## What is the total focal length...

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Beatsy
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### What is the total focal length...

...of a 10x Mitty on a 135mm prime (tube) lens? Is there a handy-dandy formula, or doesn't the term apply?

I ask because I found I'm getting appreciable benefit from the in-body image stabilisation when doing 4k video stacking with my Sony A7R2. Focal length for "dumb" lenses must be entered manually for best results, and does make a noticeable difference if incorrect. Trial and error is not working too well for finding the optimum setting though - hence my question.

Thanks.

Lou Jost
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That's a good question, I'd also like to know that. I'd like to add an additional question: How to calculate the EA of a pair of coupled lenses (including microscope objective + tube lens as a special case). Thanks! Some comments on the internet claim that by coupling two fast lenses of equal focal length (one reversed on the other), the EA is actually lower than either one separately. I'd like to understand how that could be true, if in fact it is true. It seems counterintuitive.

Beatsy
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Bump.

I've done (yet) more googling and still don't really have an answer. I found a formula for "simple" lenses close together (F=focal length), 1/Ftotal = 1/F1 + 1/F2. Using that for a 10x mitty (focal length 20mm) on a 200mm prime, you get 1/0.55 or about 18.2mm. A 135mm tube lens is about 17.4mm. I suspect the real answer must be far more complicated, but do these numbers seem roughly feasible at all?

ChrisR
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There's stuff on it in this forum, but it'll need a hunt.
Or wait for Rik to come back!
It's not a very useful number perhaps - I mean what's the focal length of a microscope.. Depends how you separate things.
For a combo you can measure the FL by using Delta magnification/Delta extension.
the EA is actually lower than either one separately.
Shorter FL by combining lenses, same diameter => larger EA
Chris R

Lou Jost
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Chris, that makes sense about the EA, thanks!

Lou Jost
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I just tried reversing a 50 1.4 on a 75 1.8. EA by quick rough measurement was around 2 at m= 75/50=1.5. That's an incredible EA, the equivalent of a lens with max aperture of f/0.8 calculated from EA = f(m+1) ... I.wonder how much I will have to stop it down to eliminate aberrations.

rjlittlefield
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When the rear lens is focused at infinity, the system focal length and thus the effective aperture is not affected by separation between front and rear lenses. This is why the manufacturers of tube lenses can quote large ranges for separation between objective and tube lens. It also means that you can use the 1/(1/f1+1/f2) formula, just like thin lens with no separation.

Again assuming the rear lens is focused at infinity, changing the FL of the rear lens does not affect the subject-side NA of a microscope objective. And the camera-side NA always tracks the subject side NA by a factor of 1/magnification, and the effective f-number (camera side) is always 1/(2*NA_camera).
What is the total focal length...
...of a 10x Mitty on a 135mm prime (tube) lens? Is there a handy-dandy formula, or doesn't the term apply?
So, for that 10X Mitty (20mm) on a 135mm tube lens, that means you're looking at a total focal length of 1/(1/20+1/135) = 17.4 mm, with a subject-side NA of NA 0.28 (unchanged), a total magnification of 10*(135/200) = 6.75, giving a camera-side NA of 0.28/6.75 = 0.041481, and a camera-side effective f-number of 1/(2*0.041481) = f/12.05.

However...
I ask because I found I'm getting appreciable benefit from the in-body image stabilisation when doing 4k video stacking with my Sony A7R2. Focal length for "dumb" lenses must be entered manually for best results, and does make a noticeable difference if incorrect. Trial and error is not working too well for finding the optimum setting though - hence my question.
I doubt that the effective focal length computed above will be anything close to the number needed for in-body image stabilization.

The reason is that (I presume!) the in-body stabilization works by measuring angular rotation of the body and using that plus focal length to estimate how far the sensor has to be shifted. Unfortunately a calculation that works well near infinity focus becomes wildly wrong at close focus, due to lateral shift of the lens with respect to the subject. Not only does the camera have no idea what's actually going on, in terms of shift versus angle, but that can change significantly depending on how you have your camera supported.

I'm afraid tha,t for your application, there's no substitute for trial and error, even if that doesn't work very well either.

It would be great if the camera had some sort of calibration mode, in which it would observe the image on sensor and compare that against the movement data to just figure out what the relationship is for any particular setup. Maybe some future version...

--Rik

Beatsy
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Thanks for taking the time to answer Rik. At least I was on the right track with the formula.

As it happens, I now don't believe IBIS is contributing to image sharpness as much as I thought initially. After several more comparative tests I see no consistent difference in motion blur between shots taken with IBIS on or off. It's impossible to introduce "consistent shake" to reliably tell the difference with my manual setup. I got the initial impression from just two samples (one with IBIS, which was good, and one without which showed motion blur). I've since taken more samples and got a couple where it's the other way around. So it seems the most important thing is just how carefully I wind the rail out!

Lou Jost
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Rik, can I infer from your formula that the effective aperture of a pair of coupled lenses is (nominal aperture of front lens)*m?
Edit: Ah, but since the front lens is reversed, the nominal aperture of that lens would have to be corrected by some pupillary magnification factor?

rjlittlefield
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Lou Jost wrote:Rik, can I infer from your formula that the effective aperture of a pair of coupled lenses is (nominal aperture of front lens)*m?
Yes, that's correct, assuming of course that the limiting aperture is only the one in the front lens, and assuming that the rear lens is focused at infinity.
Edit: Ah, but since the front lens is reversed, the nominal aperture of that lens would have to be corrected by some pupillary magnification factor?
No, a correction is not needed when the rear lens is focused at infinity and the front lens is reversed.

A good way to think about this is that correcting for pupil factor is something you have to do when you drag a lens away whatever focus arrangement its f-number or NA was specified for. When the rear lens is at infinity focus, and the front lens is reversed, then effectively the front lens is still focused at infinity also, and that's the point for which its f-number was specified.

Additional insight may come from considering the details of some example. Suppose that you have a 25mm f/2.8 lens with pupil factor 2, reversed in front of a 100 mm rear lens focused at infinity. In its normal orientation, the 25mm f/2.8 lens will accept parallel rays in an entrance beam that's about 8.929 mm wide (=25/2.8), producing a cone of light whose angle corresponds to NA 0.1786 (=1/(2*2.8)). Reversed, the 25mm f/2.8 lens will accept the same cone of light and produce the same parallel beam, which then stops down the rear lens to an effective f-number of 100/8.929 = 11.2 (also =2.8*(100/25)). Notice that there was no need to consider pupil factor in this analysis.

If the rear lens were not focused at infinity, then the reversed front lens would focus at some different point. Focusing at that different point, it would accept some different angle of cone, and computing that angle would indeed require accounting for the pupil factor. In addition, the magnification would then depend on separation between the principal planes of the lenses, so doing the calculation accurately would be much more difficult. Keeping the rear lens focused at infinity makes things so much simpler!

--Rik

Lou Jost
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