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Fix time dependence $e^{-i\omega t}$ and use pressure amplitude for $\psi$. Write the incident acoustic plane wave as ${\psi }_i=A{e}^{i\alpha x-i\beta z}$, where $\alpha =k\cos \theta$, $\beta =k\sin \theta >0$. “Perfectly reflecting” specifies zero absorbed energy but does not uniquely select a boundary condition: a rigid wall has a Neumann boundary condition, whereas a pressure-release surface has a Dirichlet boundary condition. We first give the rigid-wall result, and then the alternative pressure-release result.

For the rigid surface the normal-velocity condition, without an immaterial normalizing factor, isExpand $\psi ={\psi }^{[0]}+{\psi }^{[1]}+{\psi }^{[2]}+\cdots$ in the height amplitude, scaling a fixed differentiable profile. The flat total field is ${\psi }^{[0]}=A{e}^{i\alpha x}(e^{-i\beta z}+e^{i\beta z})$. Taylor expansion givesLet $\hat{h}(q)=\int h(x)e^{-iqx}dx$ and $\beta (q)=\sqrt{k^2-q^2}$ on the outgoing sheet: it is positive real for propagating modes and positive imaginary for evanescent modes. Fourier transforming this boundary derivative gives $2A(k^2-\alpha q)\hat{h}(q-\alpha )$. The outgoing angular spectrum therefore yields the small-height Neumann scattering fieldFor an infinite stationary surface the Fourier quantities may be interpreted as random distributions or through a finite illumination window and its limiting mean. The correction is linear in $h$ and $h^{\prime }$, so zero mean height and stationarity giveOnly the rough correction has zero mean at first order; the flat reflected wave remains.

The first nontrivial coherent correction depends on correlation, not just the r.m.s. height. Define $C_h(s)=⟨h(x+s)h(x)⟩$, its power spectrum of surface height $S_h(q)=\int C_h(s)e^{-iqs}ds$, and ${\sigma }^2=(2\pi {)}^{-1}\int S_h(q)dq$. At second order the flat-field third normal derivative and mixed derivative vanish at zero, leavingFor each first-order outgoing Fourier component, its derivative data are $g_1(q)=2A(k^2-\alpha q)\hat{h}(q-\alpha )$, its second normal derivative is $i\beta (q)g_1(q)$, and its tangential derivative is $q{g}_1(q)/\beta (q)$. Averaging their products with $h,h^{\prime }$ gives the mean derivativeDividing by the specular outgoing derivative $i\beta$ therefore gives oblique coherent reflection from a stationary rigid rough surface:This second-order expression requires the relevant spectral moments to exist and the boundary expansion to remain controlled. Small $|kh|$ alone does not control arbitrarily large slopes or high-frequency roughness. The given stationarity and $\sigma$ determine the first-order mean above, but do not determine this spectrum-weighted correction numerically.

For a pressure-release surface, instead impose $\psi (x,h)=0$. The flat reflected coefficient is $-1$, and Taylor expansion gives ${\psi }^{[1]}(x,0)=2i\beta Ah{e}^{i\alpha x}$. Thus small-height Dirichlet scattering givesThe zero-mean first correction again vanishes on averaging. At second order ${\psi }^{[2]}(x,0)=-h{\partial }_z{\psi }^{[1]}(x,0)$, givingIf the roughness spectrum is concentrated near $q=\alpha$ after the incident tangential shift, the additional long-correlation approximation yields $r_N\simeq 1-2{\beta }^2{\sigma }^2$ and $r_D\simeq -1+2{\beta }^2{\sigma }^2$. A local Gaussian random-height phase model would then give $r_N=e^{-2{\beta }^2{\sigma }^2}$ and $r_D=-e^{-2{\beta }^2{\sigma }^2}$, but Gaussian statistics and that local approximation were not supplied. They must not be inferred from stationarity and r.m.s. height alone.

Work above the surface, with $e^{-i\omega t}$ time dependence. Define the scattered field by $\psi ={\psi }_i+{\psi }_s$, so it includes the reflection from a flat surface. For small-height Dirichlet scattering, write $h=\epsilon \eta$ and expandThe zeroth-order total field ${\psi }^{[0]}={\psi }_i+{\psi }_s^{[0]}$ satisfies the Dirichlet boundary condition ${\psi }^{[0]}(x,0)=0$. Taylor expansion at the perturbed boundary givesThus the first-order rough-surface scattered field has mean-plane boundary data $g_1(x)=-h(x){\partial }_z{\psi }^{[0]}(x,0)$.

Use the outgoing angular spectrum to solve this boundary-value problem. For $\hat{g}(q)=\int g(x)e^{-iqx}dx$, letThe branch ensures upward propagation or upward evanescent decay. Each component solves the Helmholtz equation, and $\mathcal{E}g$ has trace $g$ at $z=0$. ConsequentlyThis gives the scattered field through first order by adding the two contributions. If one reserves “rough scattered field” for the non-specular correction, it is ${\psi }_s^{[1]}$ alone; the convention here keeps the flat reflection as well.

The expansion is in height for a fixed sufficiently regular profile. The condition $|kh|\ll 1$ controls the incident wave's height expansion, but very short spatial scales can create large evanescent normal derivatives. The surface regularity and relevant spectral moments must also control the subsequent boundary expansions; small amplitude alone is not a uniform guarantee for arbitrarily fine roughness.

Small-height Dirichlet scattering expands the acoustic boundary condition about a mean plane for a fixed regular surface profile. The zeroth-order field includes the flat reflection; higher orders solve outgoing Helmholtz equation problems with induced boundary data. Small $|kh|$ controls the incident-field Taylor expansion, but rapidly varying roughness can require additional control of normal derivatives and spectral moments.