ATM Re: Wavefront arithmetic

[Bratislav Curcic <epabcc@epa.ericsson.se>]

> I stand corrected.  However, most the mirrors I make are in the order of
> f/3.7 or so and the ratio of the length from the primary to the secondary
> compared to the length from the secondary to the focus is smaller than most
> (not 12:1, but more like 6:1).  By this a 1/10 wv p-v primary with a 1/10
> wv p-v secondary would produce a (1/6)(1/10) + 1/10 error = 1/60 + 1/10 =
> 7/60 error p-v = 7/30 wv error on the wavefront which is a little better
> than 1/4 wv error on the wavefront.  Please correct me if I am in error. 

Jeff, 

the wave arithmetic still doesn't work that way. First, without knowing
EXACT shape and magnitude of error, we can do very little. Do we add or
substract ? Consider optician refiguring a secondary to match erroneous
primary. He works until he gets a perfect null. Both primary and
secondary are 'off' on their own, yet together they provide perfect
wavefront (i.e. 0 wave error at focus).

If we don't know the distribution of errors (but know that sphericity
of the secondary is zero), safest assumption is that all errors are
truly random. In that case, some errors will add, some will cancel, and
resulting error will most likely be sqrt((1/10)^2 + (1/10)^2) or
about 1/7 wave.

But if secondary was mostly convex or concave (i.e. perfect long focus 
sphere) the resulting astigmatism will far overpower the primary's errors
and resulting error will be MUCH worse than 1/5 wave.

In reality we get a bit of both, and the only way to really find what
happens is put them together and test in the interferometer.

Bratislav