From: Aplanatic{at}aol.com
To: jlerch1{at}tampabay.rr.com ("James Lerch"), atm{at}shore.net
Reply-To: Aplanatic{at}aol.com


Howdy,

I've written a VB6 program to manually capture fringe locations.  The
program then calculates the Zernike coefficients (up to 8th order) from the
fringe location information by performing a least squares fit.

I've applied this program to the raw interferogram supplied to us by James:

http://lerch.no-ip.com/atm/2ndTry/results/intgram1.JPG

I've checked the program several times and can't find anything wrong with it.

Here are the Zernike coefficients that I calculate for "intgram1.JPG":

Z0   5.576  (n=0 m=0)
Z1  -0.707  (n=1 m=1 X)
Z2  -5.828  (n=1 m=1 Y)
Z3  0.0112  (n=1 m=0)
Z4   0.248  (n=2 m=2 X)
Z5  -0.075  (n=2 m=2 Y)
Z6  -0.054  (n=2 m=1 X)
Z7   0.058  (n=2 m=1 Y)
Z8   1.598  (n=2 m=0)
Z9   0.003  (n=3 m=3 X)
Z10  0.047  (n=3 m=3 Y)
Z11 -0.062  (n=3 m=2 X)
Z12  0.044  (n=3 m=2 Y)
Z13 -0.046  (n=3 m=1 X)
Z14   .080  (n=3 m=1 Y)
Z15  0.134  (n=3 m=0)

The above coefficients are in waves of wavefront deviation from spherical
(as long as the interferogram was taken with a spherical reference
wavefront, as is stated by James).  Positive numbers represent longer path
lengths.  In the above, m is the azmuthal order.  The Zernike numbering
corresponds to that of James C. Wyant.  The RMS deviation for the LSF was
0.055 waves wavefront.  Not too bad.  I used 187 points from the
interferogram.

My problem is the following:  I can't seem to match these coefficients to
the report that James received from Mr. Royce:

http://lerch.no-ip.com/atm/2ndTry/results/testdata.JPG

One would assume that the second-order coefficients could be changed by the
operator to present a more easily visualized synthethic interferogram.  In
addition, the azmuthally symmetric (m=0) terms could also be changed so
that the synthetic interferogram was referenced to a paraboloidal surface.
Unfortunately, I can't make the other terms match.  Could it be that the
operator also rotated the frame of reference of the mirror?  It's also
possible that my program is not working properly.

Any help wound be appreciated.  If your interested I can supply the
remaining coefficients from Z16-Z24.

James, can you tell me again what is the RoC of this mirror?  --Thanks.

Dave Rowe


James wrote:
> Greetings All,
>
> I spoke with Mr. Royce on Tuesday, and have the following short update:
>
> #1 The interferometry was done with the full aperture
> #2 He was concerned the numeric results were influenced negatively by the TDE
> #3 He said he would get us the 10th and 12th order Zernike polynomials.
>
> So, in the mean time I've been pondering the idea on how to manually reduce
the
> interferometry fringes.  We have two images to play with,
>
> #1 the first is the Raw fringe he supplied with the results:
> http://lerch.no-ip.com/atm/2ndTry/results/intgram1.JPG (75Kb)
>
> #2 the second is from the synthetic fringe on the test data report here:
> http://lerch.no-ip.com/atm/2ndTry/results/testdata.JPG (86Kb)
>
> As I'm told, Image #1 was taken at RoC, and would show straight fringes only
for
> a SPEHRICAL mirror.   Image #2 is a synthetic fringe produced by the
> intereferometry software and a perfect PARABOLIC mirror would show straight
> fringes. 

--- BBBS/NT v4.01 Flag-5