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The Rytov approximation at first order writes the total profile as $f_R=f_i{e}^{{\chi }_1}$ and identifies the first logarithmic perturbation with the Born relative field:Hence, in the upper half-space,Expanding the exponential function to first order shows that its reflected component isThe exponentiated expression is the Rytov approximation to the total field; only its term linear in $\alpha$ is a single specular reflected plane wave.

The Taylor expansion of the square root isSubstitution into the exact reflection coefficient givesConsequentlyThis is exactly the reflected field furnished by both the Born approximation and the term linear in $\alpha$ in the Rytov approximation. Thus the exact, Born, and Rytov fields agree through first order. If the Rytov exponential is retained without re-expansion, its higher powers generate spatial harmonics $e^{i(px+(2m-1)q_{0}z)}$; those terms are part of the approximation and should not be confused with the exact interface's single reflected wave.

Since $n=1+W$ with small variance,To leading order, $V$ is therefore a centered stationary Gaussian random field. Let its autocorrelation function of a random field beThe $W^2$ contribution gives higher-order mean and non-Gaussian corrections and is consistently omitted at this order.

The far-field Rytov approximation from part ii has unit incident intensity andThe real random variable $X={\varphi }_1+{\varphi }_1^{\ast }$ is centered Gaussian. Its moment-generating function givesDefineThenand henceThis expression depends only on the two-point autocorrelation of the scattering potential. In the weak-fluctuation expansion it becomes

Use the sign convention in the question,Then the total field satisfiesThe outgoing free-space Green function iswith $({\nabla }^2+k_0^2)G_0=-\delta$. Hence the Lippmann-Schwinger equation isReplacing the unknown interior total field by the incident field gives the first Born approximation

For the Rytov approximation, put $\psi ={\psi }_i{e}^{\varphi }$. After division by $\psi$, the wave equation givesNeglecting the quadratic term $(\nabla \varphi {)}^2$ makes ${\psi }_i\varphi$ obey the same inhomogeneous equation as the first Born scattered field. ThusandIts power series beginsso the two approximations agree through first order in the scattering potential.