# Playing around with depth of field and I think I finally get it?

Started 4 months ago | Discussions thread
Re: Playing around with depth of field and I think I finally get it?
1

LillyRoseAvalos wrote:

So according to the calculator I found at a subject distance of 10 centimeters at f9 on a 1.6 crop sensor the total depth of field is 0.46 centimeters. Which is 4.6 millimeters. Shooting at F2 at a subject distance of 20 centimeters gives me almost the same depth of field at 0.46. I tried to find some kind of formula for this but its inconsistent. Is there a formula so you can easily calculate the dof by multiplying the magnification? This far and near distance is making my head in though.

Depth of field is mathematically far from simple and there are many different mathematical formulae for DoF that all give the same result but using different sets of variables to start with.

On way of expressing it that I like, is as follows:

Nearside DoF = Ds/(1 + D)

Farside DoF    = Ds/(1 - D)

where D = c/mA .     If D>1 then the far side DoF is infinite.

c is the circle of confusion diameter

m is the image magnification

A is the entrance pupil diameter (focal length divided by f-number)

s is the distance from subject to entrance pupil.

These formulae are quite different from the ones you have probably seen already, but they give the same values for DoF and from them you can easily see that the difference between nearside and far side DoF is in the denominator (1 + D) or (1 - D) and D varies inversely with the magnification.

If m = 1 then D= c/A  is usually very small (c = 0.019mm and A = 2.7 mm for the example you show in the DoF calculator, so D = 0.007) and the nearside and far side DoF are about the same.