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Artificial star distance?

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#1 tag1260

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Posted 12 November 2019 - 11:14 AM

I have an artificial star and want to test a 6" f8 refractor. What sort of distance do I need? I'm seeing about 80 feet? Is this even remotely correct.  20x the f ratio.

 

Thanks



#2 davidc135

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Posted 12 November 2019 - 12:44 PM

That sounds OK. I get an error of 1/10 wave at that distance.

 

My calculation is based on:

Mladen (MKV) posted that for max error of 1/20 wave the allowable departure from plane of a spherical 'flat' used to test an objective = 3.6x(focal ratio)^2 so in your case 230 fringes or 115 waves green light or 57 microns. The distance you place your point source is the radius of curvature of the sphere whose sag over 6'' should be no more than 57 microns

 

Edmund Optical have a sag/ roc calculator so feeding in 80 feet for your F/8 I get 115 microns so not good enough for 1/20 wave accuracy but 1/10 wave should be OK. I don't know if that's over or under corrected. MKV/anyone?

 

David

 

PS I think the max departure from a plane was given in fringes, not waves


Edited by davidc135, 12 November 2019 - 12:58 PM.


#3 BGRE

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Posted 12 November 2019 - 04:08 PM

It will depend somewhat on the optics being tested. Raytracing the setup works best. Otherwise halve the distance and see if any changes are detectable. Repeat for double the distance.

 

A formula that depends on the focal ration alone is unlikely to be correct. 

 

The rule will fail if the refractor objective diameter is made large enough for any given focal ratio.

The same is true for a reflector. 

 

Its very easy to derive the correct formula for a Newtonian, for a refractor it is somewhat more difficult.



#4 BGRE

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Posted 12 November 2019 - 09:35 PM

The proof of incorrectness of the formulae quoted is as follows:

1) If one has a telescope for which the SA due to the finite distance of the star is say 0.1 wave

2) Scale everything (apart from the wavelength) by a factor of 10, the F# is unchanged but the SA is now 1 wave (it scales along with everything else)

Hence any formula that predicts that the minimum artificial star distance purely depends on the image space F# is incorrect.



#5 BGRE

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Posted 13 November 2019 - 02:28 AM

Example #1

F/8 125mm Newtonian Wavefront error is about 0.0.08 waves ptv for a point source 10m away

 

Example #2

F/8 250mm Newtonian Wavefront error is about 0.16 waves ptv for a point source 20m away

 

Example #3

F/8 250mm Newtonian Wavefront error is about 0.08 waves ptv for a point source 40m away. 

 

Example #4

F/4 250mm Newtonian Wavefront error is about 0.08 waves ptv for a point source 160m away.

 

In practice the closest usable distance may be limited by focuser travel and possibly by vignetting in the focuser drawtube or by the secondary.

 

Thus the minimum distance is proportional to D^2 for a fixed F# and proportional to (1/F#)^2 for a fixed diameter.

 

ie minimum distance is likely proportional to ((D^4)/(F^2))/wavelength for a Newtonian.

 

The minimum distance will be different for a refractor.


Edited by BGRE, 13 November 2019 - 02:52 AM.


#6 MKV

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Posted 13 November 2019 - 03:41 AM

That sounds OK. I get an error of 1/10 wave at that distance.

 

My calculation is based on:

Mladen (MKV) posted that for max error of 1/20 wave the allowable departure from plane of a spherical 'flat' used to test an objective = 3.6x(focal ratio)^2 so in your case 230 fringes or 115 waves green light or 57 microns. The distance you place your point source is the radius of curvature of the sphere whose sag over 6'' should be no more than 57 microns

 

Edmund Optical have a sag/ roc calculator so feeding in 80 feet for your F/8 I get 115 microns so not good enough for 1/20 wave accuracy but 1/10 wave should be OK. I don't know if that's over or under corrected. MKV/anyone?

 

David

 

PS I think the max departure from a plane was given in fringes, not waves

It is fringes of power, not waves. C. R. Burch, who published the article in 1938 for the Royal Astronomical Society, standardized the formula for #fringes (of power) = 3.6*(f-number)2 for a 1/20 wave criterion.

 

You're correct that for an f/8 the limit of tolerance for a 1/20 wave OPD residual will be 230 fringes or 115 waves of green light of 0.00055 mm wavelength. That = 115 waves*0.00055 = 0.06325 mm or 63 microns, not 57 microns.

 

Placing a light source at a relatively short distance from a concave mirror will cause over-correction. With a light source placed at 80 feet or 24,384 mm from the mirror, raytracing confirms that a 150 mm f/8 paraboloidal mirror will show a "best focus" residual optical path difference of +λ/21 (or 0.046) waves,.

 

150_f8_80 ft conj.jpg , which is predicted by the Burch formula.

 

Mladen

 

PS C.R. Burch is a very well known scientist and is unlikely that he offered a wrong equation. His RAS article is linked below

 

http://adsabs.harvar...MNRAS..98..670B


Edited by MKV, 13 November 2019 - 03:43 AM.


#7 BGRE

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Posted 13 November 2019 - 04:03 AM

Sure, but when you misuse his results you get nonsense. 

His results only apply to the tolerance for the RoC of an autocollimation flat

Do some raytracing of the actual setup for a test with an artificial star (point source) at a finite distance and you will see that your reasoning is entirely false.

If the wavefront at the objective is spherical refocusing will compensate the defocus leaving just the SA due the finite object distance when using an objective designed for an infinite conjugate.

 

For a 6"F/15 Fraunhofer doublet objective the object distance of an artificial star can be surprising close before the wavefront error exceeds 0.1 waves ptv.



#8 MKV

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Posted 13 November 2019 - 04:35 AM

Do some raytracing of the actual setup for a test with an artificial star (point source) at a finite distance and you will see that your reasoning is entirely false.

My raytrace example of a point source placed at 80 feet shows a residual of +0.046 waves for a 150 mm f/8 parabooid. What reasoning?



#9 BGRE

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Posted 13 November 2019 - 04:53 AM

Citing Burch's paper as relevant was your first mistake.

 

You don't appear to have understood the proof of the incorrectness of any formula that depends only on the F#.

 

A single example is inadequate you need to determine the effect on the source distance required to achieve the same wavefront error when you double the aperture etc.



#10 Gleb1964

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Posted 13 November 2019 - 05:05 AM

Bruce, cited formula is relevant, because it related to a sag of wavefront, and sag related to the object distance including mirror diameter, not only F# ratio.

 

correction: Of course it relevant for mirror, for refractor can be another story.

 

Gleb


Edited by Gleb1964, 13 November 2019 - 05:29 AM.


#11 BGRE

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Posted 13 November 2019 - 05:43 AM

The quoted formulae by the OP is still incorrect.

No one seems to have worked out that Burch's result is actually:

R is proportional to D^4/F^2

not:

R is proportional to (F/D).



#12 BGRE

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Posted 13 November 2019 - 06:20 AM

Bruce, cited formula is relevant, because it related to a sag of wavefront, and sag related to the object distance including mirror diameter, not only F# ratio.

 

correction: Of course it relevant for mirror, for refractor can be another story.

 

Gleb

A quasi autostigmatic setup with a reflective convex sphere of radius R has

double the SA that results when a point source located at a distance R is used.

Thus mindless application of Burch's results predicts a distance larger than actually needed for a given wavefront error.

This matters because the larger predicted distance may be longer than the space available whereas the actual minimum distance required (~0.5 the value predicted by using Burch's result) may fit in the available space.


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#13 Gleb1964

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Posted 13 November 2019 - 06:22 AM

Burch equation is about mirror sag, and it is proportional to (F/D)^2. Mirror power makes a spherical wavefront, similar to finite distance object.

Wavefront sag on aperture D connected to object distance by simple formula.

And yes, Bruce, correction is needed for coefficient compare to AC setup.

 

Burch Tolerances permissible for AC flat.jpg


Edited by Gleb1964, 13 November 2019 - 06:26 AM.


#14 BGRE

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Posted 13 November 2019 - 06:28 AM

I've read the paper.

The sag is approximately (1/4)*D^2/R

hence the result R proportional to D^4/F^2.


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#15 davidc135

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Posted 13 November 2019 - 10:05 AM

So, I was being twice as stringent as needed. An artificial star at 40 feet could be fine for the op's example. Tag1260, would you be using a Ronchi? As has been said maybe keep reducing the 'star' distance until the Ronchi (or knife edge) appearance changes. I'd be interested to read your results.  David



#16 MKV

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Posted 13 November 2019 - 02:39 PM

I've read the paper.

The sag is approximately (1/4)*D^2/R

hence the result R proportional to D^4/F^2.

The sag ≈ D²/8R



#17 MKV

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Posted 13 November 2019 - 03:53 PM

So, I was being twice as stringent as needed. An artificial star at 40 feet could be fine for the op's example. Tag1260, would you be using a Ronchi? As has been said maybe keep reducing the 'star' distance until the Ronchi (or knife edge) appearance changes. I'd be interested to read your results.  David

No you weren't being too conservative. Placing the light source at 40 feet will show up on a 133-lpi Rionchi test as an obvious overcorrection.  To show straight bands, you'll need at least 60 feet. 

 

6inf8 doublet w 40  ft conj.jpg



#18 BGRE

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Posted 13 November 2019 - 04:25 PM

The sag ≈ D²/8R

No, the wavefront sagitta after reflection from a sphere with an RoC of R, is D^2/(4*R)


Edited by BGRE, 13 November 2019 - 04:26 PM.


#19 BGRE

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Posted 13 November 2019 - 04:46 PM

The correct formula for the minimum distance for the artificial star appears to be of the form

 

minimum distance = (constant/epsilon)*(D^4/(Lambda*F^2))   ………………… (1)

 

where 

           lambda is the test wavelength

           D is the entrance pupil diameter

           F is the focal length of the test system 

           epsilon is the desired upper limit to the wavefront  error (measured in waves) due to finite distance from the entrance pupil to the artificial star.

 

 

and the value of the constant depends on the optical system being tested.

with the caveat that the system being tested may actually have zero SA for a particular distance from the artificial star and the system entrance pupil.

In the case of a concave mirror being tested an ellipsoidal mirror will have zero SA for a particular artificial star to mirror distance.

 

Raytrace results for paraboloids and Fraunhofer doublets are consistent with (1).



#20 MKV

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Posted 13 November 2019 - 06:53 PM

No, the wavefront sagitta after reflection from a sphere with an RoC of R, is D^2/(4*R)

No, the mathematical relationship sag ≈ D²/8R remains the same, and is applicable to any D and R. What changes on reflection is R, which in turn gives a different sagitta.



#21 tag1260

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Posted 13 November 2019 - 07:08 PM

OK. So you guys all lost me quite a few posts ago!!!  I flunked math as soon as letters decided to take over!!!!   lol.gif    

 

Anyways, I'm just looking to check and collimate my refractor. I would normally use a star but those are few and far these days.

 

So it sounds like I should be OK at 80 feet.

 

Thanks


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#22 BGRE

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Posted 13 November 2019 - 07:15 PM

If and only if the R is taken to be the radius of curvature of the wavefront.

 

In any case the exact value of the multiplier has no effect on the conclusion:

 

 minimum distance = (constant/epsilon)*(D^4/(Lambda*F^2))   ………………… (1)

where
           lambda is the test wavelength
           D is the entrance pupil diameter
           F is the focal length of the test system
           epsilon is the desired upper limit to the wavefront  error (measured in waves) due to finite distance from the entrance pupil to the artificial star.

 

This is just the constraint imposed by the dependence of SA on the object conjugate due to the test system properties, there are other constraints to be met as well:

 

2) The minimum distance to the artificial star should be such that an unvignetted real image of it can be formed by the optics within the range of the focuser.

 

3) The diameter of the artificial source should be smaller than half the object space Airy disk diameter 

 

i.e. source diameter < 1.22*(source distance/D)*lambda or 

source distance > ((source diameter)/1.22)*(D/lambda)



#23 BGRE

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Posted 13 November 2019 - 07:31 PM

OK. So you guys all lost me quite a few posts ago!!!  I flunked math as soon as letters decided to take over!!!!   lol.gif    

 

Anyways, I'm just looking to check and collimate my refractor. I would normally use a star but those are few and far these days.

 

So it sounds like I should be OK at 80 feet.

 

Thanks

 

Test at some distance greater than the minimum distance you can focus at, double the distance if the aberrations change then double it again until the difference in aberrations between successive distances is undetectable, use the larger distance. This doesn't require any raytracing but the artificial star diameter has to be small enough that any features are unresolved at any test distance used.

 

if the test wavelength is say 550nm (green near peak response of human photopic (colour) vision) and the telescope aperture is 6" then the maximum source diameter are

 

DISTANCE(feet)   DIAMETER(inches)

 

20                         0.001

 

40                         0.002

 

80                         0.004

 

160                       0.008



#24 MKV

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Posted 13 November 2019 - 07:32 PM

You don't appear to have understood the proof of the incorrectness of any formula that depends only on the F#.

All the equations involved contain the aperture test optic's diameter (D) and radius of curvature ®. From that one can express the aberrations and other parameters in terms of those, including the F#, which is simply R/2D.

 

Here are some examples:

 

1) Long. spherical aberration of a mirror SA = D²/32f where f = R/2, hence SA = D²/16R

 

2) The sine of the marginal ray's incidence angle sinU = 0.5/F# = D/R

 

3) max. OPD (in waves) = (D²/16R)*(sin²U/4λ) = D4/64R3λ, where λ is the wavelength (0.0005461 mm, mercury green light)

 

etc.

 

So, whether you use the F# quantity of D/R relationship should not change the outcome, if done right.



#25 BGRE

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Posted 13 November 2019 - 07:54 PM

Yes the aberration does not depend solely on the F#, the diameter is also important.

Why introduce the RoC of the test surface ? This only leads to confusion with the RoC of the return "flat" in Burch's analysis.

Its also not applicable to a refractor whereas formulating the equation using the focal length and diameter is, albeit with a different constant, of the same form for both a refractor and a Newtonian telescope.

A more general equation applicable to refractors and paraboloids is more useful.

 

Also your equations are confusing/incorrect for the case of an object at a finite distance.

In image space one should use the effective F# for the object with a finite conjugate, this is larger than the F# for an object with an infinite conjugate. 

In this case 0.5/F# is not equal to D/R.




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