With the <scattering potential> $V(\mathbf r)=k_0^2(1-n^2(\mathbf r))$, the total field obeys
$$
(\nabla^2+k_0^2)\psi=V\psi.
$$
The outgoing Green function is $G_0(\mathbf r)=e^{ik_0|\mathbf r|}/(4\pi|\mathbf r|)$. The <Lippmann-Schwinger equation> and the first <Born approximation> give
$$
\psi_s(\mathbf r)simeq
-\int_DG_0(\mathbf r-\mathbf r')V(\mathbf r')
e^{ik_0\widehat{\mathbf x}_0\cdot\mathbf r'},d\mathbf r'.
$$
For $r\to\infty$,
$$
G_0(\mathbf r-\mathbf r')
\sim\frac{e^{ik_0r}}{4\pi r}
e^{-ik_0\widehat{\mathbf r}\cdot\mathbf r'},
$$
so
$$
\boxed{
\psi_s(\mathbf r)\sim\frac{e^{ik_0r}}r
f_\infty(\widehat{\mathbf x}_0,\widehat{\mathbf r}),
\qquad
f_\infty=-\frac1{4\pi}
\int_DV(\mathbf r')e^{-i\mathbf q\cdot\mathbf r'}d\mathbf r',}
$$
where $\mathbf q=k_0(\widehat{\mathbf r}-\widehat{\mathbf x}_0)$ is the <momentum transfer>. Thus the <far-field pattern> is $-1/(4\pi)$ times the Fourier transform of $V$ at the measured transfer vectors. If sufficiently many incident directions and frequencies supply all $\mathbf q$, the formal reconstruction is
$$
\boxed{V(\mathbf r)=-4\pi\mathcal F^{-1}
\{f_\infty(\mathbf q)\}(\mathbf r),}
$$
with constants adjusted to the chosen Fourier convention.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 335 / 2 / i / Solution

{title2=$V$}

For a refractive index $n(\mathbf r)$ and background wavenumber $k_0$, one sign convention defines the scattering potential by $V=k_0^2(1-n^2)$. The total field then satisfies $(\Delta+k_0^2)\psi=V\psi$.


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