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From Wikipedia's depth of field article,

$$\mathrm{DOF} \approx {2u^2NC \over f^2} $$ for a given circle of confusion (\$C\$), focal length (\$f\$), F-number (\$N\$), and distance to subject (\$u\$).

According to the depth of field (DOF) formula, the DOF is linearly proportional to the circle of confusion (\$C\$), in which \$C\$ is also linearly proportional to the pixel size. Thus DOF is linearly proportional to the pixel size.

If that assumption is correct, then does that means some of the DSLR with APS-C sensor size and low pixel number would have a shallower DOF than some of the full frame DSLR with a high pixel number, assuming using full frame equivalent lens by considering the crop factor and same distance to the capturing subject?

Now calculating the ratio between DOF of Full frame (FF) and that of crop body (c), also applying the crop factor to the focal length and f-number, we get:

$$\begin{align*} {\mathrm{DOF_{FF}} \over \mathrm{DOF}_c} &= \frac{{P_\mathrm{FF}\cdot cN / (cf)^2}}{P_c N / f^2} \\ &= {P_\mathrm{FF} \over c P_c}\,, \end{align*}$$

for: \$P\$ = pixel number; \$f\$ = focal length of the lens on the crop body; \$N\$ = f-number of the lens on the crop body; and \$c\$ = crop factor.

Thus, C would have a shallower DOF than that of a FF when using FF equivalent lens, when:

$$ {P_\mathrm{FF} \over P_c} > c\,. $$

For example the Pentax K100D has 6 megapixels with an APS-C sized sensor, comparing to Pentax K-1 with a full frame sensor and 36 megapixels.

As such, [![enter image description here][4]][4]

$$ {\mathrm{DOF_{K-1}}\over\mathrm{DOF_{K100D}}} = {36\over6}\cdot{1\over1.5} = 3.733. $$

The DOF of K-1 would be about 4 times deeper than that of K100D.

Please correct me if I my logic is not sound. Thank you for your time and effort!

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  • \$\begingroup\$ Pixelcount is not relevant, but yes, if you compose the same framing of a picture with a aps-c and ff and e.g. a 50/1.8 lens. the ff will have a shalllower dof and a stronger bokeh effect. If you have the same distance the dof should be the same, but the framing is different. \$\endgroup\$
    – Horitsu
    Commented Feb 11, 2019 at 6:13
  • \$\begingroup\$ to be clear: pixelcount is not that relevant, sure more elements seems to be sharp if the count is really low, but this is just a side effect with no real impact. \$\endgroup\$
    – Horitsu
    Commented Feb 11, 2019 at 6:18

1 Answer 1

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the DOF is linearly proportional to the circle of confusion (C), in which C is also linearly proportional to the pixel size.

Actually, this isn't right. The CoC criterion is the largest blur that will be perceived by the viewer as a point. At low resolutions, this may be limited by pixel size, but generally in real world use other factors are dominant ­— display size, distance of viewer, etc.

Keep in mind also that we usually assume a same size print regardless of recording media size (so, greater enlargement for APS-C than full-frame), and when doing comparisons, we generally mean for the same framing, not for the same focal length. (More on this at Why do depth of field calculators show *more* DOF for larger formats with the same lens parameters?)

Since this is one of your basic assumptions, and it is not correct... everything after that falls apart from there.

Basically, pixel size has nothing to do with the depth of field calculation for most practical use, except in cases where there is so little resolution that this itself is a limit on image sharpness. But even on APS-C DSLRs from a decade ago, the pixels are so much smaller than a typically-selected CoC limit that they're effectively irrelevant. (See How do depth of field and the circle of confusion relate to pixel size on the sensor?)

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  • \$\begingroup\$ Thank you for your answer. The explanation on the CoC criterion is quite helpful! \$\endgroup\$
    – Dong Dong
    Commented Feb 11, 2019 at 11:10

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