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determin the negative element of an achromat

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#1 BabyPepper

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Posted 09 September 2019 - 07:00 PM

does anybody know the best way to determin the focal length of the plano concave element inside prebuilt telescopes?

 

Example,  the meade etx80.   The focal length of the first singlet lens is easy to figure out just by using a ruler,  but the second element is negative and not so easy.

 

I need a quick and easy method to do this for any telescope objective using an acromat.

 

 

The reason i need this ability is to locate a very specific negative plano concave optic,  because nobody on the market sells them in the size i require;  

 

That being said, i need to cannibalize the negative element from a premade manufactured achromat such as from a meade or celestron telescope.

 

 

Specifically i need an 80mm x -800mm plano concave  or a 100mm x -1000mm plano concave.

 

 

Obviously if i can have a custom lens made that is the best way to go, but i dont think anybody out there is able to do this.

 

PLEASE HELP!



#2 howardcano

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Posted 09 September 2019 - 07:18 PM

Most cheap achromat objectives are Fraunhofer.  The negative element is not plano-concave.  Here is a description that might help you:
https://www.telescop...t/achromats.htm


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#3 NinePlanets

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Posted 09 September 2019 - 08:09 PM

Make or buy a spherometer then sharpen up your calculator pencil. smile.gif



#4 BGRE

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Posted 09 September 2019 - 08:23 PM

Make or buy a spherometer then sharpen up your calculator pencil. smile.gif

You still need to know the refractive index of the glass used.

 

However none of this is necessary for a doublet:

 

1) measure focal of  the doublet

 

2) move the focal length of the positive element

 

3) calculate the focal length of the negative element from the results of 1 and 2 


Edited by BGRE, 09 September 2019 - 08:35 PM.

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#5 sg6

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Posted 09 September 2019 - 08:33 PM

As it is easy calculate the first one, then as you know the combined focal length use the simple lens formula to get the second.

 

On an ETX 80 the final focal length is 400

So in the simplest term you have 1/400 = 1/f1 +1/f2

Where f1 is the front positive lens.

 

Better is to estimate the seperation and use the formula that includes d for the seperation of the 2 lens. As said extimate that by measuring between the centers of the edges of the 2 lens.

 

Off the top of my head 1/400 = 1/f1+1/f2 - d/(f1*f2)

d being the distance between the lens centers - sort of good/best guess/approximation.

 

f1 being determined by the radaii and the refractive index - guess at BK7 glass for that.

 

Then you have the final (400) and the/a value for f1.


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#6 mashirts

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Posted 09 September 2019 - 08:40 PM

80mm f5 objectives are plentiful either from 80mm st refactors or you can buy just the objective from China on ebay.

https://www.ebay.com...r-/273513526135

#7 BabyPepper

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Posted 09 September 2019 - 09:39 PM

80mm f5 objectives are plentiful either from 80mm st refactors or you can buy just the objective from China on ebay.

https://www.ebay.com...r-/273513526135

i do not want an 80mm f5 objective.

 

i want an 80mm x -800mm plano concave , or similar negative meniscus :)   

 

the 80mm f5 was just a reference.

 

 

thanks to the rest of the posts with the math, but it looks like i am SOL on finding what i need.

 

 

I require this for a collimating lens and nothing more.

 

 

if any of you can suggest a lens manufactuing optiician i would love the information.  This lens is very important!



#8 BabyPepper

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Posted 09 September 2019 - 09:59 PM

As it is easy calculate the first one, then as you know the combined focal length use the simple lens formula to get the second.

 

On an ETX 80 the final focal length is 400

So in the simplest term you have 1/400 = 1/f1 +1/f2

Where f1 is the front positive lens.

 

Better is to estimate the seperation and use the formula that includes d for the seperation of the 2 lens. As said extimate that by measuring between the centers of the edges of the 2 lens.

 

Off the top of my head 1/400 = 1/f1+1/f2 - d/(f1*f2)

d being the distance between the lens centers - sort of good/best guess/approximation.

 

f1 being determined by the radaii and the refractive index - guess at BK7 glass for that.

 

Then you have the final (400) and the/a value for f1.

the first formula   1/400 = 1/f1 +1/f2       as a reference  1/400 = .0025     how do i apply that to the focal length?  i am just going to assume the negative in this instance is -400.  so .0025 + -.0025 = 0?   

 

lol. I am not looking for someone to do the math for me, but something about 0 is not right obviously.   The spacer is 1mm.



#9 MKV

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Posted 09 September 2019 - 10:05 PM

BMV Optical Tehcnologies (Canada)


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#10 BabyPepper

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Posted 09 September 2019 - 10:14 PM

another thing on these formulas, they dont appear to be for actual focal lengths. this just appears to be the measurements of the curviture. 

 

because looking at my etx60 objective taken apart, the focal length of the bi-convex is certainly not .00285.     (1/350....)   it is about 150mm for the moon.



#11 BabyPepper

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Posted 09 September 2019 - 10:18 PM

BMV Optical Tehcnologies (Canada)

thanks, just sent them an email requesting a custom lens.  

 

*Crosses fingers*



#12 TOMDEY

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Posted 09 September 2019 - 10:24 PM

does anybody know the best way to determin the focal length of the plano concave element inside prebuilt telescopes?

Example,  the meade etx80.   The focal length of the first singlet lens is easy to figure out just by using a ruler,  but the second element is negative and not so easy.

I need a quick and easy method to do this for any telescope objective using an acromat.

The reason i need this ability is to locate a very specific negative plano concave optic,  because nobody on the market sells them in the size i require;  

That being said, i need to cannibalize the negative element from a premade manufactured achromat such as from a meade or celestron telescope.

Specifically i need an 80mm x -800mm plano concave  or a 100mm x -1000mm plano concave.

Obviously if i can have a custom lens made that is the best way to go, but i dont think anybody out there is able to do this.

PLEASE HELP!

Hi, BabyP! Sure, I just thought about it and solved it here >>>

 

Pretty self-explanatory: You get a bright pin-light (A MagLite with the reflector part screwed off works great!) and position it pretty far from the lens. The lens is in a hole in a baffle (sheet of cardboard or even paper will do). You position a screen (e.g. a wall) downstream, so you get an enlarged circle of light there. Just measure A, B, d, D there, and (negative) focal length F pops out of my equation.

 

e.g. If A = 2000 mm

          B = 500 mm

          d = 100 mm

          D = 200mm

 

Then, we get F = -667 mm   That's all there is to it!

 

[No, I didn't "look it up" --- always derive stuff from scratch, keeps the little grey cells from going vestigial.]    Tom

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#13 MKV

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Posted 09 September 2019 - 10:26 PM

If you want a plano concave rear refractor lens (presumably flint glass, such as F2, with ref. index n = 1.62408  in the green of 550 nm) whose focal length f = -800 mm, then its radius of curvature R = (n-1)f

 

R = (1.62404-1)(-800) = -499.264 mm

 

Mladen

 

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edit typo should read (1.52408-1) 


Edited by MKV, 09 September 2019 - 10:35 PM.


#14 TOMDEY

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Posted 09 September 2019 - 10:31 PM

PS: My technique up there has the advantage that all you need is the negative lens and a pinlight --- nothing else.    Tom



#15 howardcano

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Posted 09 September 2019 - 10:36 PM

Using the numbers on the web site I gave, you would need the negative element from one of the following Fraunhofer doublets using BK7 and F2:
100x1280mm
80x1025mm
The 100mm is probably rare, but you might be able to find an 80x1000mm.


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#16 BabyPepper

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Posted 09 September 2019 - 10:59 PM

Hi, BabyP! Sure, I just thought about it and solved it here >>>

 

Pretty self-explanatory: You get a bright pin-light (A MagLite with the reflector part screwed off works great!) and position it pretty far from the lens. The lens is in a hole in a baffle (sheet of cardboard or even paper will do). You position a screen (e.g. a wall) downstream, so you get an enlarged circle of light there. Just measure A, B, d, D there, and (negative) focal length F pops out of my equation.

 

e.g. If A = 2000 mm

          B = 500 mm

          d = 100 mm

          D = 200mm

 

Then, we get F = -667 mm   That's all there is to it!

 

[No, I didn't "look it up" --- always derive stuff from scratch, keeps the little grey cells from going vestigial.]    Tom

dude that is perfect,, but unfortunatley i will not be able to buy all the telescope objectives to find the required needle in the haystack.

 

so i gotta figure this out on paper, that way i just buy the right one the first time around ;)

 

 

Using the numbers on the web site I gave, you would need the negative element from one of the following Fraunhofer doublets using BK7 and F2:
100x1280mm
80x1025mm
The 100mm is probably rare, but you might be able to find an 80x1000mm.

looks like you just saved me alot of time on that mystery.   100mm x 1200mm are pretty common, but looking at $300+ which is not too bad.   I think id rather throw that at the manufacturing however.

 

80mm x 1000mm is likely the more rare one ;)   but i will try to find it.

 

thank you all for contributing to this mission,   I am trying to optimize a huge hydrogen alpha telescope for the solar mercury transit.   Specialized Negative collimating elements are not something off the shelf, so some sort of compromise is a little bit acceptable.

 

BMV looks like the best option at this point because they can coat it and optimize the lens to the required wavelength of 656nm, and if they can doit for less than $300.00 then its done.

 

again, thank you for all your help.



#17 MKV

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Posted 09 September 2019 - 11:48 PM

The price will depend on your specs. A high precision lens (say λ/10 surface) will cost more than $300. You can ask them what precison you can get with coating and all that (including homogenity grade glass) for USD 300 + shipping.


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#18 BabyPepper

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Posted 10 September 2019 - 12:07 AM

The price will depend on your specs. A high precision lens (say λ/10 surface) will cost more than $300. You can ask them what precison you can get with coating and all that (including homogenity grade glass) for USD 300 + shipping.

it will be worth whatever they ask for it smile.gif

 

$500 upgrade to maximum performance and also save $7000.00 on a sub angstrom solar telescope is worth every penny

 

this is what im seeing now with my non-optimized setup,  and it can get ridiculously better with one lens which is crazy to think about.

 

48667841026_197dfdc3be_o.gif

 

48686969737_5489903fb0_o.gif


Edited by BabyPepper, 10 September 2019 - 12:09 AM.

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#19 MKV

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Posted 10 September 2019 - 01:41 AM

Awesome!



#20 BabyPepper

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Posted 11 September 2019 - 10:07 AM

OUCH!  Just got my quote back on the 80mm x -800mm single element unmounted commercial grade planoconcave lens.  .....  $1500.00

 

Quite laughable considering i can get 150mm h-alpha optics pre mounted with .98 strehl at that cost.

 

 

Looks like BMV is a "never gonna happen" option.

 

 

 

Back to cannibalizing the negative meniscus from a $25.00 meade telescope.



#21 starcanoe

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Posted 11 September 2019 - 12:08 PM

Okay....OP got luck in that I am motivated, mobile, in a good mood and the excrement seems to not  hitting the fan today ( though a possible hurricane appears to be lurking now).

 

I have a Jaeger's 5 inch lens set. F5 ish. So, I got the lens out in the sun. I measured the focal length of the set at 23.5 inches. I measured the lens separation (center of element to center of element) at about 0.5 inches. Measured the positive element at 11.5 inch focal length. Focal lengths probably within an inch and maybe more like 0.5 inches. Rather that do any algebra I just used "Astroboys" online calculator for DIY eyepieces to solve for the focal length of the negative lens.

 

I came up with -21.5 give or take.  That is about a minus 540mm. Not what the OP needs.

 

However, going with the assumption that a stereotypical achromat (using bog standard glass and design..which I am fairly sure describes most of Jaegers mainstream products). I did the following calculations.

 

What if you had a 4 inch f8 Jaegers lens set? So, lets scale that....32/23.5 would give you a factor of 1.36.

 

Or if you had a 4 inch f10 Jaegers lens set? 40/23.5 would give you a factor of 1.7 or so.

 

1.36 times 540 is about 730mm.

 

1.7 times 540 is about 920mm.

 

So, at first glance it sounds like a 4 F8 Jaegers lens might work...and an f10 is right in the ballpark.

 

Of course more careful measurements would be needed....

 

Good news is that an f8 lens set is for sale right now here on CN.

 

Or another way to look at it. There is a good chance that any run of the mill bog standard achromat that would work needs to be in the 800 to 1000mm focal length for the negative lens to be in the ball park. Are there 80 to 90 mm achros that have those kind of focal lengths? (I have an old Meade 90/f8.8/800 but THAT lens set is glued together...though one supposes they could be separated given the right know how).



#22 starcanoe

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Posted 11 September 2019 - 12:15 PM

Oh....another option would be for the OP to drop Fred an email at Surplus Shed...there is always the chance that he has a lens or two that fits the bill but he has not bothered to put it up online because it wouldn't be worth the time for such a small sales volume.



#23 starcanoe

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Posted 11 September 2019 - 01:36 PM

Also looks like Surplus Shed had some standard 80, 90, 100 mm diameter coated airspaced objective lens achromats in the right ballpark focal length wise....and at pretty good prices....and Fred often runs major sales where stuff is 40 to 60 percent off on top of that !




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