View Full Version : DOF Calculations for Telephotos

neil poulsen

15-Aug-2004, 16:11

If I divide the focal length by the f-stop for a telephoto lens, do I obtain the actual aperture diameter? This works for standard lens designs like a plasmat, but does it work for telephotos?

I've been making dof tables, and I'm not sure whether the same forumlas work for telephotos.

Dan Fromm

15-Aug-2004, 16:26

Why shouldn't they work?

neil poulsen

15-Aug-2004, 17:43

I only know enough to be dangerous. But, telephoto lenses are closer to the image plane then regular lenses, so for the same f-stop, their aperture diameters must be smaller. But, I'm not sure how f-stops are calculated for telephotos, hence my question.

Dan Fromm

15-Aug-2004, 18:38

I think you're confusing flange to film distance with rear nodal point to film distance.

Cheers,

Dan

sanking

15-Aug-2004, 22:45

As far as DOF is concerned lenses are lenses.

There are some pretty neat programs on the net. For example, see http://dfleming.ameranet.com/dofjs.html

Some of the programs, including this one, allow you to set the size of your circle of confusion, a feature that I particulary like.

You can also download the program Palm Pilot OS or Windows.

Graeme Hird

15-Aug-2004, 23:21

"....do I obtain the actual aperture diameter?"

My understanding is that you will get the effective aperture rather than the physical size of the opening.

Anyone is welcome to dispute this.

Cheers, Graeme

Bob Salomon

16-Aug-2004, 06:13

If you do not know the actual focal length of the lens then your answer would be close but not exact.

Chris Gittins

16-Aug-2004, 11:06

Graeme is correct. Dividing focal length by f-number gives you an effective aperture diameter which may or may not correspond to physical size of the opening you're using to stop the lens.

Chris

Struan Gray

16-Aug-2004, 14:50

The f-number is defined as the focal length divided by the diameter of the entrance pupil, not the aperture as such. The entrance pupil is the aperture as seen through the glass of the front lens cell, so if that cell has a high power the size of the entrance pupil can be fairly different from that of the hole formed by the aperture blades.

Depth of field is determined by the position and size of the *exit* pupil. This is the aperture as seen through the *back* of the lens. The exit pupil's apparent size and position fixes a base and angle for the cone of light converging onto the focal plane from the lens, which in turn determines the DOF.

If you look at a telephoto lens from both ends you'll see that the aperture looks smaller when seen from the back of the lens than from the front: i.e. the entrance and exit pupils have different sizes. The asymmetry reflects the asymmetry of the telephoto design, which in its simplest form has a converging element in front of the aperture and a diverging one behind - these necessarily give different appearances to the aperture sandwiched between them.

Crucially, the same asymmetry in the optical design that makes the aperture look different is also responsible for moving the rear nodal point forward, allowing you to use less bellows extension. For things like bellows factor the asymmetry does appear in correct formulea as a 'pupillary magnification ratio', but for DOF calculations two effects cancel out and you don't have to worry. The aperture is indeed closer to the film, but it looks smaller, so you end up with the same cone of converging light, and DOF is the same as for a regular, non-telephoto lens.

Jeff Conrad

18-Aug-2004, 00:12

Because the DoF is determined by the exit pupil, and the lens <var>f</var>-number is

related to the entrance pupil, don’t you need to consider the pupillary

magnification in any DoF equation that involves the lens <var>f</var>-number?

It’s common in elementary DoF equation derivations to eliminate the

diameter of the stop by the substitution

<var>d</var> = <var>f</var> /<var>N</var>

where <var>N</var> is the <var>f</var>-number, <var>f</var> is the focal

length, and <var>d</var> the diameter of the stop (assuming, in essence

that the lens is symmetrical). However, for an asymmetrical lens for which

<var>N</var> = <var>f</var> /<var>D</var>

and

<var>p</var> = <var>d</var>/<var>D</var>

where <var>p</var> is the pupillary magnification, <var>d</var> is the

diameter of the exit pupil, and <var>D</var> is the diameter of the

entrance pupil, the corresponding substitution

<var>d</var> = <var>p</var><var>f</var> /<var>N</var>

necessarily includes the pupillary magnification <var>p</var> in the resulting

equation.

Struan Gray

18-Aug-2004, 14:11

In the usual derivation of DOF formulea you imagine a cone of light extending behind the lens and narrowing down to some focus before spreading out again to form a second, inverted cone. The DOF limits are defined by how far from the focus point you can move before the cone gets too thick, where 'too thick' means larger than whatever circle of confusion you have chosen.

Jeff is right that the pupillary magnification changes the diameter of the base of the cone. But it also changes the height of the cone, and by the same factor. This means that the cone *angle* remains the same and the crossing rays either side of the focus point don't change their geometry. This is what I tried to say in my final paragraph above.

Where you don't get cancellation is in exposure calculations. So the pupillary magnification does crop up when calculating bellows extension factors or when using a telephoto lens reversed for macro applications.

Struan Gray

19-Aug-2004, 15:57

Jeff has tactfully pointed out in email that I am contradicting Jacobsen's lens tutorial (www.photo.net/learn/optics/lensTutorial) when I assert that telephotos have the same DOF as symmetrical lenses. Jacobsen does agree with me that the exit pupil lies behind the rear principle plane / nodal point, but not by as much as I would like.

In the interests of full disclosure, I should also point out that Emmanuel Bigler, who also usually gets these things right, did also mention this in passing in a thread I took part in on photo.net:

www.photo.net/bboard/q-and-a-fetch-msg?msg_id=003xqY

I have found out the hard way that contradicting Jacobsen and Bigler simultaneously is a waste of time so for now I'll eat crow and admit I was wrong. Telephotos do have a different depth of field to symmetrical lenses. Mea Culpa.

In a pathetic attempt to redeem myself, I'll try to figure out how much one actually needs to worry about this in LF. I have a 36" telephoto myself, so the question isn't entirely academic.

Emmanuel BIGLER

24-Aug-2004, 05:29

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Hi Guys

I'm back to this forum after a summer break. By a coincidence, I have re-computed on the back of an envelope DOF formulae for asymmetric lens designs. The formulae I find have to be double-checked before I publish them, I would have preferred to find them in a good ol' textbook of the 1930's but I did not find anything yet. Contact me off-list if you are interested to verify and re-compute them.

My first impressions after playing briefly with the formulae are

<UL>

<LI>the classical hyperfocal distance H is unchanged in asymmetric lens designs. More precisely, if you compute as usual H = f*f/(N*c) where f is the focal length and N the effective f-stop defined from the entrance pupil according to the rule for asymmetric designs, you find that setting a distance between the front focal point and the subject equal to H brings you a DOF range from infinity (this for sure) to something like H/2 (I have to check in the formaule) as usual.

</UL>

To me this is the most important result since it does not affect most photographic situations where telephotos are used i.e. an object located at a large distance. Not that I want to argue against friends who insist on using telephotos at 1:1 ratio, LF optics to me means total freedom, but...

<UL>

<LI>telephotos used in LF photography exhibit a pupillar magnification ratio which does not depart enough from unity so that classical DOF formulae are plain wrong. This is not a quantitative result, just an impression. Conventional DOF limits should be considered somewhat fuzzy, so the slight changes induced by real formulae for a non-unit pupillar magnification ratio do not bring anything really interesting for LF telephotos. The situation is probably different for retrofocus lens designs, and I would not be surprised if very asymmetrfic wide-angle lenses used for digital sensors do not obey classical DOF formulae. But in LF we do not care except for the 35mm to 50mm LF wide-angle lenses which are slighty retrofocus ;-);-)

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So the conclusion is simple, Neil, Struan and LF friends, you can sleep quietly and continue to use conventional DOF formulae as an approximation, and you can continue to use the Hyperfocal formulae without approximation. For those you demand the utmost precision, the distance between the front principal plane and the object for which DOF extends to infinity is H+f. The position of the front principal plane is documented in manufacturers datasheet. The front focal point is located one focal length ahead, from there you add one Hyperfocal distance H to determine where to place your object for infinite DOF.

Incidentally, going off-topic, the front principal (or nodal) point is not the place where you should rotate the camera for proper panoramic stitching without parallax effects. The proper rotating point is the centre of the entrance pupil. This error has been eventually corrected in Hasselblad's datasheet.

A final remark if you want to apply tilts to your lens, whether the lens is quasi-symmetrical or not, this does not change anything to the fact that the depth of focus planes in image space are parallel to the actual plane of ideal sharpness. Often, tilts with a telephoto are considered as something frightening, impossible to control, mysterious, etc... No way ! Nothing extraordinary ! Hence the slanted DOF limits in object space are given by the same 3 ray tracing like in Leonard Evens paper.

Leonard Evens, "Some Thoughts on View Camera Calculations",

http://www.math.northwestern.edu/~len/photos/pages/dof_essay.pdf

I have updated the French version of my DOF paper here where you can find the explanation, see figures 7 and 8, valid even for a strongly asymmetric lens provided that tilt angles are reasonable and not extreme

http://www.galerie-photo.com/profondeur-de-champ-et-scheimpflug.html

I still need to translate it into English to upgrade the version which is downloadable from here.

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Jeff Conrad

24-Aug-2004, 14:38

I went through a similar exercise last week, and arrived at similar conclusions. The effect of asymmetry isn't all that great even at unity magnification: at 1:1, a lens with pupillary magnification of 0.7 has about 21% greater DoF than a lens with p = 1. At m = 0.1, the difference decreases to about 4%.

I've discussed my results with Struan Gray, and we seem to be in agreement. I've also sent a copy to Emmanuel for his comments.

Struan Gray

24-Aug-2004, 15:05

I have always said that the best way to get good information online is not to ask intelligent, well-formulated questions; it is to be publically, emphatically wrong. This thread only proves the point.

Thanks Emmanuel, for hammering the final nails into my coffin. Thanks Jeff for holding my hand through something I should have seen for myself.

21% may be small (you don't need to stop down much to get the same increase), but I always prefer to know that I don't need to worry, rather than hope, or merely believe. I suspect pupillary magnification is only going to be an issue for people using, say, a reversed 100 mm macro from a 35 mm camera at high magnifications, but I'll bet that someone, somewhere will be do 1:1 with an aero lens and three meters of bellows, if only to make a point.

Finally, I think technical articles would be more widely admired if they were all written in French. "Profondeur de champ" sounds so elegant. "Depth of field", or "Skärpedjup" are downright plodding in comparison.

Emmanuel BIGLER

26-Aug-2004, 05:13

Jeff sent me his calculations yesterday and I'm happy to tell that, independently, we find exactly the same formulae. Those formulae, hopefully, perfectly fit with David Jacobson's, which suggest me that we should have asked him whether he had found them in good reference book or re-computed himself. Nevertheless,

now we are all set and anybody interested will soon be able to read Jeff's calculations, the best being here as a reference .pdf document if Tuan agrees.

Practical conclusions are that view camera users can continue to use classical DOF forumale, but those who insist on bringing a personal electronic assistant or other electronic gadget in the field have no reason to use anything but the True General Asymmetric Lens DOF Formulae ;-);-);-) Ahem. With no tilt of course ;-)

Emmanuel BIGLER

27-Aug-2004, 10:13

I cross-checked with the formulae given in the most excellent DOF paper by Paul van Walree here :

http://www.vanwalree.com/optics/dofderivation.html

Both Jeff's formulae as well as mine do correspond exactly to PvW's calculation. Together with Jacobson's this makes at least 4 totally independant calculations based on the same model.

Now are the hypothesis of the model actually right, or are we all wrong ? ;-);-);-)

Simon Clement

28-Aug-2004, 04:05

DOF calculation, in french only, here http://www.galerie-photo.com/profondeur_de_champ_avec_calcphot.html

The focusing distance is measured from film or from lens to the subject.

Emmanuel BIGLER

30-Aug-2004, 10:22

Simon's software is both free and one of the few DOF software based on general DOF formulae for assymetric lenses according to Jacobson's.

It comes with a database of Zeiss MF lenses.

Last but not least, the software runs under Linux, thanks to the Wine emulator !

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