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The normal acoustic impedance of a progressive acoustic wave at a planar boundary is the pressure amplitude divided by the normal-velocity amplitude. For propagation angle $\theta$ from the normal,The distinction from the acoustic impedance $\rho c$ matters at oblique incidence: interface reflection depends on the ratio of normal impedances. For an evanescent outgoing wave $k_n$ is imaginary and the normal impedance is reactive, giving no mean normal acoustic energy flux.

For a progressive plane wave, writeThe momentum equation gives ${\rho }_0\omega U=kP$, while the dispersion relation is $\omega =c_{0}k$. Therefore the pressure and longitudinal velocity amplitudes obeyThe constant $Z$ is the acoustic impedance.

Let $Z_j={\rho }_j{c}_j$, $k_j=\omega /c_j$, and $\varphi =k_{1}L$. Use the harmonic convention $e^{-i\omega t}$ and write the pressure and velocity amplitudes asContinuity of pressure and normal velocity at both interfaces is equivalently the layer transfer relationSubstituting $p(0)=1+R$, $u(0)=(1-R)/Z_0$, $p(L)=T$, and $u(L)=T/Z_0$, then adding the two resulting equations, givesHence the acoustic transmission through a uniform layer isFor either $\lambda \ll 1$ or $\lambda \gg 1$, the strong acoustic impedance mismatch makes transmission small except near $k_{1}L=n\pi$, where $|T|=1$. Away from those resonances,At the antiresonant points $k_{1}L=(n+\frac{1}{2})\pi$, the minimum is $2/(\lambda +{\lambda }^{-1})$.

No reflection requiresUsing $\lambda =\rho c_P^2$, ${\lambda }^{\prime }={\rho }^{\prime }(c_P^{\prime }{)}^2$, and
$\sin {\theta }^{\prime }=(c_P^{\prime }/c_P)\sin \theta$, this reduces to the normal acoustic impedance matching conditionAfter squaring and eliminating ${\theta }^{\prime }$, the required incidence angle satisfiesSuch a no-reflection angle exists only when the right-hand side lies in $[0,1]$ and the transmitted wave is propagating.