Originally Posted by
Leonard Evens
Jeff Conrad's discussion does a good job of explaining what is going on, and anyone who really want to understand the subject should read his article.
But I'm firmly of the belief that there is no way to really understand the subject without going through the mathematics, amply described in Jeff's article.
I notice that no one made any effort to go into the mathematics, and, perhaps, that is the wisest course, but I will take a chance and do so.
DOF is a mathematical concept in geometric optics, and if you are willing to make the effort, a lot that seems mysterious becomes clear. Let me illustrate my point with the formula for hyperfocal distance.
H = f + f^2/(Nc)
where f is the focal length of the lens, N is the f-number used, and c is the diameter of the largest acceptable circle of confusion. (Since usually f is much smaller than f^2/(Nc), one can usually get by with the approximation
H ~ f^2/(Nc)
)
If you know the hyperfocal distance, you can calculate everything you may want to know about depth of field.
The quantity c is what is missing from most of the previous responses. It is a measure of how sharp you need the image to be in the negative in the camera. Unless one is making contact prints, one is more interested in what the viewer of the final print sees, and the final print will be enlarged from the image in the camera. Thus, typically, to make an 8 x 10 print, a 35 mm negative must be enlarged about eight times, while a 4 x 5 negative must be enlarged twice. That means you need to use a value of c about one fourth as large for a 35 mm negative as for a 4 x 5 negative. That in turn means that if the focal length f and f-number N are the same, you will get a hyperfocal distance about four times larger for the same size final print with a 35 mm format image than with a 4 x 5 format image.
When you focus at the hyperfocal distance, everything (in the final print) between half the hyperfocal distance and infinity will be in focus. So, in that case, the larger the hyperfocal distance the less the depth of field, which means that for the same focal length lens and the same f-number, you will have significantly less depth of field (in the final print) with 35 mm format than with 4 x 5 format.
If you look at the formulas which apply when focused at any other specific distance, you see that the same is true. With the same focal length and the same relative aperture, you have significantly less depth of field if the original format is 35 mm than if it is 4 x 5.
But you have to be careful about generalizing this. Similar calculations show that if you change the focal length so that the angle of view is the same in the two cases, e.g., so the 35 mm format focal length is about one fourth of the 4 x 5 format focal length, then the depth of field (in the final print) is significantly greater if the original image is in 35 mm format than if the original image is in 4 x 5 format.
This is no a contradiction, if you keep in mind that while, for example, 100 mm is a relatively long lens for 35 mm format, it is relatively short lens for 4 x 5 format. Of course if the format stays the same, longer lenses have less depth of field than shorter lenses, but to see how this plays out when you change formats, you need to look at the mathematics.
(In particular, if in the formula f^2/(Nc) you multiply c by 4 but keep f the same, you divide the total fraction by 4. But if you multiply c by a factor of 4 and at the same time multiply f by the same factor, because f^2 appears in the numerator, the total effect is to multiply f^2/(Nc) by 4. In other words, just multiplying c by 4 has the opposite effect to multiplying c by 4 and at the same time multiplying f by 4.)
Note that there is no way to understand how these different factors play out in specific situations without doing the mathematics. A purely verbal description can't do it.
Bookmarks