T3D Re: P3D negative lens for distance

[john bercovitz <bercov@xxxxxxxxxxx>]

> I was wondering if the fact that the new focus point, using 
> the .25 diopter lens, is 27 1/4 feet (8.3M), instead of a 
> focus point of infinity, would be a problem?

There are a lot of f/number vs. depth of field formulas 
floating around to help determine whether this situation 
would be acceptable to you or not.  But rather than getting 
into all that, I think it might be more fun to take an 
intuitive approach.  Let's say we have a 35 mm focal length 
lens which is wide open and let's say that wide open in this 
case is f/3.5 so we can easily calculate that the diameter 
of the entrance pupil is 10 mm.  So roughly speakling, for 
an infinite object point we have a cone of light converging 
on an image point behind the lens and the base of the cone 
is 10 mm in diameter and the height of the cone is 35 mm.

The camera was originally focussed at 2.7 meters and if I 
assume that the 2.7 meters is measured from the front focus
(almost true), then from Newton's formula, the film plane was 

x * x' = f*f
2700 * x' = 35 * 35
x' = .45 mm

.45 mm back from the back focus of the lens.  In other words, 
the film plane was at 35.45 mm.  When we put the -1/4 diopter 
lens at the front focus, we shifted the back focus .31 mm 
toward the film plane so now the film plane is only .14 mm 
farther back than the infinity focus.

A cone which is .14 mm high and has the proportions of a cone 
which is 10 mm in diameter and 35 mm high will have a base of 
(10/35)*.14 = .04 mm.  So the fuzzy image point from an 
infinitely-distant object point would be 1/25th of a mm in 
diameter.  This is 1 part in 900 of the focal length of the lens.
This is not too bad and the lens is wide open.  In actual use,
it would very likely be much better.

John B


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