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


Hi All

It was definitely NOT the intention to approximate a curved surface. The intention
was to see if the result from some other weird surface shape with
comparable error would give a "bad image". Granted this is a very
simple model (and unrealistic)
but it tends to indicate that a one wave error (for want of a better word) gives
significant cause for worry.

This is a very different physical situation, and for this reason, it can be
used to add
support to the conclusions obtained by modelling more realistic situations.

Cheers
Frank Q

>
>
> The above is an interesting model. It would be fun see exactly what kind
of pattern results.  But if the intention is to approximate a slightly
curved flat, it's a nice try that doesn't convince.
>
> It's easy to see why not.  Suppose the diagonal is a concave hyperboloid,
correctly figured and collimated for the job.  With the principles of
geometric optics in charge, the image of an on-axis point star would be a
perfect point. Or if you prefer, wave theory would soften the point to a
perfect Airey disk.
>
> The 2-plane approximation that intends to show the messed up image caused
by a slightly spherical diagonal then would do exactly the same messy job
on the perfect image.
>
> Martial Thiebaux
> Rawdon Hills, Nova Scotia
>

----- Original Message -----
From: "MLThiebaux" 
To: 
Sent: Monday, March 10, 2003 8:22 PM Subject: Re: ATM Flats


>
> At 09:20 AM 3/10/2003 +1000, Frank Q wrote:
> >
> >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
> >

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