The shielding efficiency of an aperture depends on e−αd, where α is the frequency-dependent attenuation constant of the aperture and d is the thickness of the material (or the cutoff waveguide) at frequencies below cutoff.
Calculate α for an air-vent (unshielded) of dimension 0.5×0.3×0.8 m3.
The solution given is
Cutoff frequency = 300MHz (cutoff wavelength = 1 m), α = 5.92
The Attempt at a Solution
I've tried to apply the formula above, with
ωc = 2π(300×106), μ (free space) = 4π×10−7, ϵ (free space) = 8.85×10−12, f = 100×106, fc = 300×106
But I get α = 37.25, not the value given. What am I missing?