I'm attempting to understand magnification when using an infinitycorrected objective, tube lens and bellows for macro.
With a 20X objective attached, the FOV @20X on sensor is, by my calculations, 1.1mm.
*Based on an approximate FOV on an APSC (Canon 50D, 22.3X14.9mm).
22.3/1.1=20.27
When a dedicated Nikon tube lens or '2nd objective lens' Part # MXA20696 is positioned @ 200mm from sensor as outlined in the linked thread (below), the the FOV decreases to 0.95mm and magnification thereby increases to 23X (using the simple calculation above).
http://www.photomacrography.net/forum/v ... hp?t=12543
So, either my calculation is too simple or there is something else happening which I do not fully understand.
Craig
Magnifcation and infinitycorrected objectives.
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 Craig Gerard
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Magnifcation and infinitycorrected objectives.
To use a classic quote from 'Antz'  "I almost know exactly what I'm doing!"
 rjlittlefield
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I'm unclear what calculation you're doing. Let me summarize the few calculations I would do, and maybe you can see how they compare with yours.
1. The objective has a certain focal length, let's call it FL_obj.
2. The tube lens also has a certain focal length, let's call that FL_tube.
3. The objective is designed to be used with the tube lens focused at infinity. In this case, the calculation is simple:
4. If you add extension behind the tube lens, so that the tube lens no longer focuses at infinity, then you increase the magnification of the combo.
But in this case the magnification does depend on separation between objective and tube lens, and it's not easy to predict because the calculation depends on a number you don't know: the separation between rear principal plane of the objective and front principal plane of the tube lens.
I'm guessing that you're trying to calculate the magnification with added extension by either
(1) using the basic lens equation that 1/f = 1/f1 + 1/f2, or
(2) assuming that magnification scales as (FL_tube+added_extension)/FL_tube.
But in fact, when there is separation "d" between the principal planes, the governing equation is what's described at Wikipedia, HERE:
1/f = 1/f1 + 1/f2  d/(f1*f2)
An alternate form, paraphrased from page 28 of "Practical Optical System Layout..." by Warren J. Smith, is:
f = (f1*f2) / (f1 + f2  d)
If d is zero (no separation between the principal planes), then the basic lens equation does apply (because 1/f1 + 1/f2  d/(f1*f2) collapses to just 1/f1 + 1/f2). In this unlikely case, calculation (1) above gives a good prediction.
If d is exactly equal to the focal length of the shorter lens, then the effective focal length of the combo becomes exactly equal to that of the shorter lens. In this unlikely case, calculation (2) above gives a good prediction.
But if d is anything else, which it probably is, then the change of magnification with added extension is not accurately predicted by either (1) or (2).
Yes, it is possible to determine d and then use that value for further calculations. But this is really a lot of trouble for very little gain.
The key observation is that you get the most added magnification by putting the objective as close as possible to the tube lens (thus making d small).
Is this helping?
Rik
1. The objective has a certain focal length, let's call it FL_obj.
2. The tube lens also has a certain focal length, let's call that FL_tube.
3. The objective is designed to be used with the tube lens focused at infinity. In this case, the calculation is simple:
In this case only, the separation between objective and tube lens will not change magnification.magnification = FL_tube / FL_obj
4. If you add extension behind the tube lens, so that the tube lens no longer focuses at infinity, then you increase the magnification of the combo.
But in this case the magnification does depend on separation between objective and tube lens, and it's not easy to predict because the calculation depends on a number you don't know: the separation between rear principal plane of the objective and front principal plane of the tube lens.
I'm guessing that you're trying to calculate the magnification with added extension by either
(1) using the basic lens equation that 1/f = 1/f1 + 1/f2, or
(2) assuming that magnification scales as (FL_tube+added_extension)/FL_tube.
But in fact, when there is separation "d" between the principal planes, the governing equation is what's described at Wikipedia, HERE:
1/f = 1/f1 + 1/f2  d/(f1*f2)
An alternate form, paraphrased from page 28 of "Practical Optical System Layout..." by Warren J. Smith, is:
f = (f1*f2) / (f1 + f2  d)
If d is zero (no separation between the principal planes), then the basic lens equation does apply (because 1/f1 + 1/f2  d/(f1*f2) collapses to just 1/f1 + 1/f2). In this unlikely case, calculation (1) above gives a good prediction.
If d is exactly equal to the focal length of the shorter lens, then the effective focal length of the combo becomes exactly equal to that of the shorter lens. In this unlikely case, calculation (2) above gives a good prediction.
But if d is anything else, which it probably is, then the change of magnification with added extension is not accurately predicted by either (1) or (2).
Yes, it is possible to determine d and then use that value for further calculations. But this is really a lot of trouble for very little gain.
The key observation is that you get the most added magnification by putting the objective as close as possible to the tube lens (thus making d small).
Is this helping?
Rik
How so ? You should be getting 1.115mm.With a 20X objective attached, the FOV @20X on sensor is, by my calculations, 1.1mm.
Unless the difference is just because you meant to say "about 1.1mm" then there's something basic going wrong.
Dunno where your 0.95x comes from
If [your (200mm)tube lens is focused at infinity] then
 a 20x objective will give you 20x magnification.
Separation between tube lens and objective doesn't change that
So,
*Based on an approximate FOV on an APSC (Canon 50D, 22.3X14.9mm).
Your field of view will be (22.3/20) x (14.9/20)
=1.115 x 0.745mm
That works because the objective is designed for a tube lens focal length of 200mm, there's nothing intrinsically "20x" about the objective.
If you use a tube lens of say 160mm, your magnification will be
160/200 of what's marked on the objective, here giving 16x.
They could have just marked the (20x) objective with its focal length. (For a 20x it's 10mm)
Then the magnification would be whatever you'd get from
TubeLens_FL/Objective_FL.
 Charles Krebs
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A simple way to test that the tube lens is focused 'at infinity' it to point it and the bellows/tubes/helicoids and camera out the window and check you can get an infocus image of some distant objects (where distant is not especially distant; think of the marked focus scales on a 200mm lens. "More than 20 metres away" is likely good enough).Charles Krebs wrote:Remember that the Nikon tube lens is slightly "telephoto" so its actual location when focused to "infinity", will be less than 200mm from the sensor.When a dedicated Nikon tube lens or '2nd objective lens' Part # MXA20696 is positioned @ 200mm from sensor
Note that for some lenses, the edges may well look terrible at infinity. Ensure you have a distantish subject in the centre.
I did this as a check for an apogerogon 150/9, but didn't note the distance from the sensor. I was just checking the lens 'worked' and had no obvious defects.
 Joaquim F.
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In several Nikon diagrams and doing the "mountain test", the distance between Nikon tube lens and sensor plane is about 150mm, which is good for reducing the length of the assembly!ChrisLilley wrote:A simple way to test that the tube lens is focused 'at infinity' it to point it and the bellows/tubes/helicoids and camera out the window and check you can get an infocus image of some distant objects (where distant is not especially distant; think of the marked focus scales on a 200mm lens. "More than 20 metres away" is likely good enough).Charles Krebs wrote:Remember that the Nikon tube lens is slightly "telephoto" so its actual location when focused to "infinity", will be less than 200mm from the sensor.When a dedicated Nikon tube lens or '2nd objective lens' Part # MXA20696 is positioned @ 200mm from sensor
Note that for some lenses, the edges may well look terrible at infinity. Ensure you have a distantish subject in the centre.
I did this as a check for an apogerogon 150/9, but didn't note the distance from the sensor. I was just checking the lens 'worked' and had no obvious defects.
The Nikon lens tube image quality alone focused to infinity is not good, with many chromatic aberrations, but Pau told me that also happens with other lenses designed for microscopy. When mounted with the infinity objective with some distance the tube lens aberrations noticeably decrease (I read some post from Charles Krebs commenting that), the objective always seems dominate the combo looking the experience of many forum participants with many "tube lens"!
regards
We saw this with the "morfanon".The Nikon lens tube image quality alone focused to infinity is not good, with many chromatic aberrations
It improves significantly if you put a black paper aperture over the front end, with a hole about the size of the back end of an objective.
If you point the lens "out of the window" a lot more front element is exposed.
it wasn't for x power just scale for sizing
enricosavazzi wrote:Perhaps I miss the point, but shouldn't it be something like this?SONYNUT wrote:I made this to give me an idea of what i'm seeing....i print it out the size of the sensor used...it's not exact but it does the trick
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