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echo: atm
to: ATM
from: burrjaw{at}earthlink.net
date: 2003-03-10 21:02:42
subject: Re: ATM cancelling astigmatism from zernike coefficients

To: atm{at}shore.net
From: Jim Burrows 
Reply-To: Jim Burrows 


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At 11:16 2003-03-10 -0600, Stephen C. Koehler wrote:

>Can anyone tell me the proper way of cancelling astigmatism from Zernike
>coefficients?  Alternatively can anyone tell me how Quick Fringe cancels
>astigmatism from the Zernikes coefficients?

If you use the true Zernike polynomials, you don't have to "cancel=20
astigmatism", since the polynomials are orthogonal over the unit
circle,=20 i.e., the presence of one on the surface doesn't affect the
evaluation of=20 any of the others.  Using your interferograms you have the
surface height=20 z(x, y) as a function normalized coordinates (divide the
true coordinates=20 by the mirror semi-diameter), then the second symmetric
coefficient

         A20 =3D int{z(x,y)(2r=B2 - 1) dx dy}*(3 / pi), r=B2 =3D x=B2 + y=B2

(I think that factor 3/pi is right, but confusion has reigned over the=20
years.)  This number is independent of the values of, for example, the=20
coefficients for the two primary astigmatism polynomials

         x=B2 - y=B2 and 2xy

You can get formulas for enough of the Zernike polynomials from

http://wyant.opt-sci" target="new">http://wyant.opt-sci.arizona.edu/zernikes/zernikes.htm>http://wyant.opt-sci=
.arizona.edu/zernikes/zernikes.htm=20


or Born & Wolf, "Principles of Optics", p. 524.

         -- Jim Burrows
         -- mailto://burrjaw{at}earthlink.net
         -- http://home.earthlink.net/~burrjaw
         -- Seattle N47.4723 W122.3662 (WGS84)=20
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At 11:16 2003-03-10 -0600, Stephen C. Koehler wrote:
Can anyone tell me the proper way
of cancelling astigmatism from Zernike coefficients? 
Alternatively can anyone tell me how Quick Fringe cancels
astigmatism from the Zernikes coefficients? If
you use the true Zernike polynomials, you don't have to "cancel
astigmatism", since the polynomials are orthogonal over the unit
circle, i.e., the presence of one on the surface doesn't affect the
evaluation of any of the others.  Using your interferograms you
have the surface height z(x, y) as a function normalized coordinates
(divide the true coordinates by the mirror semi-diameter), then the second
symmetric coefficient
        A20
=3D
int{z(x,y)(2r=B2 - 1) dx dy}*(3 / pi), r=B2 =3D x=B2 +
y=B2 (I think that factor 3/pi is right, but confusion
has reigned over the years.)  This number is independent of the
values of, for example, the coefficients for the two primary astigmatism
polynomials
        x=B2
- y=B2
and 2xy
You can get formulas for enough of the Zernike polynomials from
http://wy" target="new">http://wy=">http://wyant.opt-sci.arizona.edu/zernikes/zernikes.htm">http://wy=
ant.opt-sci.arizona.edu/zernikes/zernikes.htm

or Born & Wolf, "Principles of Optics", p. 524.

        --
Jim=
 Burrows
        --
mailto:%2F%2Fburrjaw{at}earthlink.net"=
 eudora=3D"autourl">mailto://burrjaw{at}earthlink.net;
        --
http://home.earthlink.net/~burrjaw"=
 eudora=3D"autourl">http://home.earthlink.net/~burrjaw;
        --=
 Seattle N47.4723 W122.3662 (WGS84)


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