Infinity

[Bernard Negrin]

Hello. I would like to confirm the following in regards to "infinity": If I a m working with a 210mm (8.26"inch lens)and I stretch the bellows out to a length of 8.26" inches, is this infinity? I have it understood that when you stretch the bellows to the length of the lens (8.26") you open up by 1 stop. At what po int with any LF lens do you have to start compensating for exposure? Thanks for any feedback...Bernard

[Alec Jones]

Yes. No. See here: http://www.cs.berkeley.edu/~qtluong/photography/lf/bellows-factor.html

[Huw Evans]

Generally speaking, you're right. An ordinary lens brings light from an infinitely distant source to focus at a point about one focal length behind it. I say 'about' because strictly speaking this is for a 'thin' lens, but it is more or less accurate for conventional multi element/group photographic lenses. It does not apply to telephoto designs, though.

When the lens is set one focal length from the film plane, the nominal aperture designations are correct and no compensation is necessary. If the lens is racked out to about 1.4 x focal length (or the square root of 2, to be precise) then you will need +1 stop compensation, 2 stops compensation at 2 x focal length, and so on. You may, or may not, find it simpler just to multiply the nominal f-number by the ratio of bellows extension to focal length. If you want to know what this is in stops the exact formula gives the compensation as the log of this ratio divided by the log of the square root of 2.

As for when you actually need to compensate, that is really up to you, considering the way you work and what sort of film you use, and so on. Personally I tend not to worry about 1/3 of a stop or less, but others are much more fastidious.

In general, the longer the focal length the more compensation is necessary, for any given subject distance. For example, your 210mm lens on a 4x5 might be said to be roughly equivalent to a 60mm lens on a 35mm camera. If you used the 210, focussed at 1m, that would require a bellows extension of around 266mm, giving about 2/3 of a stop compensation. However, the 60mm lens on the 35mm camera would need less than 4mm of extension, and only 1/6 of a stop compensation, which is hardly worth bothering with.

[tim atherton]

For easy calculation of bellows factor, go here

http://www.salzgeber.at/disc/index.html

download and print off, stick to some thin card, and read the instructions

quick, easy, free and so far, I have found, does a great job.

tim a

[Jim Galli]

Hi Bernard, The rough and reddy way to do this in the field has never failed for me. I also have a 210mm lens. I was inches away from a morning glory last Sun morning and the bellows was (guessed) at 14 inches. I didn't have a ruler. Round your 8.26 off to 8 and square it for 64, =8X8. Square the 14 that is the actual lens length now that you've stretched the bellows out that far for a macro shot. 14 squared is 196. Divide 64 into 196 for a factor of 3. Whatever shutter speed I would use for a normal shot at infinity will be multiplied by 3 for this shot. In my case I was looking at 1/4 at f16. That now becomes 3/4 at f16. I used 1 sec and stopped down to f16 1/3. When you use big rounded off inches instead of mm you can usually do these in your head. The shot came out perfect.

[Pete Andrews]

If you think in terms of f-stops, then you don't have to square anything. Just multiply the f-number by the bellows extension as a multiple of the 'normal' infinity length. For example: Total bellows extension is 12"; infinity bellows extension is (say) 8". 12 divided by 8 = 1.5. Now, if your aperture scale says f/16, then your real aperture is 16 x 1.5 = f/24.If the bellows extension is 10", then that's 1.25 x f/16 = f/20.With 16" of bellows, the effective aperture is f/32. Etcetera.Most people can do these simple sums in their sleep, let alone first thing in the morning.

BTW, the f-number that you arrive at by the above method is the real effective f-stop that you should use for calculating depth-of-field.

[Struan Gray]

Every time I go shopping I am reminded of my painful schoolboy discovery that there is a difference between mathematics (which I can do) and arithmetic (which I can't).

Just divide your focal length by ten enough times that you get a number between 1 and 50. Divide your bellows extension by the same amount. Look at your aperture scale, find approximately where the two numbers lie, and estimate the number of stops between the two.

For example, an 8 inch lens gives us 8 straight away. With 11 inches of bellows extension we get 11 for our second number. From f8 to f11 is one stop, so that's the compensation we need to apply. It's harder in metric, but bear with me. A 150 mm lens gives us f15 when divided by ten. 300 mm of extension gives us f30 when divided by that same ten. f15 to f30 is near-as-dammit f16 to f32, which is two stops.

This won't get you to within 1/10th of a stop, but dawn around here is at about 4 am right now, and at that time of day nothing gets me to within 1/10th of a stop. If you're really talented you can memorise the standard f-number series and avoid the need to focus on your aperture scale before that first cup of coffee has done it's work.

[Chad Jarvis]

Wow. Check out the big brains in this place...

[Huw Evans]

Well, everyone has their own quick and easy way of getting the answer. Then again, if you use a lens regularly enough it shouldn't be too long before you can estimate the compensation well enough without doing any arithmetic at all.

Alternatively, I presume that if you meter off the ground glass (I've never done it myself) then you could forget about the problem altogether.

Finally, a not terribly serious suggestion: as I implied above, the need for compensation only occurs, for a smaller format, at much closer distances, so maybe this is why we should all trade in our antiquated LF behemoths and use APS!

[Pete Andrews]

"the need for compensation only occurs, for a smaller format, at much closer distances,"Not quite so, Huw. For a given magnification ratio on film, and with a lens/film format giving the same angle of view, the exposure compensation necessary is the same, regardless of format size. It's probably just that small format users don't care if they're 1/2 a stop underexposed. ;^)