= Solution
Let $Z_j=\rho_jc_j$, $k_j=\omega/c_j$, and $\phi=k_1L$. Use the harmonic convention $e^{-i\omega t}$ and write the pressure and velocity amplitudes as
$$
\begin{array}{lll}
x<0:&p=e^{ik_0x}+Re^{-ik_0x},&u=Z_0^{-1}(e^{ik_0x}-Re^{-ik_0x}),\\
0<x<L:&p=Ae^{ik_1x}+Be^{-ik_1x},&u=Z_1^{-1}(Ae^{ik_1x}-Be^{-ik_1x}),\\
x>L:&p=Te^{ik_0(x-L)},&u=Z_0^{-1}Te^{ik_0(x-L)}.
\end{array}
$$
Continuity of pressure and normal velocity at both interfaces is equivalently the layer transfer relation
$$
\binom{p(0)}{u(0)}=
\begin{pmatrix}
\cos\phi&-iZ_1\sin\phi\\
-iZ_1^{-1}\sin\phi&\cos\phi
\end{pmatrix}
\binom{p(L)}{u(L)}.
$$
Substituting $p(0)=1+R$, $u(0)=(1-R)/Z_0$, $p(L)=T$, and $u(L)=T/Z_0$, then adding the two resulting equations, gives
$$
2=T\left[2\cos\phi-i\left(\lambda+\lambda^{-1}\right)\sin\phi\right],
\qquad
\lambda=\frac{Z_1}{Z_0}.
$$
Hence the <acoustic transmission through a uniform layer> is
$$
\boxed{|T|=\left[\cos^2(k_1L)+\frac14(\lambda+\lambda^{-1})^2\sin^2(k_1L)\right]^{-1/2}.}
$$
For either $\lambda\ll1$ or $\lambda\gg1$, the strong <acoustic impedance> mismatch makes transmission small except near $k_1L=n\pi$, where $|T|=1$. Away from those resonances,
$$
|T|\sim\frac{2\lambda}{|\sin k_1L|}\quad(\lambda\ll1),
\qquad
|T|\sim\frac{2}{\lambda|\sin k_1L|}\quad(\lambda\gg1).
$$
At the antiresonant points $k_1L=(n+\tfrac12)\pi$, the minimum is $2/(\lambda+\lambda^{-1})$.
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