View Full Version : Bellows ext. exposure compensation for telephoto

Charles Forde
9-Aug-2005, 16:57
For telephoto lenses, are the calculations the same as for a standard lens?


9-Aug-2005, 18:40
I want to make sure, too.

Is it the physical bellows extension or the distance between the lens’ rear nodal point and the film that matters? My understanding is the latter.

Steve Feldman
9-Aug-2005, 18:45

I have a 10" (254mm) Tele Wolly. I've never used any compensation when focused at 'bout 10' to infinity. Normal exposure calculations for f stop and shutter apply. The lens actually focuses at infinity at less than 10". My guess is 8" +-. I've never tried to pull it out to twice that. But if you have enough bellows draw - go for it. I'd guess that the compensation would be 'bout an extra stop and a half to two and a half stops. As always - YMMV.

Hmmmmm. An interesting idea. Think I'll try later on.

Bernard Languillier
9-Aug-2005, 22:48
My understanding is that the intensity of the light decreases per the square of the distance.

This means that if you double the distance between the lens and the film plane, then you need to expose 4 times more, meaning an extra 2 stops.

This should be the same thing whether the lens is a tele design, or a conventional design.


Daniel Geiger
10-Aug-2005, 01:15
I got the Nikor-T 360-500-720 series and do a bit of close-ups. Rather than risking any odd calculation errors, I simply taped the Calumet exposure calculator target to the post of the back porch, focussed at various distances so that I would get 1/3, 1/2, 2/3 etc f-stop exposure compensation. Then I measured the rail extension with a short tape measure and made a table that I glued to my clipboard where I keep my shoot log. [I actually made that table for all lenses I have].

The table approach works well in the field. It's quick and mindless, so just one thing less to worry about. I shoot chromes and all shots up to 1:3 have come out fine. I shoot both available light as well as with flash. With my base tilt the measurement of the rail extensions = magnification/exposure compensation are a bit off, but not so much as to ruin any shots.

It's a more pragmatic answer to your question, but it work.

Ole Tjugen
10-Aug-2005, 07:32
It's the magnification which matters. And that again is controlled by the distance from the film to the front nodal point, which is one focal length in front of the film when the camera is focussed at infinity. A little bit in front of the front element of a tele lens.

So for a 360mm Tele lens with a flange focal distance of 210mm, you reach 1:1 (and 2 stops compensation) at 470mm (210+360mm) extension, not 420mm (2x 210mm). Don't trust me, check it yourself.

ronald moravec
10-Aug-2005, 08:07
Illuminate a white board in your studio with cross light. Focus the lens at infinity and put a flat disc incident meter where the glass was removed. Set the lens to 11 or 16 and take a reading. Start racking the lens out measuring the bellows and increasing the aperature to achieve the same light intensity.

No formulas, no nothing. Just perfect results. This method takes into consideration all variables and works every time.

I used a Minolta Flash meter IV in incident/continuous mode. You will need to put the darkcloth over the back to keep out stray light .

Write the correction factors on the front of the lens board. cover the paper with clear tape.

10-Aug-2005, 14:27
I think I got it now. I (maybe Charles, too) was a bit confused because books, tutorials, etc. use the term “lens to film distance” vaguely, and often interchangeably with “bellows extension.” It works until you hit a snag with, for example, a tele-design lens.

I didn't start this thread, but thanks everyone for helpful comments.

Charles Forde
10-Aug-2005, 15:36
Thank you all very much for your help, I was assuming that the flange focal distance was doubled for 1:1.

I do like the method using a light meter and writing the correction factors on the front of the lens board.

Struan Gray
11-Aug-2005, 00:24
I too think the practical approach is best, but for those that want the background, David Jacobson's Lens Tutorial at photo.net (www.photo.net/learn/optics/lensTutorial) has the full info. What has been missing in this discussion so far is the 'pupillary magnification', which is a polysyllabic way of saying that for telephoto lenses the aperture has different sizes when seen from the front and back of the lens. The 'correct' formula becomes:

effective f-stop = marked f-stop * (1 + magnification/pupillary magnification)

Note that other asymmetric designs - like Tessars - also have this factor to include.

This is more likely to be significant if you are using a 35 mm macro lens reversed at large magnifications, in which case omitting the pupillary magnification will lead to over exposure, but if you care about 1/3 stop exposure corrections, and don't have time to measure or test, it's worth getting the numbers right.