From: "Nils Olof Carlin" 
To: "John Sherman" , 
Reply-To: "Nils Olof Carlin" 


hi John

> So I point my scope at a star, and it does not make an image of a
star. It makes a diffraction pattern. Right there at the focus point of the
scope is a diffraction pattern containing rings and a disc. I can measure
that pattern, and determine that the size of the disc is dependent on the
f/# of  the scope, and is independent of the aperture (assuming green light
and a scope without aberrations). So now I can put an eyepiece into the
scope, and  look at that diffraction pattern. Now, all of a sudden, for no
reason that I have been able to determine, the size of the disc has nothing
to do with the f/# of the scope.

With a low-f/# telescope, the diffraction pattern at the focal plane is
indeed smaller - so that's why you need a shorter f-l eyepiece to magnify
it more! I find the concept of exit pupil to be particularly helpful here -
the apparent size of the diffraction pattern is a function *only* of the
exit pupil (regardless of aperture). The magnification is the aperture
divided by the exit pupil, so with the same aperture and same exit pupil,
the magnification is the same, regardless of f/#. One simple way of
calculating the exit pupil is dividing the f.l. by the f/#. If you pick an
eyepiece with a f.l equal to the f/# of the telescope, you get an exit
pupil of 1 mm (for any telescope), and the *apparent* angular radius of the
Airy disk will be (1.22*lambda * f/#) / (eyepiece f.l.), that is
1.22*lambda/(exit pupil) e.g. 1.22*0.00055=2.3 arcmin for 1 mm exit pupil
with any instrument.

Hope this helps (there's always hope)

Nils Olof

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