz you are safe.

NOTE: cutoff means cutoff. There is NO penetration into the waveguide, no matter how short it is, i.e. how thin your shield is.

Attenuation for freq. ABOVE cutoff are a function of the metal's resistivity (and dielectric if not air) but those formulas do NOT apply to frequencies below cutoff.
Source(s):
Ramo & Whinnery, 'Fields and Waves in Modern Radio', Wiley, 2nd ed. pp76