= Solution
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 $h=\varepsilon\eta$ and expand
$$
\psi_s=\psi_s^{[0]}+\psi_s^{[1]}+O(\varepsilon^2).
$$
The 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 gives
$$
0=\psi^{[0]}(x,0)+h(x)\partial_z\psi^{[0]}(x,0)+\psi_s^{[1]}(x,0)+O(\varepsilon^2).
$$
Thus 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 $\widehat g(q)=\int g(x)e^{-iqx}dx$, let
$$
\beta(q)=\begin{cases}\sqrt{k^2-q^2},&|q|\leq k,\\i\sqrt{q^2-k^2},&|q|>k,\end{cases}\qquad (\mathcal E g)(x,z)=\frac1{2\pi}\int\widehat g(q)e^{iqx+i\beta(q)z}dq.
$$
The branch ensures upward propagation or upward evanescent decay. Each component solves the <Helmholtz equation>, and $\mathcal E g$ has trace $g$ at $z=0$. Consequently
$$
\boxed{\psi_s^{[0]}=-\mathcal E[\psi_i(\cdot,0)],\qquad \psi_s^{[1]}=-\mathcal E[h\,\partial_z\psi^{[0]}(\cdot,0)].}
$$
This 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 $|kh|\ll1$ 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.
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