Originally posted by Anvh Actually how did you came with this formula.

DOF = 2cN(1+m/p)/m^2/(1-(d/d*)^2)

It seems to me you simply added the hyperfocal distance and the question is why?

I added it because it should be there.. Here's the detail on how it gets there

The previous equation in the wikipedia article is: first factor fm/Nc out of the denominator, then factor 1/m from the numerator

DOF =2f(1/m+1/p)/(fm/Nc -Nc/fm) = 2f(1/m+1/p)/((fm/Nc)(1-(Nc/fm)^2) = 2cN(1+m/p)/m^2/(1-(Nc/fm)^2)

Using the fact that m= f/(d-f) =(f/d)/(1-f/d), the denominator becomes (1-(Nc/fm)^2) = (1-(Nc/(f(f/(d(1-f/d))))^2) =(1-(d/(f^2/(Nc(1-f/d))))^2)

(1-(d/(f^2/(Nc(1-f/d))))^2) = (1-(d/d*)^2) where d* = f^2/Nc + f - the "hyperfocal distance".

so DOF = 2cN(1+m/p)/(1-(d/d*)^2)

But it is just as useful for macros to leave the denominator in terms of magnification (so you don't have to calculate hyperfocal distance).

DOF = (2cN(1+m/p)/m^2)/(1-(Nc/fm)^2)

Say a focal length effect less than 1% DOF can be ignored, and choose some fairly extreme parameters for normal macro use, like: N < f:22, c~ .02mm, f>10mm, then 1/10 > Nc/fm for m ~ 1/2. That is,

*focal length effect on macro DOF is less than 1% for magnifications greater than about 1:2. *
Dave in Iowa