This is not a true answer, but an expansion on calculating diffraction patterns from @whuber's answer.
First, we have the diffraction integral. The function Up describes the complex amplitude in the observation plane at a distance (xp, yp) from the optical axis, and a distance Lz from the source (some kind of diffractive object, e.g. pinhole, camera aperture, etc.) Us is a function that describes the complex amplitude in the source plane; for an extremely small pinhole you could use a dirac delta function. The third variable in Us is 0 because for convenience we say the diffractive object is the origin of the coordinate system. The variables xs and ys in its arguments bookkeep for that fact that the object may have some size in the x–y plane.
This may not look like such an awful integral, but k and rsp are both just notation for something bigger:
Integrating a function with a radical with square terms in it both in the numerator of e and in the denominator is a very nasty integral indeed.
One simplifies the integral by removing the square roots by using the binomial series representation and truncating higher order terms. The Fraunhofer integral holds when one needs 2 terms; the Fresnel integral is for when one needs 3 terms. There is some nuanuce to the proof of that, but it is outside the scope of this.
When we start manipulating these things to get the Fresnel and Fraunhofer diffraction integrals, we get three quantities.
If Nfd * (θd)2 << 1, the Fresnel integral is valid. If that is true and Nfs << 1, the Fraunhofer integral holds.
The two integrals are:
and νx and νy are the size of the source in a given dimension divided by the wavelength of light times the distance to the source. Normally it would be written νs = d / (λxs).
To answer @whuber's question as to why you may need one or the other, despite what Wikipedia states, requires a bit of thought.
The "at the focal plane of an imaging lens..." comment is probably lifted from a textbook, and the implication is that the source of the diffraction (i.e. the pinhole, slit, whatever — these equations are agnostic as to the geometry of the source) is very far away. Unfortunately, not only could the lens be at any distance and closer than the Fraunhofer integral allows, but the diffraction also originates inside the lens system for a camera.
The correct model for diffraction from a camera's aperture is an n-sided aperture (n is the # of aperture blades in the lens) illuminated by a point source at the location of the thing in the image that produces the starburst pattern.
When the objects are really far away (a few meters would be fine), the point sources behave as if they are plane waves and the derivations carried out on Wikipedia are fine.
For example, the aperture for a double gauss 50 mm lens is on the order of 40 ~ 60 mm from the image plane. It is imaged by a couple of lenses behind the physical stop to a distance greater than that (this is the location of the exit pupil), but the exit pupil is not where the Us(xs,ys,0) function is centered!
For 500 nm and a 1 mm radius aperture light, we can check if the Fraunhofer integral is valid. It is equal to (0.001)2 / (500*10-9 * 50*10-3), or 40, which is >> 1 and the Fraunhofer integral is invalid. For visible light, as long as the aperture stop is on the order of millimeters from the detector, Nfs will never be anywhere near 1, let alone much smaller.
These equations may differ somewhat from the ones on Wikipedia; I would reference OPT 261, Interference & Diffraction at the University of Rochester Institute of Optics taught by professor Vamivakas. The equations in Optics by Hecht should be fairly similar. The equations are for the complex amplitude, to get the Irradiance (a.k.a. intensity or brightness), you would take the magnitude squared of the result.