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Beatsy
Joined: 05 Jul 2013 Posts: 858 Location: Malvern, UK

Posted: Tue Aug 16, 2016 4:40 am Post subject: What is the total focal length... 


...of a 10x Mitty on a 135mm prime (tube) lens? Is there a handydandy formula, or doesn't the term apply?
I ask because I found I'm getting appreciable benefit from the inbody 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. 

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Lou Jost
Joined: 04 Sep 2015 Posts: 1666 Location: Ecuador

Posted: Tue Aug 16, 2016 5:50 am Post subject: 


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. _________________ Lou Jost
www.ecomingafoundation.wordpress.com
www.loujost.com 

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Beatsy
Joined: 05 Jul 2013 Posts: 858 Location: Malvern, UK

Posted: Sat Aug 20, 2016 5:03 am Post subject: 


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? 

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ChrisR Site Admin
Joined: 14 Mar 2009 Posts: 7131 Location: Near London, UK

Posted: Sat Aug 20, 2016 6:01 am Post subject: 


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.
Quote:  the EA is actually lower than either one separately.  Shorter FL by combining lenses, same diameter => larger EA _________________ Chris R 

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Lou Jost
Joined: 04 Sep 2015 Posts: 1666 Location: Ecuador


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Lou Jost
Joined: 04 Sep 2015 Posts: 1666 Location: Ecuador

Posted: Sat Aug 20, 2016 10:25 am Post subject: 


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. _________________ Lou Jost
www.ecomingafoundation.wordpress.com
www.loujost.com 

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rjlittlefield Site Admin
Joined: 01 Aug 2006 Posts: 18087 Location: Richland, Washington State, USA

Posted: Sat Aug 20, 2016 8:52 pm Post subject: 


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 subjectside NA of a microscope objective. And the cameraside NA always tracks the subject side NA by a factor of 1/magnification, and the effective fnumber (camera side) is always 1/(2*NA_camera).
Quote:  What is the total focal length...
...of a 10x Mitty on a 135mm prime (tube) lens? Is there a handydandy 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 subjectside NA of NA 0.28 (unchanged), a total magnification of 10*(135/200) = 6.75, giving a cameraside NA of 0.28/6.75 = 0.041481, and a cameraside effective fnumber of 1/(2*0.041481) = f/12.05.
However...
Quote:  I ask because I found I'm getting appreciable benefit from the inbody 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 inbody image stabilization.
The reason is that (I presume!) the inbody 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 

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Beatsy
Joined: 05 Jul 2013 Posts: 858 Location: Malvern, UK

Posted: Sun Aug 21, 2016 6:03 am Post subject: 


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! 

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Lou Jost
Joined: 04 Sep 2015 Posts: 1666 Location: Ecuador

Posted: Mon Aug 22, 2016 6:06 am Post subject: 


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? _________________ Lou Jost
www.ecomingafoundation.wordpress.com
www.loujost.com 

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rjlittlefield Site Admin
Joined: 01 Aug 2006 Posts: 18087 Location: Richland, Washington State, USA

Posted: Mon Aug 22, 2016 8:40 pm Post subject: 


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.
Quote:  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 fnumber 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 fnumber 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 fnumber 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 

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Lou Jost
Joined: 04 Sep 2015 Posts: 1666 Location: Ecuador

Posted: Mon Aug 22, 2016 9:23 pm Post subject: 


Thanks so much, Rik, for explaining this to me, and for confirming that the formula for EA of coupled lenses (f*m) is different from the formula for EA of single lenses with extension (f* (m+1)). Then for low m, there are big potential benefits to using coupled lenses rather than extension. _________________ Lou Jost
www.ecomingafoundation.wordpress.com
www.loujost.com 

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