How does on work out the minimum focusing distance given the maximum extenstion of the camera, the focal lenght of the lens and the flange distance?

As an example for a 450mm lens with flange distance of 428.4mm and a 8x10 camera with a maximum extension of 570mm, what would be the minimum focusing distance and magnification?

Thank you.

-- Natasha Sekha (natasha@buycomputer.com), October 01, 1997

Answers

The extension past infinity focus, e, is the max extension minus the flange-focal distance. Here e = 570 - 428 mm = 142mm.

The image distance from the lens, i, is about the focal length, f plus the extension e. So here i = 450 + 142 mm = 592 mm.

The object distance from the lens, o, is then given by the lens equation: 1/f = 1/i + 1/o.

For this example, 1/450 = 1/592 + 1/o. This gives o = 1876 mm, or about 1.9 meters from the lens (not from the film).

There will be slight deviations (possibly larger for lenses of telephoto design).

-- Ben Weiner (bweiner@muon.rutgers.edu), October 20, 1997.