Trying to understand DOF in

Started Jul 24, 2009 | Discussions thread
Anonimo Contributing Member • Posts: 583
Long answer (better DoF calculator, math, etc.)

Hi Ed!

Links

Excelent article about lens equivalence:
http://www.josephjamesphotography.com/equivalence/

Better DoF calculator.
http://www.cambridgeincolour.com/tutorials/DOF-calculator.htm

The answer:

can_ka_no_rey wrote:

Example:
24mm f/1.4 10 feet distance:
5DII: total DOF = 4.69ft.
1DmkIII = 3.52ft.
50D = 2.88ft.
I'm lost. This means "the smaller the sensor, the shallower the DOF"?

  • The image formed by the lens in the focal plane is the same for the 3 cameras but each sensor will make a crop of it, resulting 3 pictures with the same deep of field but different angles of view.

  • If you enlarge the 3 pictures to the same print size you will be in enlarging the 50D image more than the 5D image and this will make the DoF of the 50D print be shallower than the DoF of the 5D print. This is the scenario that you are pluging in the DoF Master calculator.

The math

From the optics, if s is the focusing distance, f is the focal lenght, N is the f-number, C is the circle of confusion at the print and E is the enlargement factor (i.e. the ratio between print size and sensor size) then the DoF is given by the distances Dn (near) and Df (far) from the lens, calculated as follows:

1/Dn = 1/s + (CN/Ef)(1/f - 1/s)
1/Df = 1/s - (CN/Ef)(1/f - 1/s)

Now the trick when comparing croped versus full frame is to understand that:

  • for the same print-size, (enlargement factor for the croped sensor) = (enlargement for full-frame sensor) x (crop factor);

  • for the same field of view, (focal lenght for a croped sensor) = (focal length of full-frame sensor) / (crop factor);

  • from above, for same print size and angle of view the product Ef will be the same in croped and full-frame.

Now let's compare the DoF of prints of the same size in different scenarions:

a) 50mm f/10 shoot from a 50D versus 50mm f/10 shoot from a 5D croped to the same angle of view as the 50D. We have the same s, f, C, N , E. There is no difference .

b) 50mm f/10 shoot from a 50D versus 80mm f/10 shoot from a 5D. We have the same s, C, N and Ef but f is different. The 5D has a shallower DoF .

c) 50mm f/10 shoot from a 50D versus 80mm f/16 shoot from a 5D. We have the same s, C and Ef. We stick with N(1/f - 1/s) = N/f - N/s, where N/f is the same but N/s is different. The 5D has a shallower DoF. However : if s is large compared with f (i.e. if you are not doing macro) then 1/f - 1/s ~ 1/f; once N/f is the same the difference is neglicenciable. This is the lens equivalence .

Notes about DoF Master:

DoF Master adopts an arbitrary circle of confusion C = 0,030mm for full-frame sensors and E=1, correcting that for the crop factor of the selected camera. This number comes from an international norm for the calculation of DoF scales in lenses. Carl Zeiss claims that this norm was set long before the WW-II and was based on the resolution provided by film emulsions at the time (check page 3 of newsletter below).

http://www.zeiss.de/C12567A8003B8B6F/EmbedTitelIntern/CLN01e/ $File/CLN1.pdf

For comparision: the pixel-pitch of an EOS 5DMkII is 0,0064mm.

I recommend the DoF calculator from Cambridge-in-Colour, which uses print size, viewing distance and visual accuity to estimate a circle of confusion.

Regards,
Anonimo

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tko
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