LF Macro Lens (...and understanding bellows extension for tele lenses)

[EdSawyer]

I do have the smaller Nikkor-T set. As mentioned there's some math and theory that goes into it but the Telephoto design by nature needs less extension. The only trade off is probably smaller coverage for the Tele vs. symmetrical lens of similar focal length. (e.g. a 360 plasmat may cover larger film than a 360 tele). That, and movements like tilt or swing become a bit weird due to the location of nodal point. But really no trade-off in IQ, as far as normal shooting goes, esp. for portraits. If anything the Nikkor-T teles should have better IQ than standard plasmats of the same length, since they use more exotic glass.

[Steve Goldstein]

Bellows factor X times = (1 + M / M_p )²

Just to clarify, this is really (1 + (M/M_p))², right?

[Dan O'Farrell]

Hi!

Regarding the shorter bellows draw of a telephoto lens, the gain is interesting only for far-distant objects. When you come close to the 1:1 ratio, the additional bellows draw beyond the focal point to get a sharp image is the same, about one focal length, for all lens designs.

For example, consider a lens of focal length 360 mm.
If the lens is like an apo ronar, quasi-symmetrical, the distance between the shutter, in the middle of the lens and the focal point is about 360 mm.
In order to reach the 1:1 configuration, you have to add an other 360 mm, total bellows draw = 720 mm.
Now take a 360 Schneider Tele Arton. The flange focal distance is only 210 mm. But if you want to reach the 1:1 magnification ratio, you need to add the same additional 360 mm bellows draw, total draw = 210 + 360 = 570 mm.
Hence you have a gain of 360-210 = 150 mm which is somewhat helpful when the image is close to the focal point i.e. for far-distant subjects, but you cannot avoid the additional bellows draw required by close focusing (see below how to easily compute this).


Won't the bellows compensation be the same with tele and non-tele lenses?


In principle, bellows correction factors differ between a quasi-symmetrical lens design and a telephoto design.
It is quite simple to compute, the only additional parameter that you need to know is the pupillar magnification of your lens M_p = (diameter of exit pupil)/(diameter of entrance pupil).
Some manufacturers like Schneider Kreuznach, do provide the value of this parameter in their detailed technical data-sheet, but unfortunately, old archives on the German Schneider-Kreuznach web site are no longer directly available, however, they are stored in the Web archive "wayback machine".

The general formulae for a telephoto of pupillar magnification factor M_p, of a given (image)/(object) magnification M, is very simple.
First compute the magnification factor "M" vs. the additional bellows extension "ext"
M = ext / f
ext = additional bellows extension beyond the focal point; f = focal length.
This formula M = ext / f for the additional bellows draw "ext" beyond the focal point in order to reach a given magnification M, is universal and valid for all lenses even very asymmetric.

Bellows factor X times = (1 + M / M_p )²

The origin of this somewhat cryptic formula is explained in detail here in this article in French,
http://www.galerie-photo.com/telecha...t/pupilles.pdf
The maths are here (again, in French, sorry)
http://www.galerie-photo.com/annexe-pupilles.pdf
You can just have a look at the graph attached here in pdf
This graph is a good summary of the differences between a retrofocus (M_p > 1) a quasi-symmetrical lens (M_p ~= 1, all standard LF lenses) and a telephoto (M_p < 1). For the Schneider 360 mm Tele-Arton, M_p ~= 0.57. This yields a difference of approx one f-stop at 1/1 ratio.
But in principle an assymetric lens should never be used at 1:1 ratio!!
For use at 1:1 ratio, symmetrical lens formulae are preferred, at least for view camera lenses of fixed focal length; modern macro lenses with floating elements are another issue, those lenses are a kind of a zoom lens with focal lenght and a pupillar magnification factor changing throughout the focusing range!

Hence in most usual cases of ordinary, non close-up shots like e.g. M = 1:5 = 0.2 or smaller (i.e. object is located further away than 6 times the focal length) you can safely ignore the additional correction with respect to a quasi-symmetrical lens.
For M_p ~= 0.57 like in the 360 Tele-Arton, at M = 1:5 = 0.2, the Tele-Arton would in principle require only 1/2 f-stop of additional exposure with respect to a quasi-symmetrical lens formula.
The basic formula for M_p = 1, quasi-symmetrical lens like many standarc LF lenses plus apo-repro lenses is simply :
M = ext / f
Quasi-Symmetrical Lenses Bellows factor X times = (1 + M )²

Wow! Interesting !

Now, let us say that we have a 360mm Tele-Xenar designed for a 4X5 frame ( at infinity ).
If we expand the draw to fill a(n) 8X10 frame, what would the additional ( or, preferrably, total) draw...lensboard to film.. be?

We're not expecting a full and finished answer to this, but how would we calculate ( or measure) it?

[Dan Fromm]

360 mm.

The diameter of the image circle at 1:1 is twice the diameter of the image circle at infinity. Extension from the infinity position to 1:1 is 1 focal length.

[Pere Casals]

Hi!

Regarding the shorter bellows draw of a telephoto lens, the gain is interesting only for far-distant objects. When you come close to the 1:1 ratio, the additional bellows draw beyond the focal point to get a sharp image is the same, about one focal length, for all lens designs.

Well, IMHO from what Dan says we can conclude: if at infinite the tele shortens X the bellows draw (compared to a regular lens), then at 1:1 the telephoto will also save the same X bellows draw compared (again) to the regular lens.

[Corran]

The problem I was trying to solve for is that I primarily shoot portraits indoor using natural light. In order to get head and shoulders on 5x7 or 8x10 my bellows draw is so long (on a longer lens) that the exposure is also long and the subjects move resulting in blurry photos.

Add more light or use a shorter lens. The "traditional" longer lens for head and shoulders goes out the window once you start using larger formats due to magnification. Try a normal focal length and see what you get. If you still want the additional compression of a longer lens, well, see my first point. There's no free lunch unfortunately.

[Dan O'Farrell]

360 mm.

The diameter of the image circle at 1:1 is twice the diameter of the image circle at infinity. Extension from the infinity position to 1:1 is 1 focal length.

Ah... much simpler than I had expected.

Thank you, Dan.

[Emmanuel BIGLER]

Hello all!

From Steve Goldstein
Just to clarify, this is really (1 + (M/Mp))2, right?

Yes, exactly, with the additional () inside to avoid any ambiguity.
An equivalent formula would be ((M+M_p)²) / ((M_p)²)
but definitely (1 + (M/Mp))² is much simpler


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from Dan O'Farrell
If we expand the draw to fill a(n) 8X10 frame, what would the additional ( or, preferrably, total) draw...lensboard to film.. be?

I agree with Dan's answer. However the rule of thumb regarding doubling the image circle at 1:1 ratio is just an estimate, not something very precise.
But even for a telephoto design, this should also work, but this property is not easy to explain without doing some ray tracing.


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From Pere Casals
if at infinite the tele shortens X the bellows draw (compared to a regular lens), then at 1:1 the telephoto will also save the same X bellows draw compared (again) to the regular lens.

Yes, exactly, so you cannot avoid the additional extension of exactly one focal length beyond the focal point in order to reach the 1:1 ratio with all lenses.
With a telephoto, you keep the gain in distance between the lens and a sharp image throughout the whole focusing range.
For example with the 360 tele Arton, the gain in flange focal distance with respect to a quasi symetrical 360 lens (like most standard lenses, plus all apo-repro lenses) is 150 mm at any magnification ratio.

In fact the only possibility to have less distance between the lens and the image at 1;1 ratio is ... to use a telephoto with a bigger asymmetry, or a lens of shorter focal lenght!


Actually if you have an unknown unmarked positive lens, i.e. allowing you to get a sharp image on a ground glass for a distant object, by measuring the additional extension beyond focal point needed to reach the 1:1 ratio, you get a good estimate of the focal length, even if the lens is a telephoto or whichever thick and complex compound lens design.
In other words, with this old method due to MM. Davanne & Martin at the end of the XIXst century, you can measure the focal length without any knowledge of where the image principal plane H' is located, no need to guess where an hypothetical "optical center" would be located

[Pere Casals]

In other words, with this old method due to MM. Davanne & Martin at the end of the XIXst century, you can measure the focal length without any knowledge of where the image principal plane H' is located, no need to guess where an hypothetical "optical center" would be located

hmmm, this is interesting...

[Emmanuel BIGLER]

from Pere Casals
hmmm, this is interesting...

I realize that Davanne & Martin's method to measure the focal length of a photographic lens has been explained here by Leigh
http://www.largeformatphotography.in...=1#post1144455

The attached diagram below explains the principles. I have chosen an hypothetical lens with principal planes H and H' "crossed" and located far from each other.


The method appears in « Traité encyclopédique de photographie, » by Charles Fabre, Paris, Gauthier-Villars et fils, 1889-1906, 1st supplement A, page 19
http://cnum.cnam.fr/CGI/fpage.cgi?8K...100/402/11/381
(thanks to Dan Fromm for finding this golden treasury of XIXst century photographic literature)

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Regarding what happens in real life with a real asymmetric tele lens, the other diagram (as an external link next line) shows the 360 Schneider Tele-Arton (effective focal length = 353 mm) with its principal planes H & H' and its pupils, entrance and exit.
http://bigler.blog.free.fr/public/do...ton-360-EN.pdf

The pupils are not located at the principal planes, H and H' are not crossed but H & H' are floating in air in front of the first lens vertex.

Regarding the increase in image circle at magnification 1:1 or so, looking for the distance requested to fill the 8x10" format, after thinking about it twice, I think that we have to count distances from the exit pupil of the lens and not the principal plane H'. It is when the distance between the exit pupil and the image point is doubled, that the illuminated image circle is doubled.

If we plot extreme rays crossing the exit pupil and corresponding to the absolute limit of the illuminated circle, we define a cone which is fixed with respect to the lens. Hence the maximum illuminated circle for any image position is defined by the intersection of the image plane and the extreme cone of rays propagating from the center of the exit pupil and blocked by some parts of the lens mount.
Again in all questions related to image illumination, or limits of illuminated image field, principal planes play no role, only the position of the main iris (stop) and other obstructing diaphragms, and the position of their images in image space actually count.

However F & F' the focal points; H and H' the principal (or nodal points N & N', they are identical to F & F') of course, tell us where to look for a sharp image for a given object distance, and what the magnification will be, regardless of pupils.