To: atm{at}shore.net
From: MLThiebaux 
Reply-To: MLThiebaux 


A little more on flats that needn't be so flat:

Last week ( Fri, 07 Mar 2003 09:19:37 -0400) I posted a numerical example
showing that a 1 wavelength error in a Newtonian diagonal was equivalent to
a substantially smaller astigmatic error in the primary.  A range of
responses followed, some understandably skeptical.

Here is an alternative explanation (mostly simpler) that looks directly at
the wavefront error caused by the diagonal.  It avoids dragging the primary
into the melee, an approach that probably only confused the issue.  I'll
use the same dreamed-up 16" telescope as before (lengths in inches):

        f = primary focal ratio = 5
        d = diagonal minor diameter = 4
        L = primary axis to image plane distance = 9
        e = sagittal error across minor diameter = 1 lambda
        E = sagittal error across major diameter = 2 lambda

Note that e and E distinguish between minor and major sagittal errors. The
overall peak-to-valley deviation from a plane is actually 2 wavelengths.
Let's ignore the small departure of the major:minor diameter ratio from
sqrt(2):1.

There are 3 effects that reduce the error:

1. Only part of the diagonal is used in forming an on-axis star image.
Let's call this area the "small" ellipse.  The minor diameter of
the small ellipse is L/f = 1.8, so the sagittal error across the MAJOR axis
of the small ellipse is reduced to E(L/fd)^2 = .2025*E, which is the
maximum deviation of the small ellipse from the plane.  This is the most
important effect and the easist to understand.

2. The slightly spherical diagonal should be compared to the nearest
HYPERBOLOID, not to the perfect plane.  The required hyperboloid has two
principal curvatures differing by a factor of 2, say k and 2k.  One minimax
solution (hard to explain without a picture) for fitting such a hyperboloid
to the sphere is found to have the following properties:

        (a)  k = 3/4*curvature of the sphere,
        (b)  equal and opposite maximum deviations relative to the minimax
hyperboloid occur at the endpoints of the axes of the small ellipse,
        (c)  the peak-to-valley error in the small ellipse is HALF the maximum
deviation of the small ellipse from the plane.

3. The optical path difference at 45 degrees incidence is sqrt(2) times the
normal separation (not a factor of 2 as in normal incidence).

Combining the 3 effects, we find the wavefront error is

               .2025 * 1/2 * 1.414 * E = .286 lambda,

not great but it's just an example and the point should be clear.  Change
the focal ratio to 6, keep everything else the same, and the wavefront
error drops to a more acceptable 1/5 lambda.


Martial Thiebaux
Rawdon Hills, Nova Scotia

--- BBBS/NT v4.00 MP