Bellows Extension for a Given Lens to Subject Distance

[mike_lyons]

Hi all,

what is the formula (there is an equation for everything,or so it

seems)to calculate the amount of bellows extension for a particular

lens to subject distance.I understand that infinity focus is regarded

as equivalent to the focal length of the lens being used,(with the

subject to lens distance being about 200 times the focal length)and

that "life size" is twice that,presumeably 1/2 life size 1.4 the

focal length or there abouts.

However,if a 150mm lens is being used,and the subject is 1metre

away,what would be the amount of bellows extension needed?Or 1.5m?

Thanks in advance-Mike

[tom_westbrook]

Take a look at: <a href="http://www.photo.net/bboard/q-and-a-fetch-msg?msg_id=004H3Q">http://www.photo.net/bboard/q-and-a-fetch-msg?msg_id=004H3Q</a>

[robert_a._zeichner1]

(M+1) x F = D

This is a great optical formula that can be used to calculate bellows extension required in macro work, how big a print you can make in your enlarger or even screen size in slide projection.

M = magnification

F = focal length

D = distance

[philip_sweeney]

As a general rule: at 8X the lense focal length or greater, one can forget about the bellows factor. formula: bellows length squared divided by focal length squared.

[brian_ellis3]

"Life size" is 1-1 (i.e. the image on the film is equal in size to the subject size) and is attained when the bellows is extended to twice the focal length of the lens.

 

I've never heard that infinity is 200 times the focal length of the lens. For a 100mm (4 inch approx) lens, this would mean that infinity was approximately 800 inches or about 66 feet away. I always thought "infinity" was much farther than that but I'm not into formulas or math very much so maybe I'm wrong. I do know that when setting infinity stops on a large format camera the recommendation I've read is to focus on something at least a half mile away.

[leonard_evens]

The basic formula you asked for is

 

1/u + 1/v = 1/f

 

where u is the lens to subject distance, v is the lens to film distance (also called bellows extension), and f is the focal length. You can also solve for v as in

 

v = uf/(u - f)

 

In using these formulas you have to make sure you use a calculator with enough significant digits. Otherwise rounding errors can lead to significant errors in the answer.

 

What subject to lens distance is effectively the same as infinity depends on what you are doing. You have to choose what is called a circle of confusion, which is a disk you can't distinguish from a point. A typical such choice for 4 x 5 photography would be 0.1 mm, but some people might insist on half that. With that value. you can calculate the hyperfocal distance as

 

H = f^2/(Nc)

 

where N is the f-number. If you focus at the hyperfocal distance, anything from half it to infinity will appear in focus. The hyperfocal distance with the lens wide open would be one reasonable estimate of effective infinity. For example, with my f/5.6 150 mm lens and coc of diameter .1, it would be about 40 meters or about 132 feet. But with my 90 mm lens at f/6.8 it would about 12 meters or about 39 feet.

 

These values might be overestimates as far as viewing the ground glass is concerned because the coc for viewing purposes would ordinarily have to be less than what would be acceptable for viewing the final developed image.

 

However, if you are focusing with a high power loupe, that effectively reduces the circle of confusion appropriate to your viewing the ground glass. Picking 2 X as normal, a 4 X would double the effective infinity distance and an 8 X would multiply it by 4.

 

Scale of reproduction, the ratio of image size to subject size, denoted M, is related to subject distance by

 

u/f = 1 + 1/M

 

M is 1, i.e., subject and image are the same size when subject distance and film distance are both equal to twice the focal length.

 

If the subject is one meter (1000 mm) away, and the focal length is 150 mm,

then the film to lens distance is

 

v = 150*1000/(1000-150) = 176.5 mm

 

The scale of reproduction would be

 

M = 1/(u/f - 1) = 1/(1000/150 - 1) = 0.1765

 

For 1.5 m = 1500 mm, the film distance would be 166.7 mm and the scale of reproduction would be 1/9.

[michael_briggs2]

Take a look at the Lens Tutorial on Photo.Net: http://www.photo.net/learn/optics/lensTutorial. The first section has the equations relating focal length f, image distance Si, object distance So and magnification M.

 

To help you verify your understanding of the equations, for your example with f=150 mm and So = 1000 mm, the equations give Si = 176.47 mm as the bellows extension and the magnification M = 0.18. The image on the negative is 18% of lifesize. Be sure to use the same units for all quantities, e.g., mm for everything rather than mixing mm and m.

[dave_whitehead1]

Mike, here is the method I use. It's a bit fiddly, but you will soon get a feel for it. Once you have done, you will probably be able to estimate the amount of extra exposure you will need to give for a given extension. (In terms of stops). Anyway, you need a tape measure and a pocket calculator. At infinity focus, you should find that the distance between the lens and film standards is the same as the focal length of the lens you are using - no compensation is required. As you focus closer, you have to rack the bellows out. Let's assume you are using a 150mm lens. Focus on your subject, then measure the distance between your lens and film standards. Again, let's assume this distance is 200mm. MULTIPLY this by the APERTURE indicated on your light meter. So, if the indicated aperture is f22, multiply 22 X 200. This gives a figure of 4400. DIVIDE this by the focal length in use - in our example 150. You now have a figure of 29. F29 is, in fact, f22 and one half. This is the exposure that is going to reach the film - as if you underexposed by one half stop. So you must compensate - by opening up the lens - by half a stop. So your bellows extension has taken half a stop of light off you. Your new aperture is now f16 and one half. Sometimes though, the new aperture is no good to you because of your depth of field requirements. If you can't achieve adequate d.o.f. using movements, don't overlook the fact that you can simply re-cock the shutter and fire it again on the same sheet of film, to achieve adequate exposure. This is something I do time and again for commercial studio photography. It is not uncommon for me to give 4 or 8 "hits" onto the same sheet of film.

[george_jiri_loun]

DAVE, am I missing your point? Why do you use 4-8 hits instead of a longer exposure?

[ralph_barker]

George: Dave is talking about using studio strobes, where the shutter speed doesn't actually relate to exposure time. The duration of the flash is much shorter, so multiple pops (hits) solves the problem by effectively adding real exposure time.

[george_jiri_loun]

Did he mention any strobes??? Studio photography is also done with flood lights...