ATM - Re Field Curvature

[Richard Schwartz <richas@idt.net>]

To determine what you can get away with, you need to know your film
grain size, g, your focal ratio, and the distance from the center to the
farthest edge of your film.   The idea is that the spot size can be no
bigger than the film grain.

Now, the spot size is just the saggita of the curved surface divided by
the focal ratio.

This lets us write      (r^2)/(2*R*f) = g

Solving, we get R = (r^2) /  (2*f*g).

Example: a 35mm film frame has r = 1.7" at the corner.   If the system
is f/4 and g=0.001", we get R = 361".   For most of our telescopes, the
value of R is less than this, so the CURVATURE, k, is greater.
Generally, k = 1/R.  In newtonian telescopes, the field curvature
(sometimes called the "Petzval") is the same as the mirror curvature.
But in telescopes with more than one optical element, the field
curvature is a sum of the curvatures of the individual surfaces.  Each
of these terms depends on the surface curvature and the index of
refraction.

All of this assumes that the image is perfect except for field
curvature.

. . . Richard


Date: Thu, 21 Jan 1999 09:53:41 -0500
From: Michael Davis <mdavis19@ix.netcom.com>
Subject: ATM Field curvature

Hello all,

Someone asked a question similar to mine a week or two ago, but I don't
remember seeing any answers.

Assuming that other aberrations have been reduced to the point that they

are not a problem. At what radius does field curvature become so extreme

that film must be bent to get good star images across the whole field.
Is there a rule of thumb or some conventional wisdom on this point?

Thanks,

Michael Davis