OpenStudy (anonymous):

If a 135 mm telephoto lens is designed to cover object distances from 1.15 m to infinity, over what distance must the lens move relative to the plane of the film? Could someone help?

3 years ago
OpenStudy (whpalmer4):

At infinity focus, the lens is going to be separated from the film plane by exactly 1 focal length. At 1:1 magnification, the lens is separated from the film plane by exactly 2 focal lengths. This I remember from my view camera usage. However, I never had to calculate the distance between film plane and lens plane because you just move the lens and watch the ground glass to see when you're in focus! Let me see if I can find or derive the formula...

3 years ago
OpenStudy (whpalmer4):

Assuming we can use the thin lens model, \[\frac{1}{f} =\frac{1}{b}+\frac{1}{v}\]where \(f\) is focal length, \(b\) is object distance and \(v\) is distance from lens principal plane to film plane. At \(v = 1.15\text{ m}\), \(f = 0.135\text{ m}\), plug in values and solve for \(b\).

3 years ago