Re: (ATM) Wire Spiders

[cds@peanut.sarnoff.com (Clay Spence x3039)]

rhill@lpl.arizona.edu (Rik Hill) wrote:

> It was my understanding from my old optics texts that diffraction
> is caused by edges (like the aperture and spider vane edges) and 
> not by obstruction. Is this not so? -Rik

Hmm... You can't have an edge without an obstruction, or an obstruction
without an edge.

I think you're asking what difference does it make how thick the vanes
are (for practical thicknesses).  It's just a matter of the shape of
the aperture, or apertures in this case.  With a four-vane spider we
have four apertures through which the light is passing, and it
interferes with itself after passing through.  I don't have an
intuitive explanation for how things change with vane thickness, except
that the farther the apertures are from each other, i.e., the thicker
the vanes, the more like an intereferometer it becomes.  The
calculation for an infinitely long vane in an infinite plane aperture
with a plane wave traveling perpendicular to the aperture and incident
on a distant screen in the Kirchhoff approximation is not too hard,
depending on your opinion of what's too hard.  You get some error
integrals.  Right now I'm at work, so I shouldn't take the time to do
it.

You could also do something like taper the reflectivity of the primary
mirror gradually to zero (or near zero) at the edge, and so you would
have something close to an aperture without an edge.  (The diffraction
rings would nearly go away, leaving a smooth central spot, still with
an infinite surround, just no rings.  To completely eliminate rings, I
think you'd have to have an infinite mirror with reflectivity tapering
to zero at infinity.)

Clay Spence
cspence@sarnoff.com