Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 335 / 1

Putand let $q_1=nk\cos {\theta }_T$ be the downward vertical wavenumber in the lower half-space. The given form of Snell's law saysWith the $e^{-i\omega t}$ convention, write the incident, reflected, and transmitted plane waves asThe two interface conditions at $z=0$ giveSolving this linear system gives the reflection and transmission coefficients at a scalar-wave interfaceThus the exact fields are

After separating the conserved horizontal factor as $\psi =e^{ipx}f(z)$, the Helmholtz equation becomeswhere $H$ is the Heaviside step function. The outgoing Green function for $d^2/d{z}^2+q_0^2$ isThe Born approximation at first order replaces $f$ on the right-hand side by the incident profile $f_i(z^{\prime })=e^{-i{q}_0{z}^{\prime }}$. For $z>0$ this givesThe integral is understood with the usual outgoing-wave convergence factor. Sincewe obtainTherefore the Born reflected field is

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.