Does Fraunhofer diffraction also automatically imply that the Fresnel approximation is simultaneously satisfied? I am confused about the regimes of validity for the Fresnel and Fraunhofer diffraction approximations, and would appreciate some clarification. Let's say we are interested in calculating the field $U_2(x,y)$, given a known input field $U_1(\xi,\eta)$ in the following coordinate system:

The Rayleigh-Sommerfeld diffraction integral is a general solution, and is good as long as we are assuming scalar diffraction theory and considering distances much greater than the wavelength of light ($r_{01}\gg\lambda$):
$$
U_2(x,y) = \frac{z}{i\lambda} \iint_\Sigma U_1(\xi,\eta)\frac{\textrm{exp}(ik\,r_{01})}{r_{01}^2}\,d\xi\, d\eta\;, \tag{1}
$$
$$
\textrm{where}\hspace{0.5cm}r_{01} = \sqrt{ z^2 + (x-\xi)^2 + (y-\eta)^2 } \tag{2}
$$
is the distance from point $P_1$ to $P_0$.
Fresnel Approximation
This is done by applying a binomial expansion to $r_{01}$, and retain only first two terms to approximate $r_{01}$ in the exponential to be
$$
r_{01} = \sqrt{ z^2 + (x-\xi)^2 + (y-\eta)^2 } \;\approx z \;\Bigg[ 1 + \frac{1}{2}\bigg(\frac{x-\xi}{z}\bigg)^2 + \frac{1}{2}\bigg(\frac{y-\eta}{z}\bigg)^2 \Bigg]. \tag{3}
$$
We also approximate $r_{01}^2\approx z^2$ in demoninator of Eq. (1), to obtain the Fresnel integral
$$
\begin{split}
U_2(x,y) &= \frac{\textrm{exp}(ikz)}{i\lambda z} \iint_\Sigma U_1(\xi,\eta)\; \textrm{exp}\Bigg( \frac{ik}{2z} \bigg[ (x-\xi)^2 + (y-\eta)^2 \bigg] \Bigg)\,d\xi\, d\eta \hspace{2.4cm} (4)\\
&= \frac{\textrm{exp}(ikz)}{i\lambda z} \textrm{exp}\bigg( \frac{ik}{2z}\big[x^2+y^2\big]\bigg)\times\; ... \\ 
&\hspace{1.5cm}\iint_\Sigma U_1(\xi,\eta)\; \textrm{exp}\bigg( \frac{ik}{2z}\big[\xi^2+\eta^2\big]\bigg) \; \textrm{exp}\bigg(-\frac{2\pi i}{\lambda z}\big[x\xi+y\eta\big]\bigg)\,d\xi\, d\eta \hspace{1cm} (5)
\end{split}
$$
which should be valid as long as
$$
z^3\gg \frac{\pi}{4\lambda}  \big[(x-\xi)^2+(y-\eta)^2\big]^2_{\textrm{max}}. \tag{6}
$$
Fraunhofer Approximation
If we further assume that
$$
z\gg\frac{k(\xi^2+\eta^2)_{\textrm{max}}}{2}, \tag{7}
$$
then the first exponential inside the integration in Eq. (5) is $\approx 1$, which leads to the more simplified Fraunhofer integral
$$
U_2(x,y) = \frac{\textrm{exp}(ikz)}{i\lambda z} \textrm{exp}\bigg( \frac{ik}{2z}\big[x^2+y^2\big]\bigg)  \iint_\Sigma U_1(\xi,\eta)\; \textrm{exp}\bigg(-\frac{2\pi i}{\lambda z}\big[x\xi+y\eta\big]\bigg). \tag{8}
$$
My questions:
I always have read that "Fraunhofer corresponds to the far-field regime", whilst "Fresnel corresponds to near-field regime". However:

*

*In obtaining the Fraunhofer formula in Eq. (8), I have first had to pass through the Fresnel approximation of Eq. (5). Does this mean that Fraunhofer and Fresnel are not two distinct individual regimes, but that Fraunhofer automatically implies Fresnel simultaneously?

*Both conditions on the two approximations, in Eq (6) and (7), require large value of $z$ - how can I reconcile this with the idea that "Fresnel is near-field", if one of its requirements is large distance $z$?

If I have made a mistake in any of the mathematics, please point it out, but I would also appreciate an intuitive picture/explanation as well. Thank you!
 A: Here is your equation for large $z$
$$
\begin{split}
U_2(x,y) &= \frac{\textrm{exp}(ikz)}{i\lambda z} \iint_\Sigma U_1(\xi,\eta)\; \textrm{exp}\Bigg( \frac{ik}{2z} \bigg[ (x-\xi)^2 + (y-\eta)^2 \bigg] \Bigg)\,d\xi\, d\eta \hspace{2.4cm} (5)\\
&= \frac{\textrm{exp}(ikz)}{i\lambda z} \textrm{exp}\bigg( \frac{ik}{2z}\big[x^2+y^2\big]\bigg)\times\; ... \\ 
&\hspace{1.5cm}\iint_\Sigma U_1(\xi,\eta)\; \textrm{exp}\bigg( \frac{ik}{2z}\big[\xi^2+\eta^2\big]\bigg) \; \textrm{exp}\bigg(-\frac{2\pi i}{\lambda z}\big[x\xi+y\eta\big]\bigg)\,d\xi\, d\eta \hspace{1cm} (6)
\end{split}
$$
Notice that this integral is not the Fourier transform of the given aperture field $U_1(\xi,\eta)$ but of its phase modulated version $U_1(\xi,\eta)\textrm{exp}\bigg( \frac{ik}{2z}\big[\xi^2+\eta^2\big]\bigg)$. This $\Sigma$ being a finite sized aperture, for large enough $z$ the phase of the exponential can be made arbitrary small and then can be neglected. When you can do so that is called the Fraunhofer zone, but when you are not that far away you have to take the quadratic phase variation into account and you are in the Fresnel zone.
If a plane wave is incident on the aperture and you are in the Fraunhofer zone then the behavior of the far field is essentially dependent on the amplitude $|U_1|$ amplitude for its phase is a linear function and can be absorbed by the Fourier kernel as an angular displacement. But that is not true in the Fresnel zone and the quadratic phase modulation is an added nasty complication when evaluating the case of oblique incidence.

In summary: Having the quadratic phase modulation beyond the Rayleigh (Fraunhofer) limit [1] $\frac{2D^2}{\lambda}$ in Equation $(6)$ is legitimate but adds nothing more than numerical/analytical complications. The Fresnel approximation is replacing the square root with a quadratic expression in the complex exponential and is resulting in a Fourier transform of the phase modulated aperture field. The Fraunhofer approximation is a further simplification of that of the "Fresnel" valid in the Rayleigh limit, in fact, is a linearization of the exponent resulting in a Foruier transform of the aperture field but without the quadratic phase modulation.
[1] https://en.wikipedia.org/wiki/Fraunhofer_distance
