locustella
Well-known member
This is strictly theoretical question, knowing the depth of field is not needed for anything practical.
But how to calculate it ?
Example 1:
f = 24 mm (focal length of the camera lens)
N = 2.8 (F - number in the camera lens)
s = 50 m (object distance)
m = 60 (scope magnification)
d = 80 mm (entrance pupil diameter for the scope)
c = 0.029 mm (the circle of confusion for 35 mm film)
P = 1 (Pupil Factor, but of what ? It strongly affects the depth of field in the macrophotography)
Example 2:
the same data, but N = 22 (aperture limited by the aperture of the lens, not exit pupil of the scope)
Maybe using these data:
f' = 24 mm x 60 mm = 1440 mm (effective focal length)
s = 50 m
N' = 1440 mm / 80 mm = 18 (Example 1)
or
N' = 22 (Example 2) (like for the camera lens alone because the exit pupil of the scope: 80 mm / 60 = 1.33 mm is bigger then the entrance pupil of the camera lens: 24 mm / 22 = 1.09 mm)
???
And the depth of field would be:
DOF = 2 * N * c * [s - (P-1)/P*f]*(s-f) / {f^2 - [N*c*(s-f)/f]^2}
DOF ≈ 2*N*c*s*(s-f)/f^2 (approx and without P)
^2 means raised to power 2, formulas based on:
http://toothwalker.org/optics/dofderivation.html
http://www.largeformatphotography.info/articles/DoFinDepth.pdf
Example 1
DOF = 2 * 18 * 0.000029 * [50 - (1-1)/1*1.440]*(50-1.440) / {1.440^2 - [18*0.000029*(50-1.440)/1.440]^2} = 1.22 m
DOF ≈ 2 * 18 * 0.000029 * 50 * (50-1.440) / 1.440^2 ≈ 1.22 m
Example 1
DOF = 2 * 22 * 0.000029 * [50 - (1-1)/1*1.440]*(50-1.440) / {1.440^2 - [22*0.000029*(50-1.440)/1.440]^2} = 1.49 m
DOF ≈ 2 * 22 * 0.000029 * 50 * (50-1.440) / 1.440^2 ≈ 1.49 m
This way of calculating the F - number is perhaps good in terms of luminance (and exposure time), but is it correct for the depth of field (the angular blur) ? This is a little suspicious ... I think that it may be wrong.
Maybe knowing the image (or object) distance of the camera lens alone is needed ?
But how to calculate it ?
Example 1:
f = 24 mm (focal length of the camera lens)
N = 2.8 (F - number in the camera lens)
s = 50 m (object distance)
m = 60 (scope magnification)
d = 80 mm (entrance pupil diameter for the scope)
c = 0.029 mm (the circle of confusion for 35 mm film)
P = 1 (Pupil Factor, but of what ? It strongly affects the depth of field in the macrophotography)
Example 2:
the same data, but N = 22 (aperture limited by the aperture of the lens, not exit pupil of the scope)
Maybe using these data:
f' = 24 mm x 60 mm = 1440 mm (effective focal length)
s = 50 m
N' = 1440 mm / 80 mm = 18 (Example 1)
or
N' = 22 (Example 2) (like for the camera lens alone because the exit pupil of the scope: 80 mm / 60 = 1.33 mm is bigger then the entrance pupil of the camera lens: 24 mm / 22 = 1.09 mm)
???
And the depth of field would be:
DOF = 2 * N * c * [s - (P-1)/P*f]*(s-f) / {f^2 - [N*c*(s-f)/f]^2}
DOF ≈ 2*N*c*s*(s-f)/f^2 (approx and without P)
^2 means raised to power 2, formulas based on:
http://toothwalker.org/optics/dofderivation.html
http://www.largeformatphotography.info/articles/DoFinDepth.pdf
Example 1
DOF = 2 * 18 * 0.000029 * [50 - (1-1)/1*1.440]*(50-1.440) / {1.440^2 - [18*0.000029*(50-1.440)/1.440]^2} = 1.22 m
DOF ≈ 2 * 18 * 0.000029 * 50 * (50-1.440) / 1.440^2 ≈ 1.22 m
Example 1
DOF = 2 * 22 * 0.000029 * [50 - (1-1)/1*1.440]*(50-1.440) / {1.440^2 - [22*0.000029*(50-1.440)/1.440]^2} = 1.49 m
DOF ≈ 2 * 22 * 0.000029 * 50 * (50-1.440) / 1.440^2 ≈ 1.49 m
This way of calculating the F - number is perhaps good in terms of luminance (and exposure time), but is it correct for the depth of field (the angular blur) ? This is a little suspicious ... I think that it may be wrong.
Maybe knowing the image (or object) distance of the camera lens alone is needed ?