Minimum thickness of radome cover

The plastic sheet has air on both sides. A slab with the same medium on both sides gives zero reflection when its thickness is a multiple of half the wavelength inside the slab (a half-wave window, or radome). The wave reflected from the first face then cancels the one reflected from the second face, and the slab behaves as if it were not there.

f = 10 GHz,  ε_r = 2.25,  μ_r = 1

λ₀ = c/f = 3×10⁸ / 10×10⁹ = 3 cm
n  = √ε_r = 1.5
λ  = λ₀/n = 3 / 1.5 = 2 cm      (wavelength in the plastic)

Condition: β₂d = mπ, i.e. d = mλ/2 (m = 1, 2, 3, ...). For the minimum thickness take m = 1:

d = λ/2 = 2 cm / 2 = 1 cm

Check: with η₂ = η₀/1.5 = 251 Ω, the input impedance seen at the first face is Z_in = η₂ (η₃ + jη₂ tanβ₂d)/(η₂ + jη₃ tanβ₂d). With β₂d = π, tan β₂d = 0, so Z_in = η₃ = η₀ = η₁ and Γ = 0.

Answer: minimum thickness = 1 cm (any multiple of 1 cm also works).