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Take $e^{-i\omega t}$ time dependence and measure angles from the upward normal. Write $p_i=k\sin {\theta }_i$, $q_i=k\cos {\theta }_i>0$ and choose incident amplitude $A_i$. Separate the flat Dirichlet reflection from the rough correction:Taylor expansion of the boundary condition about $z=0$ givesSince ${\partial }_z{\psi }^{[0]}(x,0)=-2i{q}_i{A}_i{e}^{i{p}_{i}x}$, the induced trace isUse $\hat{h}(Q)=\int h(x)e^{-iQx}dx$ and $\beta (\xi )=\sqrt{k^2-{\xi }^2}$ with the outgoing branch: $\beta >0$ for $|\xi |<k$ and $\mathrm{Im}\beta >0$ outside. The outgoing angular spectrum solves the Helmholtz equation with that trace, giving the first-order rough-surface scattered fieldIf “scattered field” includes the specular reflection, add $-A_i{e}^{i{p}_{i}x+i{q}_{i}z}$ to this rough correction. Fourier transforms of an infinite random surface are generalized objects; finite-window transforms give an ordinary integral and are also needed for its far-field intensity. The expansion assumes a fixed regular surface shape scaled to small height. Highly oscillatory profiles can require additional control of slopes and spectral moments; small height alone is not a uniform error estimate for every profile.

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.

For the incident acoustic plane wave, set $q_0=k\sin \theta$ and ${\beta }_0=k\cos \theta >0$. The flat reflected field and total field areTherefore ${\partial }_z{\psi }^{[0]}(x,0)=-2i{\beta }_0{e}^{i{q}_{0}x}$ and the first-order rough-surface scattered field isIt is linear in the height. Since $⟨h(x)⟩=0$, its mean vanishes, with the expectation interpreted through finite windows or stationary spectral distributions when needed. HenceThe coherent first-order reflection is the flat-surface reflection. If the field symbol is instead used only for the rough correction, its first-order mean is zero. Stationarity ensures the coherent reflection retains the incident horizontal wavenumber, but zero mean height already explains the vanishing linear correction. The nonzero root mean square height does not enter this mean at first order; it does enter the fluctuating reflected field and its intensity.