# DOF and Cropping take 2

Started Feb 11, 2014 | Discussions thread
Shop cameras & lenses ▾
Back to basics

Here is a quote from the site

“As sensor size increases, the depth of field will decrease for a given aperture (when filling the frame with a subject of the same size and distance). This is because larger sensors require one to get closer to their subject, or to use a longer focal length in order to fill the frame with that subject. This means that one has to use progressively smaller aperture sizes in order to maintain the same depth of field on larger sensors. The following calculator predicts the required aperture and focal length in order to achieve the same depth of field (while maintaining perspective).

As an example calculation, if one wanted to reproduce the same perspective and depth of field on a full frame sensor as that attained using a 10 mm lens at f/11 on a camera with a 1.6X crop factor, one would need to use a 16 mm lens and an aperture of roughly f/18. Alternatively, if one used a 50 mm f/1.4 lens on a full frame sensor, this would produce a depth of field so shallow it would require an aperture of 0.9 on a camera with a 1.6X crop factor — not possible with consumer lenses!”

Back to basics.

1. If you draw a ray diagram, you have two triangles with their bases at the lens aperture, and their apexes at the subject and the image. If the lens is focused at infinity the height of the triangle on the on image side is the focal length (f) We always write the aperture (a) in terms of f - we don't say we have a 50mm lens with a 20mm aperture we say the aperture is f / 2.

If the image is formed in front of or behind the sensor / film, instead of a point we get a circle. So if we move the lens so it is slightly further from the sensor than f we can calculate the distance D where it will be focused, and if we know the aperture size and we can calculate the size of the circle c we get for something at it infinity.

The formula given in all books I've seen and many websites and which I sat down and proved for myself years ago is

c = f^2/(a*D)

Now there is some size of c which is so small it is indistinguishable from a point. This size depends on the how much you magnify a film negative, or how much of the sensor it take up. AFTER ALL CROPPING HAS BEEN DONE. It's normally taken as longest side of the used image / 1000 rounded down to 1 significant figure. c= 0.03 mm for 35mm film or FF digital and c=0.02 for APSc

We can rearrange the equation above so that we calculate the distance where something at infinity just appears to still be in focus

D = f^2/a*c  - this distance is called the hyperfocal distance H

c changes when you change sensor size.
If you want to keep the same angle of view so does f, and the equation contains f SQUARED.

So if you go from a 100mm lens on APSc to a 150MM lens on FF, and change the circle size from 0.02 to 0.03 you still increase the hyperfocal distance by 50%.

So the same d.o.f has a smaller aperture on a big sensor than a small one. In the extreme case, the lens can't stop down to a small enough aperture to give the big sensor as Deep a d.o.f as a small one; and or can't open wide enough to get give the small as shallow a D.o.f as the big one.

When you work out the distances of some other focus distance D which is in the d.o.f zone H is used in the calculation (D*H)/(D+H), and (D*H)/(D-H)

Once you know the bits in bold you can do your own calculations.

Complain
Post ()
Keyboard shortcuts: