Fresnel diffraction approximation (parabolic waves) The Huygens-Fresnel principle (Introduction to Fourier Optics, Goodman),
$$ U(x,y)=\frac{z}{i\lambda}\int_\Sigma U(\xi,\eta)\frac{e^{ikr}}{r}d\xi d\eta\,, $$
where $\cos \theta=\frac{z}{r}$, shows that the field produced by a punctual source propagates as spherical waves because of the phase $ \vec{k}\cdot \vec{r}$. Is it safe to say that in the Fresnel approximation,
$$ U(x,y)=\frac{e^{ikz}}{i\lambda z}\int U(\xi,\eta) e^{\frac{ik}{2z}[(x-\xi)^2+(y-\eta)^2]}d\xi d\eta \, ,$$
the field propagates as parabolic waves, since the phase takes the form of a paraboloid?
 A: I had this doubt once. Shape of a wavefront is a surface of constant phase in space. So, when you combine $e^{ikz}$ & $e^{\frac{ik}{2z}[(x-ξ)^2 + (y-η)^2]}$, the total phase comes out to be $kz + {\frac{k}{2z}[(x-ξ)^2 + (y-η)^2]} $ and when you find the surfaces of constant phase they turn out to be ellipses. The parabolic wavefronts will be realized when you make an assumption that when z varies then variation in  ${\frac{k}{2z}[(x-ξ)^2 + (y-η)^2]} $ is much less then variation in $kz$ due to variation in z. This can be justified by taking partial derivative of both ${\frac{k}{2z}[(x-ξ)^2 + (y-η)^2]} $  & $kz$ with respect to z, you will find that variation due to z in the denominator is less, because its derivative involves $\frac{1}{z^2}$, than the variation due to z in numerator as it doesn't involve any $\frac{1}{z^2}$ kind of decreasing term. So, you can treat $z$ in the denominator of ${\frac{k}{2z}[(x-ξ)^2 + (y-η)^2]} $ as a constant, say c, and in $kz$ as variable. The surface of constant phase will have the equation as $$kz + {\frac{k}{2c}[(x-ξ)^2 + (y-η)^2]} = \gamma$$ ,where $\gamma$ is the constant phase for a surface. This gives you a paraboloid with its axis as $x=ξ$ & $y=η$ and vertex as $(ξ,η,\frac{\gamma}{k})$. As the phase changes, vertex of the paraboloid changes.
