There has been a bit of discussion about this already, however, my question arises more from the mathematical requirements to allow for Fraunhofer diffraction proposed by Born and Wolf's optics book.
First: variable definitions: Given $r'$ is the distance of the source from the aperture and $s'$ is the distance of resultant pattern, and $(\zeta, \eta)$ as the aperture coordinates, and $l_0, l$ and $m, m_0$ as direction cosines of the aperture to the source/point (these are not terribly important for the discussion).
The quadratic (Fresnel) phase can be ignored under the following condition:
$\frac{1}{2} k | (\frac{1}{r'} + \frac{1}{s'}))(\zeta^2 + \eta^2) - \frac{(l_0\zeta + m_0 \eta)^2}{r'} - \frac{(l\zeta + m\eta)^2}{s'}| << 2\pi$
There are two possibilities:
(1) The distance of the source and the resulting diffraction are observed very far away relative to the aperture size divided by the wavelength, i.e.
$|r'| >> (\zeta^2 + \eta^2)_{max}/\lambda$
$|s'| >> (\zeta^2 + \eta^2)_{max}/\lambda$
This one is very familiar and intuitive: clean, constructive interference off the small slit is successfully achieved with planar waves incident on the aperture and observed far enough away.
(2) The other possibility to satisfy this is:
$\frac{1}{r'} + \frac{1}{s'} = 0$ and the angle of the source/observation are very paraxial.
The 2nd condition is very confusing, however, it seems vital to justify the usage of a lens i.e. an ideal lens placed after the aperture will always satisfy the Fraunhofer condition. How can these distances be negative? If a point source is imaged by a lens, the image plane will reconstruct this point source. But how can this end up making $r'$ negative? From geometric optics, I know that we can interpret the image as being an additional point source, but wouldn't r' stay a positive quantity?