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How to Measure the Focal Length of Barlow (teleconverter, telecentric, Powermate, concave lens)

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#1 RickV

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Posted 04 January 2017 - 09:41 PM

I had a project where I needed to know the focal length of a Televue 2X Powermate.  I couldn’t find this information on a web search so I decided to measure it myself.   This is how I measured the focal length of a Barlow.

 

The Powemate is a diverging lens assembly and has a negative focal length.
e.g. Catch sunlight with one and there is no convergence of light into an intense focal spot as with a magnifying glass (convex lens).  Instead the light diverges with no apparent focal spot.  Thus measuring the focal length of such a ‘concave’ lens is a challenge.

 

There may be several ways to do this but this is how I measured the focal length of a 2X Powermate.

 

Here’s my Televue 2X Powemate.

Pic-1.jpg

 

Unscrew the extension barrel (black) from the lens assembly (chrome).

Pic-2.JPG

 

Here’s a close up of the lens assembly.  In use, light enters the bottom and exits at the top (black ring).

Pic-3.jpg

 

I needed a laser to perform the measurements and used a Kendrick laser collimator.
Note: Any laser with do; I was going to use my $2 dollar-store laser but couldn’t find it.

Pic-4.jpg

 

I mounted the lens assembly in a vice – very gently.

Pic-5.jpg

 

I mounted the laser in a cross-slide vice and aligned the components.
Note: The back reflection form the first lens element shows on the front of the laser so I would adjust positions to get the laser reflection onto the center of the laser.

Pic-6.jpg

 

I fixed a sheet of graph paper behind the lens assembly so that I could mark where the laser dot hit.  I began with the laser striking the center of the front lens element and marked where the laser dot hit the paper.

Pic-7.jpg

 

TO BE CONTINUED…



#2 RickV

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Posted 04 January 2017 - 09:48 PM

I traversed the laser to the left edge of the lens assembly and marked where the laser dot hit the paper.

Pic-8.jpg

 

I traversed the laser to the right edge of the lens assembly and marked where the laser dot hit the paper.

Pic-9.jpg

 

At the left traverse position and at the right traverse position I marked the cross-slide vice position.

Pic-10.jpg

 

I measured the distance from graph paper to the center of the lens assembly (I used middle of threaded section) as 136mm.

Pic-11.jpg

 

I plotted the results on the graph paper and projected along the lines the laser travelled and beyond to the find the point of convergence (focus).  In my case this indicated a focal length of -134mm from the center of the threaded section of the lens assembly.

Pic-12.jpg

 

Thanks for reading.



#3 Nils Olof Carlin

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Posted 05 January 2017 - 04:38 AM

 

The Powemate is a diverging lens assembly and has a negative focal length.
e.g. Catch sunlight with one and there is no convergence of light into an intense focal spot as with a magnifying glass (convex lens).  Instead the light diverges with no apparent focal spot.  Thus measuring the focal length of such a ‘concave’ lens is a challenge.

 

Edit: this link is better to explain the difference between the Powermate and the Barlow lens

 

http://www.televue.c...id=53&Tab=_back

 

The problem is - that is not true! The Powermates have a first negative lens group playing the part of the Barlow, then a positive group to collect the diverging light into a more (but not perfectly) parallel bundle, to make the job easier for the EP. Thus, no defined focal length, unlike a true Barlow.

(if you have one, this old webpage of mine might be of interest: http://web.telia.com...low/Barlext.htm  )


Edited by Nils Olof Carlin, 05 January 2017 - 06:52 AM.


#4 andreaconsole

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Posted 05 January 2017 - 05:48 AM

I suppose that it works, as a whole, like a negative element with an equivalent (negative) focal length 



#5 Nils Olof Carlin

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Posted 05 January 2017 - 06:47 AM

I suppose that it works, as a whole, like a negative element with an equivalent (negative) focal length 

Not really - there is a fixed combination. But I notice I had given the wrong link:

http://www.televue.c...id=53&Tab=_back

Try this link instead! It should make clearer why your supposition (and the OP's) is not valid.



#6 andreaconsole

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Posted 05 January 2017 - 07:06 AM

I see. Nevertheless, I would say that it still has a focal length, as every combination of lenses has, but it's unrelated to the rays divergence from the axis. For this reason, the measure of the focal length can't be done by measuring the off-axis displacement. Is it correct?


Edited by andreaconsole, 05 January 2017 - 08:17 AM.


#7 Starman1

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Posted 06 January 2017 - 04:42 PM

First, there is the magnification factor provided by the negative lens assembly.

This lens does have a focal length (negative, of course).

It is followed by a positive lens that bends the diverging light cone into an *almost* parallel bundle of rays.

Were the rays completely parallel, there would be absolutely no change of magnification with distance from the lens, but there is some.

See: http://www.televue.c...?id=53&Tab=_app

As you can see, the 2" 2X Powermate comes closest to having parallel rays and being afocal.

The 5X PowerMate functions more like a barlow, with increasing magnification with increased distance from the lens.

 

If the magnification doesn't change with distance, it is afocal and essentially has no focal length at all.

It's a brilliant idea, since induced aberrations like edge of field astigmatism, vignetting, and increase of eye relief do not occur in the eyepiece.

Or changes in distortion.

It is similar to a Bravais lens, used in photography and image projection.

 

Here is a description of what Telecentric means: https://en.wikipedia...elecentric_lens

 

Of course, if the distance to the lens is too long, vignetting will occur in the barrel between lens and eyepiece, similar to what happens when you back away from an eyepiece while watching the image.

 

Not all 4 element barlows, even those with a similar internal configuration, are parfocal with the eyepiece used separately.  One of the goals for the PowerMates was that, when inserted, the eyepiece

would still be in focus.  If you think focal plane, then the PowerMates move the focal plane of the scope backwards an amount equal to the length of the barrel above the resting shoulder of the PowerMate.



#8 andreaconsole

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Posted 07 January 2017 - 11:28 AM

I'm not sure you can have magnification without reducing the convergence of incoming rays. As far as I understand, telecentric is only about keeping parallel to the optical axis the rays that were parallel on their arrival, but for achieving the barlow effect I suppose you should still have an overall negative focal length. Can you elaborate a bit more on that?

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#9 Starman1

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Posted 07 January 2017 - 11:57 AM

The 4 element telecentric barlow provides magnification with a negative lens that provides divergence, followed by a positive lens that pulls the rays into approximately parallel rays.

So you have magnification but the rays hitting the eyepiece are easy to handle.

Of course, rays hit the bottom of the telecentric barlow as they are converging, so, after passage, they could be converging still, but just with the larger image scale provided by the magnification.



#10 andreaconsole

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Posted 07 January 2017 - 02:52 PM

Just to be absolutely clear, B is what I would expect from a telecentric barlow (convergent rays emerge with a minor convergence, collimated rays become divergent). Then I understand that you are saying that A (focal length is infinite, so collimated rays stay collimated) has still a "barlow effect". How can the image result magnified if the convergence angle, and thus the equivalent focal length, is unchanged? (sorry if the picture is not perfect, but I hope that the meaning is clear)

Attached Thumbnails

  • A.gif
  • B.gif

Edited by andreaconsole, 07 January 2017 - 02:57 PM.


#11 Starman1

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Posted 07 January 2017 - 05:12 PM

OK, the size of any image is determined by its image scale.

If the field has 1' per millimeter, then a 42mm field will be 42' wide.

Increase the image scale with a 2X barlow negative lens, and now 2mm = 1'.

The size of the image scale has doubled.

Now the same 42mm wide field is only 21' wide and everything in that field is twice as wide (and 4X the area).

This is what happens with a standard barlow lens at its operating distance.

 

But, the farther you are from the lens, the more the divergence, the larger the magnification and image scale.

 

Add a second lens to prevent further divergence of the beam, and the magnification will be independent of distance from the lens.

The divergence will have occurred before the second lens, like your drawing in A.  This apparently is like the 2X 2" PowerMate.

Look at the cross section of the beam.  Notice it is wider after the lens, which represents an increased image scale and greater magnification of the image.

 

This divergence, however, definitely would have a different character depending on the rays hitting the bottom lens.

If the lens is made such that there is still some divergence after the positive lens, the picture in B might pertain, and this seems to be the case

for the 5X PowerMate since the magnification continues to increase with increased distance from the lens.

 

A Focal REDUCER would have a shrinking light cone after it, so that increasing distance would result in a smaller image scale.  A good example is the f/6.3 SCT reducer, which functions more like f/5.5

or even f/5 when a large 2" diagonal is used with a greater back focus than the 1.25" diagonal and eyepieces for which it was designed.

 

Here's the thing about a negative lens--it has a distinct curvature and focal length.  So, the angular divergence of the beam after the lens depends on the angle of the rays entering it (not the %).

If all rays hitting the lens are parallel, it will expand the field at a particular rate.  If the rays hitting the lens are strongly converging, one way to think of it is that a 2X diverging lens turns f/3 to f/6 (like the first half of A) but turns f/10 to f/20 (like the first half of B). Needless to say, the slower the light cone entering the negative lens, the closer to parallel they will be after exiting. The rays will still converge, but at a much slower rate.  The % of increase of the apparent focal length would be the same, however.

 

This will explain it better than I can: http://www.brayebroo.../BarlowLens.pdf



#12 Ed Jones

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Posted 09 January 2017 - 12:49 PM

One way to determine the focal length of a barlow is to put it in a scope that you know the aperture and focal length and focus it so that light comes out parallel.  Projecting the sun onto a far wall for example would be a close approximation.  Then looking back into the scope (not at the sun of course) measure the exit pupil with calipers.  Divide the aperture of the scope by the exit pupil will give you the magnification and dividing the scope focal length by the magnification gives the barlow focal length.  Works for eyepieces too.


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#13 Nils Olof Carlin

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Posted 09 January 2017 - 05:49 PM

Ed,

the method you describe works, but one difficulty is that the exit pupil seen with a negative lens (Barlow or Galilean EP) is inside the tube - as opposed to with a "normal" EP, where the exit pupil is reachable with calipers. Looking from a good distance might work well enough though. The other difficulty remains - not so much measuring the focal length, as definning the focal length of an essentially afocal lens combination.



#14 andreaconsole

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Posted 07 February 2017 - 08:21 AM

In my effort to understand how a telecentric Barlow works, I came to this conceptual scheme (not precise, not in scale). Do you see any flaw in my reasoning?

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#15 Nils Olof Carlin

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Posted 07 February 2017 - 11:31 AM

I see it somewhat differently. To construct images with the simplified "thin lens" model, you can use 2 lines: one parallel to the optical axis, and refracted towards (or from, as the case is) one focal point. The other passes through the center of the lens, and it is not refracted. (For this construction, these rays need not be physically possible due to vignetting).
This is what I tried to do here - I hope I have done it right, but indeed the whole task was a bit messy.

Start with the real image from the objective, as a given. Now place the Barlow lens so that the real image is about midway between the lens and its (outer) focal point. Construct the Barlowed image - a real image about twice the size of the original (or what you find convenient).

To make things telecentric, you add a positive lens between the barlowed image and the Barlow itself, such as the (inner) focal points of the positive and the Barlow lens coincide - this is the condition to have a ray parallel to the optical axis emerge from the combination still parallel to the optical axis. Now construct the new (real) image, a little smaller than the Barlowed image but larger than the original.

Note that the "focal ratio" is not a physical parameter - it does not denote a new virtual position for the objective. This is why I don't like to think of the Barlow affecting the focal ratio - better to see it as simply enlarging the image (or image scale) presented to the EP or camera.

nr2.gif



#16 petmic

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Posted 07 February 2017 - 11:46 AM

I am just wondering if one can use a star drift method (first test without barlow and then second test with barlow) to establish the focal length of barlow? 



#17 Nils Olof Carlin

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Posted 07 February 2017 - 12:11 PM

I am just wondering if one can use a star drift method (first test without barlow and then second test with barlow) to establish the focal length of barlow? 

What you can establish with star drift is the actual magnification of the Barlow/EP combination you use. The nominal 2x (for instance) holds only for one position of the EP focal plane - if you move out the EP, you increase the magnification.

I fear star drift is not useful to decide the Barlow focal length.



#18 andreaconsole

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Posted 08 February 2017 - 02:35 AM

Nils, thank you for your drawing and for your explanation: I couldn't find the rule for designing a telecentric scheme, and I had only a few hints I used to prepare my drawing (a negative lens first and a positive one after and an overall enlargement).

I didn't mean to produce a precise ray tracing - even if I know the general procedure since I was 16 - for two reasons: first of all, as said, I didn't know how to properly design a telecentric Barlow; secondly, I just wanted to express the basic concept that the resulting image can be bigger even if the focal ratio is unchanged.

The focal ratio is another interesting topic I would like to dig into: as everybody knows, for a simple lens it is FR = F/D. For more complex schemes, in my understanding, it is a number related to the resulting convergence of rays coming from infinity (e.g. think about an SC telescope). This is why I extended the converging lines to the left in order to better show their convergence and, ultimately, the effective focal length, which I calculate as EFL = FR*D. I'm still wondering if I should instead consider the magnification for calculating the EFL since it's the first time that I notice that magnification and ray convergence can be uncorrelated...


Edited by andreaconsole, 08 February 2017 - 02:37 AM.


#19 Nils Olof Carlin

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Posted 08 February 2017 - 06:00 AM


It took me a fair bit of work to figure out one way of implementing a "telecentric image magnifier" - I hope it, and my drawing, is essentially correct, but this is nothing I've read about or believe I know well.

How I understand focal ratio: The light-collecting diameter of the aperture is usually known (obstruction disregarded), but what about effective focal length? in a telescope, aimed at an objective of angular size V (in radians: convert from degrees by multiplying by (2*pi)/360 or 1/57.3), you get an image of size B: then the focal length by definition is F=B/V.

For an EP this is used the other way round - for a linear image size B at its focal plane, appearing to the eye at an angle V, the focal length is F=B/V as well.
Example: The moon at 0.5 deg gives an image of 10 mm diameter, the effective focal length is 10*57.3/0.5 = 1146 mm. And if this image completely fills a 50 deg FOV, the f.l. of the EP is 10*57.3/50=11.5 mm.*)

So if you use a Barlow or Cassegrain secondary to enlarge the image of the objective alone, the effective focal length is increased, but since the aperture is the same, the effective focal ratio (for that focal plane!) must also be increased (answering your last sentence).

*) Distortion ignored. In cartography, you want to map part of a sphere (unless of course you prefer alternative facts  :grin: ) on a flat map - some distortion must be, but you choose the projection least unsuitable for your particular purpose. In a telescope you have the inverse problem - to map an essentially flat image on a virtual celestial sphere. If lines are straight, circles are deformed, etc. But my formula holds closely for "modern" EPs AFAIK.



#20 andreaconsole

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Posted 08 February 2017 - 06:25 AM

Ok, you are suggesting the alternative point of view I had in mind. It makes sense, thus it's probably the correct way to look at the problem. Moreover, I like a lot your definition of focal length, which look more general than "the distance from the lens where incoming parallel rays converge in a point" I've always had in mind. Do you have an official font for this definition? Nevertheless, I still find it disturbing that the angle of the exiting cone of light generated by incoming parallel rays can't be related to the focal ratio in every case (it's not the case for common Barlow, though)... Anyway, in your understanding, which alternative parameter could I use to define this angle?


Edited by andreaconsole, 08 February 2017 - 10:54 AM.



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