Solution

Let

ω = k c_{P}

, and let

ψ

be the transmission angle. Since

\overset{c}{^}_{P} = 2 c_{P}

, tangential phase matching gives

k sin ϕ = \hat{k} sin ψ, \hat{k} = \frac{k}{2}, sin ψ = 2 sin ϕ .

The assumption of no evanescence requires

sin ϕ \leq 1/2

With a common suppressed factor

e^{- iω t}

, take

u_{I} u_{R} u_{T} = (sin ϕ, 0, cos ϕ) e^{ik (x s i n ϕ + z c o s ϕ)}, = R (sin ϕ, 0, - cos ϕ) e^{ik (x s i n ϕ - z c o s ϕ)}, = T (sin ψ, 0, cos ψ) e^{i \hat{k} (x s i n ψ + z c o s ψ)} .

For an inviscid liquid, continuity of normal displacement and normal stress gives

cos ϕ (1 - R) = T cos ψ, Z (1 + R) = \hat{Z} T,

where

Z = λ / c_{P}

and

\hat{Z} = \hat{λ} / \overset{c}{^}_{P}

. Solving,

R = \frac{Z ^ cos ϕ - Z cos ψ}{Z ^ cos ϕ + Z cos ψ}, T = \frac{2 Z cos ϕ}{Z ^ cos ϕ + Z cos ψ} .

The condition

∣ R ∣ = ∣ T ∣

gives

∣ \hat{Z} cos ϕ - Z cos ψ ∣ = 2 Z cos ϕ .

The sign that could instead produce

Z cos ψ = (\hat{Z} + 2 Z) cos ϕ

leads to

sin^{2} ϕ > 1

and is inadmissible. Hence

Z cos ψ = (\hat{Z} - 2 Z) cos ϕ .

Using

cos^{2} ψ = 1 - 4 sin^{2} ϕ

and squaring gives

sin^{2} ϕ = \frac{3 Z ^{2} - 4 Z Z ^ + Z ^ ^{2}}{Z ^ ( Z ^ - 4 Z )} .

Solved by gpt-5.6-sol high.

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