the exact path of light leaving a telephoto lens

[Jerry Fusselman]

I hope someone is brilliant and friendly enough to help me.

My immediate interest is the path of light from the rear element of a Nikon 600T f/9 lens when focused at infinity. I need to know the diameter of the disk of light leaving the rear element at 2 inches behind the glass. Also, if this diameter is larger than I hope, then I need to know what affect it has on my 8x10 image---i.e, do I get visible vignetting or just a little insignificant loss of light?

This lens has a telephoto design, with an extended medium-sized rear element. It's P factor (i.e., its pupillary magnification factor, which the ratio of the exit pupil diameter as seen from the rear to the entrance pupil diameter as seen from the front) is approximately 0.72. That means that the apparent size of the rear aperture is 0.72*600/9 = 48mm.

Two possible answers occurred to me: First, maybe I should imagine a 48mm circle inside my Copal 3 shutter and expand it linearly to my image circle on the film plane. Second, maybe I should move that 48mm circle in front of the lens, for this is a telephoto lens, and measure 600mm to the film plane, again expanding it to the size of the coverage on the film plane. The problem is, I am not sure, in fact I even doubt, that there is room for this second shape to fit in the exit glass of the rear element.

Does anybody know?

I have studied mechanical vignetting from compendium shades enough to know that if 10% of the apparent image of the aperture is blocked looking from where the extreme corner of the film will be, then the loss of light in the corner will rarely be noticeable. However, the proper considerations near the rear element might be completely different. Maybe I don't need to worry as much in this region, for maybe a little light is lost but there would no vignetting---I cannot figure it out.

Someone is sure to ask "Why do I want to know this?" Honestly, just curiosity would be enough for me, but in this case, I have a practical reason too: In the thing I am building, I want the metal near the rear element to be cut only once, and I want to leave as much strength in the structure as possible. Furthermore, I would like to learn the general principles, because I have a similar-but-trickier project in mind for my wonderful Telomar 36cm lens.

[Struan Gray]

Do you want the diameter of the ray bundle for a single point source at infinity (like a star or distant street light), or the size the bundle ceated by all the image-forming light from all the points in the lens' field of view?

Both can be calculated, but my instinct would be to measure. If you can mount the lens on a camera, why not point it at a large diffuse object that fills the field of view (up at the sky, or a large white building wall) and just put the ground glass two inches behind the lens. Failing that, you can learn quite a bit by projecting an image onto a wall or a piece of paper.

[Emmanuel BIGLER]

Sure an experimental approach will do the job fast, however "this is not rocket science" as they say on largeformatphgotography.info ;-)

Jerry. This is extrely simple.

As surprising as it can be, when you know the pupillary pagnification factor P, the focal length and the numerical aperture, you know almost everything needed for a simple drawing on paper that will solve the question. Probably the value of HH' could help to make a complete drawing of the gaussian lens elements.

The route to follow is :

1/ place the exit principal (or nodal) plane H' at 600 mm ahead of the focal plane, in your case, infinity->focus this will be the image plane.

2/ locate the exit pupil. The trick is that pupils being conjugated throught the whole lens, there is only a single position where they can be for a given P-factor. If d_p and d_p' are the algebraic distances between the pupils with respect to the corresponding principal plane H or H', we have : -1/d_p + 1/d_p' = 1/f, like for any object/image (gaussian) connection, in general

Let d_p' be the algebraic distance betwen H' and the exit pupil, we have : -P +1 = d_p'/f
A similar formula for the entrance pupil, just in case you are in need of some parallax-free panoramic stitching ;-) : -1 + 1/P = d_p/f.

So in your case, f=600, P = 0.72, d_p'= 600*(-.72+1) = 168mm.
So the exit pupil is located 168 mm behind H', i.e. (600-128) = 432 mm in front of the focal plane.

As you mentioned, its diameter is .72 times the entrance pupil, @f/9 we get 48mm.

For panoramic stitching you can remember if needed (or forget if useless ;-) that the entrance pupil is
located 600*(-1 + 1/.72) = 233mm behind the input principal plane H,
but you need to know the value of HH' from the manufacturer's specs. HH' can be
negative.

[Emmanuel BIGLER]

I forgot to mention one key trick : pupils being conjugated like any image/object, they obey the usual coupling law between longitudinal position and lateral magnification, so we have : P = a'/a = d_p'/d_p.
P is actually positive, because the pupils do not cross the foci in reasonable photographic lenses. So the pupils are located both on the same side of their respective principal plane.

[Struan Gray]

I'd bet my beret that the lens vignettes at f9. So Jerry can make his hole smaller.

[Eric Biggerstaff]

Holy Smoke! I have NO idea what you guys are talking about but it appears impressive! :-)

I am always amazed by the knowledge on this forum.

[Jerry Fusselman]

Do you want the diameter of the ray bundle for a single point source at infinity (like a star or distant street light), or the size the bundle ceated by all the image-forming light from all the points in the lens' field of view? Primarily the latter, though the former might help too if I need to determine the vignetting effect at larger taking apertures.

Emmanuel, thanks so much for your careful description. I made the drawing you described as best I can, and it depicts a 48mm disc parallel to and 432mm in front of the film, but I cannot tell where to put the lens on my picture. Maybe I need to know where the rear principal (nodal) point is relative to the lens. Or maybe I just need to know where H' or the exit pupil is on my lens. Sorry if I am being dense.

[Dan Fromm]

Jerry, about where to put the lens.

If you can find the lens' flange-to-film distance at infinity, well, that will tell you where to put it. Surely the lens' specifications will tell you that.

Your explanation of why you're asking "In the thing I am building, I want the metal near the rear element to be cut only once, and I want to leave as much strength in the structure as possible." baffles me. In conventional cameras, the lens is in shutter, the shutter attaches a lens board that's supported by the front standard, and the back of the rear element will sit behind the front standard. So where's the problem? Why will there be an obstruction between the back of the lens and the film?

Is Jerry Fusselman another way of spelling Bill Glick?

[Jerry Fusselman]

Yes, I was being dense. The flange-to-film distance is supposed to be 409.6mm for the Nikon 600T. Don't know what it is for the Telomar 360mm f/5.5, but I easily measure it.

Don't worry about being baffled, for I did not explain it completely. It is not a conventional-view-camera situation. The back of the Nikon lens will be in front of my camera's front standard, because there is a front extension device which will be mounted on a monopod. This gives me a little more extension (sorely needed for the 1200mm) and also more stability. In the related case of the Telomar, I am making a spacer for a focusing mount for use on a 617 camera.

Sorry, to my knowledge I have never met Bill Glick.

[Dan Fromm]

Jerry, thanks for the explanation. I faced a similar problem when designing a tandem camera. It is made from 2 2x3 Graphics with a coupling between the cameras. The coupling is made from a lens board (goes on the rear camera's front standard), a film pack adapter (goes in the front camera's Graflok back), and a pair of concentric tubes. The larger tube is attached to the FPA, the smaller to the board. In this case, the obstruction that's potentially a problem is not the rear camera's front standard but the tube in the coupling.

I solved it with a model using similar triangles. If (format's diameter)/(exit pupil to film plane distance ) <= (obstruction's diameter)/(exit pupil to obstruction distance), all's well. You can use this reasoning to determine the obstruction's limiting diameter for the lens focused to infinity.

Incidentally, I got something wrong in my calculations. On my tandem camera as built, a 480 mm lens covers 2x3 at infinity even though the obstruction is somewhat smaller than I'd told the machine shop to make it. According to my model, it shouldn't work, but it does.

Cheers, good luck,

Dan

[Dominique Cesari]

Now, the experimental method:

I placed the rear element of a 600 mm Nikkor at 50 mm from the focusing glass (circa 2 inches).
The measured diameter of the cone is 126 mm (circa 5 inches)

This diameter remains more or less constant as closing down the lens, the limit of the illuminated zone is only more accurate.

[Dominique Cesari]

Now, a better understanding of the question :

you mean : inside the lens

Sorry, I will not drill the back element to experiment that