From: "Dwight K. Elvey" 
To: atm{at}shore.net
Reply-To: "Dwight K. Elvey" 


Hi
 I thought I'd jump in here. I guess the question of what
kind of surface would be correct for a non-flat secondary be for a Newton
should be answered. It has been mentioned that a concave or convex surface
could be used. My understanding of optics and focal points says one should
use an ellipsoid or use a hyperboloid. A little more thought would show
that one should make this mirror an off axis mirror. The focal point of the
system will not be the same as a true flat but the star image should come
to focus, with the right placement of things.
 Now, when one considers that a spheriod surface approximates
these other surfaces for a small area, one might suspect that one could
find that for small errors in a flat that was in fact a spherical error,
there would be no problem with getting a good image. I would also suspect
that this error could even be several wavelengths, with the proper location
of the secondary and the proper error ( hyperboloid, off axis ).
 Errors that don't fit this arrangement, such as turned edge
would be a different matter.
Dwight


>From: "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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