Subject: [interferometry] Digest Number 1319 From: interferometry@yahoogroups.com Date: 4/14/2009, 4:01 AM
 To: interferometry@yahoogroups.com

interferometry

# Messages

1a.

## Re: Relationship between movement on optical axis and Z3

### Posted by: "Stephen Koehler" s.c.koehler@gmail.com steve_koehler

#### Mon Apr 13, 2009 7:36 am (PDT)

Michael,

> Let's assume we are testing a spherical mirror with an interferometer. It doesn't care which type of interferometer.
> Let's assume the mirror is perfect, and it is aligned perfectly. After analyzing the interferogram we see that the power (or defocus) term Z3 is zero.
> Now let's move the mirror 0.1mm on the optical axis and take a new interferogram. This interferogram must show some power. How can this amount of power be calculated? I tried to derive a formula and got this one:
>
> Z3 = x / (8 N^2)
>
> where x is the movement on the optical axis, and N is the ratio R/D, with R = radius of curvature and D = mirror diameter.
> Z3 is the Zernike coefficient for the reflected wavefront (not for the surface error).
>
> I tried to check this formula with my interferometer but got a different result, too small by a factor of 2.
>
> Is the formula right or wrong, and why?

This formula looks similar to the formula in the appendix in Suiter's
book, but I don't have my copy handy to look it up. Here's a piece of
R code that does a calculation of z3 from dz. I'm quite sure that my
formula works. Perhaps you can figure out the factor of two by
simplifying my formula and applying a Taylor series approximation.

# Compute the defocus from a pv calculation of error at center. Suiter
# has a drawing of this.
pv <- ( sqrt((diam/2)^2+roc^2)
- (sqrt((diam/2)^2+(roc+dz)^2) - dz)) / lambda
z3 <- pv/2

--
Steve Koehler

1b.

## Re: Relationship between movement on optical axis and Z3

### Posted by: "Michael Peck" mpeck1@ix.netcom.com mikepeck5440

#### Mon Apr 13, 2009 8:23 am (PDT)

At 11:50 04/12/2009, Michael Koch wrote:
Now let's move the mirror 0.1mm on the optical axis and take a
new interferogram. This interferogram must show some power. How
can this amount of power be calculated? I tried to derive a
formula and got this one:

Z3 = x / (8 N^2)

where x is the movement on the optical axis, and N is the ratio
R/D, with R = radius of curvature and D = mirror diameter.
Z3 is the Zernike coefficient for the reflected wavefront (not
for the surface error).

Taking Steve's hint I checked Suiter's derivation. He gets this
formula for Peak to Valley defocus. Peak to 0 -- the Zernike
coefficient -- would be half as large.

Mike P.

------
Michael Peck
mpeck1@ix.netcom.com
http://www.wildlife-pix.com

<http://www.wildlife-pix.com/>
1c.

## Re: Relationship between movement on optical axis and Z3

### Posted by: "Oscar Gonzalez Regueira" ogrydc@gmail.com ogrydc

#### Mon Apr 13, 2009 10:37 am (PDT)

Damn, my other reply have some trouble...

I think that you have derived the correct formula for a fixed source, but Bath is behaving as a moving source system... just the half...

Saludos, OSCAR.

--- In interferometry@yahoogroups.com, "Michael Koch" <astroelectronic@...> wrote:
>
> Hello,
>
> Let's assume we are testing a spherical mirror with an interferometer. It doesn't care which type of interferometer.
> Let's assume the mirror is perfect, and it is aligned perfectly. After analyzing the interferogram we see that the power (or defocus) term Z3 is zero.
> Now let's move the mirror 0.1mm on the optical axis and take a new interferogram. This interferogram must show some power. How can this amount of power be calculated? I tried to derive a formula and got this one:
>
> Z3 = x / (8 N^2)
>
> where x is the movement on the optical axis, and N is the ratio R/D, with R = radius of curvature and D = mirror diameter.
> Z3 is the Zernike coefficient for the reflected wavefront (not for the surface error).
>
> I tried to check this formula with my interferometer but got a different result, too small by a factor of 2.
>
> Is the formula right or wrong, and why?
>
> Michael
>

1d.

## Re: Relationship between movement on optical axis and Z3

### Posted by: "Michael Koch" astroelectronic@t-online.de astroelectronic

#### Mon Apr 13, 2009 11:36 am (PDT)

Mike,

> Taking Steve's hint I checked Suiter's derivation. He gets this formula for Peak to Valley defocus. Peak to 0 -- the Zernike coefficient -- would be half as large.

Yes, I know that the Zernike coefficient is half as large as the PV value.
Suiter's formula is for single pass. In the interferometer I have double pass, so I must multiply the PV value by two. And then we have the same formula again. But it doesn't fit to my measurement.
By the way, I didn't test with a Bath interferometer but with a Fizeau interferometer. But that shouldn't change things.

Michael

2a.

## Re: Relationship between movement on optical  axis and Z3

#### Mon Apr 13, 2009 11:32 am (PDT)

Suiter formula mentioned by Stephen and Michael is derived with some
simplifications (Taylor expansion of sagita and f*f'= f^2) and is

s-s'= (f-f')/ 8 Fnr^2 where Fnr is f/D, (f-number)

Michael Koch used insted R/D (where R=2f) twice the f-number.

His resulting defocus is 4 times smaller than Suiters for same
f and D.

Regards
2b.

## Re: Relationship between movement on optical  axis and Z3

### Posted by: "Michael Koch" astroelectronic@t-online.de astroelectronic

#### Mon Apr 13, 2009 11:52 am (PDT)

> Michael Koch used insted R/D (where R=2f) twice the f-number.

Suiter's formula is for the movement of an eyepiece in a telescope. In this case the focal ratio is f/D.
My formula is for the movement of the interferometer (or movement of the mirror under test). In this case the focal ratio is R/D.
In both cases the focal ratio describes the same cone angle.

> His resulting defocus is 4 times smaller than Suiters for same
> f and D.

But the difference between theory and experiment is only a factor of two :-(

Michael

2c.

## Re: Relationship between movement on optical axis and Z3

### Posted by: "Stephen Koehler" s.c.koehler@gmail.com steve_koehler

#### Mon Apr 13, 2009 12:27 pm (PDT)

Michael,

Vladimir may be on the right track. Suiter's formula is

pv = dz / (8 * (f/D)^2)

so

z8 = dz / (16 * (f/d)^2)

putting in terms of R/D,

z8 = dz / (4 * (R/D)^2)

--
Steve Koehler

2d.

## Re: Relationship between movement on optical axis and Z3

### Posted by: "Michael Koch" astroelectronic@t-online.de astroelectronic

#### Mon Apr 13, 2009 12:37 pm (PDT)

Steve,

> Vladimir may be on the right track. Suiter's formula is

I don't think so. Suiter's formula is for single pass. Light is coming from a star and going to the eyepiece.
In my case I have double pass. Light is coming from the interferometer and is reflected back to the interferometer.
The wavefront error must be twice the geometrical error from Suiter's formula.

Michael

2e.

## Re: Relationship between movement on optical axis and Z3

### Posted by: "Michael Koch" astroelectronic@t-online.de astroelectronic

#### Mon Apr 13, 2009 1:55 pm (PDT)

Hi all,

I've uploaded my derivation for the Z3 / dz relationship to the files section, Michael Koch, Z3_derivation.gif

The interferometer is at the left side, near the center of curvature.
In case 1, the spherical mirror is aligned perfectly so that Z3 is zero.
The sagitta s1 can easily be calculated. This is only an approximation, but that doesn't care.
In case 2, the interferometer (or the mirror) was moved along the optical axis by dz. Sagitta s2 is the sagitta of the wavefront when it arrives at the edge of the mirror. It is different from the sagitta of the mirror (which is still s1).
The PV error in the reflected wavefront must be 2 * (s1-s2).
The Z3 coefficient is half as large as the PV error, Z3 = s1 - s2.

I still have no idea where the bug is.

Michael

2f.

## Re: Relationship between movement on optical axis and Z3

### Posted by: "Michael Koch" astroelectronic@t-online.de astroelectronic

#### Mon Apr 13, 2009 11:58 pm (PDT)

Hi all,

I found the bug. The formula is correct, and the measurement is correct as well. My interpretation of the measurement result was wrong. I did make this measurement with Zygo's MetroPro software. It has an output line for "Power", and I wrongly assumed that this power is the same as the Zernike coefficient Z3. But it isn't. Power is twice as large, and describes the PV error of the power term. The problem is solved.

Michael

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