If Z1=1.6 MRayl and Z2=5 MRayl, approximately what percent of the incident energy is reflected at normal incidence?

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Multiple Choice

If Z1=1.6 MRayl and Z2=5 MRayl, approximately what percent of the incident energy is reflected at normal incidence?

Explanation:
At normal incidence, how much energy reflects depends on the impedance mismatch between the two media. The amplitude reflection coefficient is Γ = (Z2 − Z1)/(Z2 + Z1), and the energy reflection coefficient is R = Γ^2. Compute with Z1 = 1.6 and Z2 = 5 (in MRayl): Γ = (5 − 1.6)/(5 + 1.6) = 3.4/6.6 ≈ 0.515. Then R ≈ 0.515^2 ≈ 0.265, which is about 26.5%. So about 26.5% of the incident energy is reflected, with the remainder (~73.5%) transmitted (ignoring absorption). The sizable mismatch explains the substantial reflection.

At normal incidence, how much energy reflects depends on the impedance mismatch between the two media. The amplitude reflection coefficient is Γ = (Z2 − Z1)/(Z2 + Z1), and the energy reflection coefficient is R = Γ^2.

Compute with Z1 = 1.6 and Z2 = 5 (in MRayl): Γ = (5 − 1.6)/(5 + 1.6) = 3.4/6.6 ≈ 0.515. Then R ≈ 0.515^2 ≈ 0.265, which is about 26.5%.

So about 26.5% of the incident energy is reflected, with the remainder (~73.5%) transmitted (ignoring absorption). The sizable mismatch explains the substantial reflection.

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