From: "Frank Q" 
To: 
Reply-To: "Frank Q" 


Hi All

A very interesting analysis.... Let me add my 2 cents' worth:

For this, I'm looking at a diagonal that consists of a flat surface with a
one-wavelength central flat depression - ie a combination of 2 flat
surfaces.

For light hitting the central section, this is the same as a diagonal which
is wavelength/cos(45) further from the mirror compared with the edge
portion.

Now to get some realistic feel for this, assume that

wavelength = 0.5 microns
primary = 200 mm (8") (f/5)
focal length = 1000 mm

The airey disk size is d = 1.22 * f * wavelength / primary_diameter ie

d = 3.06 microns

And, the central portion of the diagonal creates an image which is
(laterally) shifted by x = wavelength / cos(45) relative to the that formed
by the outer portions of the diagonal. Plugging in the numbers gives

x = 0.7 microns

So, we're now looking at 2 superimposed airey disks, each with a diameter
of 3 microns but sideways displaced by 0.7 microns - ie about 1/4 of their
diameter.

For the 16" example:

d = 3 microns
x = 0.7 microns

Some comments which have to be made are that (1) this is a VERY* simple and
artificial situation and (2) the superposition of the airey disks needs to
have the phase taken into account (after all, there is an optical path
difference (opd) introduced by the depressed zone).

opd = wavelength / cos(45) = 1.4 wavelengths (approx)

which is roughly destructive interference over about 75% of the pattern!!!

*VERY perhaps should be "ridiculuous"

FWIW - My gut feeling is that you will end up with a pretty messy airey pattern.

Apologies for metric measurements

Cheers

Frank Q

----- Original Message -----
From: "MLThiebaux" 
To: 
Sent: Friday, March 07, 2003 11:19 PM Subject: Re: ATM Flats


>
> On flats that needn't be so flat:
>
> I'll go through a numerical example based on geometric optics showing the
effect of a slightly spherical diagonal in a Newtonian.  Inches throughout
(although Canadian, I don't mind). Suppose our telescope has
>
> D = primary diameter = 16
> F = focal length = 80
> d = diagonal minor diameter = 4
> L = primary axis - image plane distance = 9
>
> Suppose the diagonal is concave spherical with a sagitta across the minor
diameter = e = 1 wavelength.  So
>
> e = 1/50000
>
> The curvature on the flat corresponds to a focal length d^2/16e = 50000.
At 45 degrees the two oblique principal focal lengths are 50000*cos45 and
50000/cos45 or 70711 and 35355.
>
> Think of the image of a star.  After a little fiddling with the lens
formula we find that if one of the oblique focal lengths is called  f, then
the point image formed in its principal plane is closer to the primary than
it would be if the diagonal were truly flat by the amount  L^2/f.  Hence
there is a separation of the two images by the amount  s =
81/35355-81/70711
= 0.00115.  This is the longitudinal astigmatic focusing error.
>
> Now we'll figure out how much astigmatism in the primary mirror would
produce this same longitudinal focusing error.  The two principal focal
lengths of the mirror would differ by  s, and so the edge of the mirror
surface has a peak error  +/- s*D^2/32F^2 = 1.4375E-6 relative to a truly
parabolic mirror.  The peak-to-valley error would be twice this, and the
surface rms error = peak error divided by sqrt(6). Choose your favorite.
>
> Summary:
> Diagonal error =  1 wavelength
> Equivalent astigmatic primary surface error in wavelengths
> Peak:  1/14
> PV:    1/7
> RMS:   1/34
>
> Martial Thiebaux
> Rawdon Hills, Nova Scotia
>
>
>
>
>

--- BBBS/NT v4.00 MP