in treating a scattering problem, the incident particles are treated as a plane matter wave in Schrödinger equation. a plane wave is:

$\phi(r') = \frac {1}{\sqrt{2 \hbar}^3} Exp( i k' \cdot r')$

and the scattered wave is combined a spherical wave with un-scattered plane wave :

$Exp( \pm i k z ) \rightarrow Exp( i k z ) + f ( \theta ) \frac { Exp ( i k r ) } {r}$

Recall from the scatter equation:

$\psi(r) = \phi(r) +\int {G(r,r') U(r') \psi(r') dr'}$

where the integral is taking inside the nucleus or the effective field from the nucleus. the Born first approximation is that, when the nuclear interaction on the incident wave is weak, that the wave within the potential does not distored, the wave function inside the integral is equal to the plane wave.

$\psi(r') \simeq \phi(r') = \frac {1}{ \sqrt{ 2 \hbar } ^3 } Exp ( i k' \cdot r' )$

since the distance is far away, the Green function can be approximated by

$G(r,r') \simeq - \frac{1}{4 \pi} \frac {1}{r} Exp( i k \cdot r - i k' \cdot r')$

combined everything,

$f(\theta) = \frac {m } { 2 \pi \hbar^2 } \int { V(r') Exp ( i (k-k') \cdot r' )dr'}$

we can see that it is a Fourier transform of the potential. and since the amplitude of the spherical wave is related with the differential cross section by:

$\frac { d \sigma }{ d \theta } = | f(\theta) |^2$

thus we can see why the differential cross section can reveal the nuclear potential.