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:
$$
\chi_1=\frac{f_{s,B}}{f_i}
=-\frac{\alpha}{4}e^{2iq_0z}.
$$
Hence, in the upper half-space,
$$
\boxed{
\psi_R(x,z)=e^{i(px-q_0z)}
\exp\left(-\frac{\alpha}{4}e^{2iq_0z}\right)}.
$$
Expanding the <exponential function> to first order shows that its reflected component is
$$
\boxed{\psi_{r,R}^{(1)}(x,z)
=-\frac{\alpha}{4}e^{i(px+q_0z)}}.
$$
The exponentiated expression is the Rytov approximation to the total field; only its term linear in $\alpha$ is a single specular reflected <plane wave>.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-335/1/ii/b/solution