I thought I had a pretty good seat of the pants understand of DOF. But I've been playing with DOF calculators and it's giving me results that don't line up with my practical experience, so I'm wondering where I'm going wrong here. This is about the relationship between film/sensor size and DOF.

Let's use some fixed parameters (keep film size variable) for the sake of argument:

focal length: 100mm, 
aperture: f/8, 
subject distance: 20 feet.

I know (from experience) that 35mm film has much more DOF than 4x5 film with those parameters. But the DOF field calculators I've tried are telling me the opposite, that the larger film/sensor will have greater DOF. The DOF calculators I'm looking at are http://www.dofmaster.com/dofjs.html and http://www.cambridgeincolour.com/tutorials/dof-calculator.htm, and both are telling me that larger sensors/film will have more DOF all other things being equal.

The Cambridge in Colour DOF calculator says DOF (with above parameters) for 35mm would be 6.29 feet (makes sense to me) and for 5"x4" film 34.6 feet (doesn't make sense to me).

So I'm missing something here in either my understanding of depth of field or the terminology or my understanding of DOF calculators. Can anyone straighten me out? What am I missing?


3 Answers 3


Here's what you're missing: that larger formats have less depth of field for the same framing, not at the same focal length. A 100mm lens is much wider on medium format than it is on 35mm film. If you keep that and the aperture constant, DoF will be identical assuming you print with the same enlargement (that is, the medium format print will be much larger). If you instead enlarge the 35mm print to the same size, you also enlarge the blur, yielding the decreased DoF the calculator is showing you. (It has these assumptions hard-coded into its chosen values for circle of confusion.)

Instead, adjust the focal length so that you have the same framing. If you hold that constant, you will find that your original understanding still holds: greater depth of field for the smaller format (even with greater enlargement). More background on this at previous question Can a smaller sensor's "crop factor" be used to calculate the exact increase in depth of field?

  • \$\begingroup\$ Your first sentence nails it. (face palm for my own sloppy thinking) \$\endgroup\$
    – obelia
    Commented Mar 26, 2014 at 5:09

The depth of field is the zone of acceptable sharpness in an image. This is a relative value; it is never absolute. Traditional depth of field numbers have been based on a fixed final image output size, namely an 8x10 print, but regardless of the target output size, the principle is the same.

For a 4x5 camera, to produce an 8x10 print, the image has to be enlarged 2x. Anything that looks sufficiently like a line or a point in the print, no matter how "fuzzy" it is on the negative in absolute terms, is deemed to fall within the depth of field. For a 35mm frame to make the same print, the image must be enlarged approximately 8x, so whatever is equally fuzzy on the negative in absolute terms will be four times as fuzzy on the reference print.

Given that the same focal-length lens used at the same aperture will produce the same image on the recording medium, the image that requires the least enlargement will result in the greatest depth of field. If you were to use a 24mmx30mm crop of the 4x5 image negative to produce an 8x10 print, it would require the same enlargement as the 35mm frame and have the same depth of field.

However (and this is a rather large however, as howevers go), it is a very rare occasion when we'd use the same focal length lens on different formats to try to achieve the same image. In this case, you would either be using a 400mm lens on the 4x5, or a 24mm lens (or a 25mm if you can find one) on the 35mm camera so that the resulting image would have the same field of view. At infinity, you would find that the two images would have as near the same DoF as makes no never mind, modulo film grain/sensor density, diffraction and aberrations. But life doesn't happen at infinity most of the time; you have to take into account bellows draw (or the change in focal length that happens when an internal-focus lens is focused closer than infinity) and the way that affects the depth of field.

Digital, too, complicates matters a bit, since there are absolute, discrete steps between things falling entirely within one sensor pixel and spilling over into the next. The pixel-peeper's circle of confusion is fundamentally a different beast from the film shooter's circle of confusion. (Personally, I'd love to see the day when resolutions are so high that pixel peeping becomes an exercise in futility; it would put an end to a lot of the utter silliness that currently pollutes photographic discussion.) As well, the size of the sensor is not directly related in any way to the pixel resolution of the sensor, so a (hypothetical) 30MP 4x5 "full-frame" digital camera would need exactly the same amount of enlargement as a D800 or an α7R to make the same-sized image, whether printed, projected or displayed on a screen.

Basically, though, it boils down to this: the less you need to enlarge an image, the more fuzziness you can tolerate before things start to look fuzzy.

  • 4
    \$\begingroup\$ "In this case, you would either be using a 400mm lens on the 4x5, or a 24mm lens (or a 25mm if you can find one) on the 35mm camera so that the resulting image would have the same field of view." This. The same focal length on each sized sensor completely changes the picture, it's not a straight swap \$\endgroup\$
    – Dreamager
    Commented Mar 25, 2014 at 9:51
  • \$\begingroup\$ +1 That's a great answer Stan but I've got to give the checkmark to mattdm for identifying my erroneous thinking so concisely. \$\endgroup\$
    – obelia
    Commented Mar 26, 2014 at 5:14

In my opinion, the question does not concern 'equivalent field of view'. Therefore, I believe what's missing are circle of confusion and hyper focal distance

Wiki DoF formula here.

The DoF calculators take into account circle of confusion and the hyper focal distance, which decreases as the sensor size increases.

As an example, 50mm @ ƒ/1.8, the hyper focal distance is:


Where: f = focal length, A = aperture, c = circle of confusion


= 92,592.6mm (92.6m) for 4/3 format, coc being 0.015


= 46,296.3mm (46.3m) for 135 format, coc being 0.03

Keeping everything else constant except format size, these are the only values that change due to the format size

You'll notice in the online calculator, these change:

4/3 format 135 format

(Yes, the values above are different. I'm not sure why but they are pretty similar)

Seeing as everyone is so caught up in 'equivalent field of view', of course... If you want to keep field of view and DoF equivalent, you must change the aperture and focal length/subject distance.

enter image description here Changing subject distance.

enter image description here Changing focal length.

  • \$\begingroup\$ If you don't keep the same field of view, then you're taking different images and thus any comparison is meaningless. Take an image from an ultra compact at 10mm and another image from a 35mm DSLR at 10mm and they will look totally different. The reality is that people using cropped sensors don't walk around taking images which are very tightly cropped because they're using a small sensor. Hence in the real world you have to conclude that sensor size does matter for DOF, you can't just say "DOF is the same, image is just cropped". \$\endgroup\$
    – Matt Grum
    Commented Mar 25, 2014 at 12:26
  • \$\begingroup\$ Thanks Matt, for totally ignoring the person's question. \$\endgroup\$
    – BBking
    Commented Mar 25, 2014 at 12:39
  • 1
    \$\begingroup\$ The person's question was, and I quote, "What am I missing?". What the questioner was missing was, if you don't keep field of view constant, then your comparison is meaningless hence the strange result posted. \$\endgroup\$
    – Matt Grum
    Commented Mar 25, 2014 at 12:43
  • \$\begingroup\$ I think he is missing the circle of confusion that gets taken into account in the formula. How else can you explain " 35mm at 6.29 and... 5"x4" at 34.6? \$\endgroup\$
    – BBking
    Commented Mar 25, 2014 at 12:50
  • 1
    \$\begingroup\$ The whole question is about Field of View, because that is the answer to the part of the question that asks, "What am I missing?" \$\endgroup\$
    – Michael C
    Commented Mar 26, 2014 at 2:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.