### Focus distance needed for lenses

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How can you figure out how much bellows extension you need to focus a lense from say 3 feet to infinity? The lense that I use most often is a 90mm. Is it as simple as at infinity it has to be 90mm away from the ground glass? If so then how do you figure out for close ups? Is there a chart or equation? Thanks in advance for any help that anyone can give on this subject.

-- Russell Johnston (naturwrk@bellsouth.net), April 05, 2000

### Answers

I,m very new to large format. so please forgive me. but I might be able to help. I think the biggest part of the answer will be what f- stop to use. I use schneider lenses and have found a really helpful chart on the schneider page. it lists the specs for the various lenses and the minimum focus distance to infinity; DOF for each stop.

http://www.schneideroptics.com/large/depth/90depth.htm

I hope I haven't made a fool of myself. I don't really care that much! looks like 3 feet may be too close to ask for infinity!

-- paul schuster (paul@schusterphoto.com), April 05, 2000.

try f32 for Infinity to 4.54 feet

-- paul schuster (paul@schusterphoto.com), April 05, 2000.

When focused at infinity, it probably is not 90mm. It depends on the optical design and the 'back focus', but it may be close to 90mm. To achieve 1:1 on the ground glass, the bellows will be racked out to twice the infinity position (with 2 stops light loss). Using tilts, you can get 3 feet to infinity with less stopping down, a real benefit in LF.

-- Ron Shaw (shaw9@llnl.gov), April 05, 2000.

Russell: You are correct that the 90mm lens needs 90mm (give or take a few) to focus at infinity. For closeups, 180mm of bellows would give you a 1-1 ratio of subject to image size. The in-between sizes will then fall between the two extremes. For extreme depth of field, you would need to focus at the hyperfocal distance for the f-stop used. Whole books have been written on this subject. Basically, you would focus about one-third of the way into the scene and let the depth of field carry the rest. It is usual to focus closer than the subject with wide angle lenses to take care of teh curvature of the projected image. It is called "focusing in" with wide angle lenses. I have probably confused you more, but maybe you can figure out what I mean.

-- Doug Paramore (dougmary@alanet.com), April 05, 2000.

Russel,

For any focal length lens at any aperture there is a "hyperfocal" distance. This is the focus point that holds reasonable sharpness from half that distance to infinity. On a 35mm camera you can find this simply by using the focus scales, at least on older manual focus lenses. Charts are available for large format lenses. For example, a 14" (360mm) lens at f/45 has a hyperfocal distance of 46 feet. Focus on something 46 feet away, stop down to f/45, and everything from 23 feet to infinity will be in good focus. I make up a little chart of HF distances for my most often used lenses and tape copies of it to the lids of my equipment cases. I think you can find useful charts at the Schneider site.---Carl

-- Carl Weese (cweese@earthlink.net), April 05, 2000.

Russell, it IS as simple, almost, as you said; ie, 90mm from lens to image plane when focused at infinity. The little trick is that the 90mm is to a certain point on the lens that you don't know exactly where it is. But you can simply focus at infinity at put a reference mark somewhere that you can measure from; then, in your own mind you treat that point as though it is 90mm from somewhere.

Now, to focus closer, you can use the fairly simple equation: 1/focal_length = 1/front_distance + 1/rear_distance where front_distance is nominal lens to subject and rear_distance is your mark plus 90mm.

If, for example, you now want to focus at 7 feet, convert the 7 feet to mm; there are 25.4 (exactly) millimeters in an inch, so 7 feet = 7x12x25.4 = 21,336mm. So 1/90mm = 1/21,336 + 1/unknown_distance, OR 1/unknown_dist = 1/90 - 1/21336 = 0.0110642. Invert that; ie, 1/0.010642 = 90.38 and you're done! The 90.38mm distance is from the film plane to the previous "certain point on the lens". However, since the reference mark you made represents 90mm, you would simply make that point move away from the lens by an additional 0.38mm.

It sounds complicated and difficult, but if you struggle through it once or twice it'll be a piece of cake. You could pre-make a table for your standard subject distances.

PS; you will encounter a similar difficulty on the subject side of the lens by not knowing the exact point to use for measurement, you can approximate this by using the formula and measuring from the subject. Feel free to email me with questions! PPS: for 3 feet, the lens should be at 99.82mm, that is, racked out 9.82mm from the infinity focus position.

-- Bill C (bcarriel@cpicorp.com), April 05, 2000.

I think the formula you need (and you can use to build charts yourself) is:

Hyperfocal distance = (Focal length)^2 / (f-number * diameter of circle of confusion)

For example, if you want your depth of field to extend from 3 feet (= 914mm) to infinity, you need the hyperfocal distance to equal 6 feet (= 1828mm). This implies, given a circle of confusion diameter of 0.15mm (standard value for 4x5, as provided by the Kodak Book of Large Format Photography), that you need to set the f-number to 32, and focus to 6 feet. Using the formula:

1/focal length = (1/focusing distance) + (1/bellows extension),

this implies you need bellows extension equal to 94.7mm (but please check my computations before taking the picture...)

-- Benjamin Croitoru (croito32@wharton.upenn.edu), April 05, 2000.

Benjamin, your calculation of the 6 ft focus is correct, but my prior example certainly is not!

I don't know if I should pretend I meant 70 (instead of 7) feet or act like it they were typographic errors. Anyway, here's a fixed up version:

>> If, for example, you now want to focus at 7 feet, convert the 7 feet to mm; there are 25.4 (exactly) millimeters in an inch, so 7 feet = 7x12x25.4 = 2,134mm. So 1/90mm = 1/2,134 + 1/unknown_distance, OR 1/unknown_dist = 1/90 - 1/2134 = 0.010642. Invert that; ie, 1/0.010642 = 93.96 and you're done! The 93.96mm distance is from the film plane to the previous "certain point on the lens". However, since the reference mark you made represents 90mm, you would simply make that point move away from the FILM by an additional 3.96 mm. <<

-- Bill C (bcarriel@cpicorp.com), April 05, 2000.

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