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It's entirely correct. ƒ relates to DoF only in that it's relative to the maximum aperture of a specific lens. Without knowing the specifics of the lens, you ca

It's entirely correct. ƒ relates to DoF only in that it's relative to the maximum aperture of a specific lens. Without knowing the specifics of the lens, you cannot know from arbitrary ƒ value if the aperture has been contracted (narrowing the flow of light, thus increasing the DoF), if it's just really dark glass in the lens, if someone has a TC mounted that steals 1-2 exposure stops, or if someone has simply used a gray filter or two attached to the end of the lens. I can produce two identical exposures, both of ƒ/8.0, where one has a narrow DoF and the other 4 aperture steps deeper DoF, by simply using a gray filter in one shot, and contracting the aperture in the other. In a third example I can use a very, very dark lens such as f.e. what Leica used to produce in the 60s, still exposing at ƒ/8.0 in wide-open aperture, reaching the same exposure and achieving a narrow DoF (contrary to what anyone would think just knowing it's ƒ/8.0), but without having stopped down a single step and without having used gray filters.

ghshephard 15y ago
I agree with the parent. F-Stop is the ratio of the Focal length to aperture diameter. Increasing that ratio, increase the depth of field. Decreasing that ratio, decrease the depth of field. Instinctively, F23 will put everything in focus (which is why we use it to identify dust spots on our CCDs, whereas F1.4 on anything reasonably close is going to have a DOF measured in inches. Can you come up with a counter example where DOF doesn't change in this direction? Apologies if I'm missing your entire point - I've been starting at it for five minutes, not sure if you are going a different direction.
- hackermom 15y ago
  "Can you come up with a counter example where DOF doesn't change in this direction?" I gave three examples above. I can elaborate: DoF changes depending on how you strangle the light by contracting the diaphragm the lights passes through (thus preventing light from dispersing), not from having a higher ƒ value. A really dark lens is a good example of this. Some can have a light throughput that begins at ƒ/6.3, or even ƒ/8.0, yet at wide-open aperture they produce as narrow DoF as f.e. my Nikkor 50mm ƒ/1.4 does, which is 5 exposure steps brighter - or, as you reason, 5 aperture steps more open. Why? Because their apertures are wide-open in both cases.
  - lutorm 15y ago
    Sorry, but you are wrong about this. Look at http://en.wikipedia.org/wiki/F-number http://en.wikipedia.org/wiki/F-number You seem to be thinking of something similar to what the Wikipedia article calls "T-stops", a calibration of the absolute transmission of different lenses. That's something different from the f/ratio.
    - hackermom 15y ago
      Are you nitpicking? You can point to aperture stop theory and the history of the f-numbers etc. indefinitely, but ƒ is and will remain synonymous with light transmission (or "T" as you keenly insist; thank you, I am familiar with it) in every single practical situation of photography, no matter how much you try to balance the needle on its tip.
      - ghshephard 15y ago
        hackermom, I mean this with earnest sincerity - so please take it in that nature; In fifteen years of photography, photography classes, photography magazines, and photography books - I have never, ever, had anyone suggest that ƒ is synonymous with light transmission. It is possible, that I, and every photographer, photography magazine, photography book, photography instructor, technical reference, manufacturer, and Wiki-Page, has been using the term incorrectly all this time - but could you stop for a second and possibly consider that perhaps you might be the one using it somewhat differently than everyone else? At the very least - look at a camera, and note that when you set the "F-Stop" to, or read it to be, say, F/3.5, it does precisely the same thing to the camera if you are in a completely pitch-dark room or outside on a very sunny day. That alone should suggest that F-Stop is not a measure of light, but instead of something else, and, perhaps also consider that it is, as I have suggested numerous times on this thread, the ratio of the focal length to aperture diameter.
  - ghshephard 15y ago
    You can't adjust your F-Stop with dark lenses. F-Stop is not a measure of light coming in, it's a measure of the focal length to the aperture. Light conditions prior to hitting the aperture (Either by flashes, sunlight, or dark lens) are entirely separate from the concept of F-Stop. For a given focal length on a subject in focus at a reasonable distance (not at infinity), increasing the F-Stop will always increase the DOF.
barrkel 15y ago
Can you explain why the ratio of aperture to focal length would change simply by using dark lenses or grey filters? The implied meaning of "ƒ only tells you how much light is passed to the film plane" is that ƒ would change merely by adding or removing a filter, rather than adjusting aperture or focal length.
- hackermom 15y ago
  ƒ is in every practical use an indicator of how much light comes out of the lens, nothing else. Two lenses of identical diameter and identical focal length can at wide-open aperture have two completely different ƒ values (due to, among other things, the brightness of the glass used in the groups of lenses inside), yet produce identical DoF. This is why ƒ tells you not the state of the aperture, which is primarily what affects your DoF, but rather how much light reaches the film plane.
  - ghshephard 15y ago
    F is not an indicator of light coming out of the lens. It is the ratio of the Focal Length to your aperture diameter. Nothing else. It says nothing about the amount of light coming out of your lens. And, indeed, cannot. Two Lenses of identical diameter [edit: aperture size] and identical focal length, by definition of the concept "F-Stop", will always have the same F Values. Period.
  - andrewjshults 15y ago
    This explanation is going to confuse non-photographers because for modern 35mm format lenses the difference between the maximum f-stop (which is purely the ratio between the focal length and the diameter of the aperture) and the maximum t-stop (the actual amount of light transmitted through a lens) are minimal (no where near full t-stops of light loss). Two lenses with the same focal length at the same f-stop with the same distance to the subject are going to produce the same DoF[1] and same exposure. [1] To really understand how to calculate DoF for the format you are shooting with and the size you are viewing at, you also need to understand Circles of Confusion (CoC). The same (effective) focal length lens at the same aperture/focal distance and printed at the same size is going to give dramatically different DoF if one camera is 35mm format and the other is large format (8x10).
lutorm 15y ago
I don't get it either. ƒ relates to DoF only in that it's relative to the maximum aperture of a specific lens. No, the f/ratio is f/D, which determines the divergence angle of the ray bundles that end up on the detector plane, and this is what determines the depth of field. You can not produce two exposures that have the same focal length and f/ratio that have different depths of field. Adding an ND filter will just make one image darker, it will not change the depth of field. You can't change the aperture without changing the f/value, because the f/value determines the aperture.
- dagw 15y ago
  You can not produce two exposures that have the same focal length and f/ratio that have different depths of field Just to nitpick, you can if you change the size of the film/sensor. f2 on a medium format camera looks very different than f2 on a digital P&S sensor. But other than that you are correct.
- hackermom 15y ago
  "You can not produce two exposures that have the same focal length and f/ratio that have different depths of field." I am not sure why we are both establishing to one another that DoF is determined by the dispersion of the light hitting the film plane when we already know this. My Mamiya Sekor 50mm ƒ/6.3 has a DoF not even tangibly different from my Nikkor 50mm ƒ/1.4, unless I stop the latter down to ƒ/6.3 which produces two identical exposures at same ƒ/ratio but with very different DoF.
  - ajkessler 15y ago
    This is simply not true. If you stop your Nikkor down to 6.3, it will have the identical depth of field as your Mamiya does at 6.3.
    - eftpotrm 15y ago
      No; the larger sensor size of the Mamiya as a medium format camera will give less depth of field for a given aperture than the 35mm Nikon. Otherwise you're right, so in the odd circumstance the poster has a Nikkor 50/1.4 medium format or Mamiya small format 50/6.3 then their depth of field characteristics will be as you said. I'm also pretty sure that even on 6x7 50mm @ f/6.3 wouldn't give DoF characteristics anywhere near 50mm @ f/1.4 on a DX sensor (as the extreme examples to artificially push them closer together), and they'd have a radically different field of view.

ghshephard 15y ago

I agree with the parent. F-Stop is the ratio of the Focal length to aperture diameter. Increasing that ratio, increase the depth of field. Decreasing that ratio, decrease the depth of field. Instinctively, F23 will put everything in focus (which is why we use it to identify dust spots on our CCDs, whereas F1.4 on anything reasonably close is going to have a DOF measured in inches. Can you come up with a counter example where DOF doesn't change in this direction? Apologies if I'm missing your entire point - I've been starting at it for five minutes, not sure if you are going a different direction.

hackermom 15y ago
"Can you come up with a counter example where DOF doesn't change in this direction?" I gave three examples above. I can elaborate: DoF changes depending on how you strangle the light by contracting the diaphragm the lights passes through (thus preventing light from dispersing), not from having a higher ƒ value. A really dark lens is a good example of this. Some can have a light throughput that begins at ƒ/6.3, or even ƒ/8.0, yet at wide-open aperture they produce as narrow DoF as f.e. my Nikkor 50mm ƒ/1.4 does, which is 5 exposure steps brighter - or, as you reason, 5 aperture steps more open. Why? Because their apertures are wide-open in both cases.
- lutorm 15y ago
  Sorry, but you are wrong about this. Look at http://en.wikipedia.org/wiki/F-number http://en.wikipedia.org/wiki/F-number You seem to be thinking of something similar to what the Wikipedia article calls "T-stops", a calibration of the absolute transmission of different lenses. That's something different from the f/ratio.
  - hackermom 15y ago
    Are you nitpicking? You can point to aperture stop theory and the history of the f-numbers etc. indefinitely, but ƒ is and will remain synonymous with light transmission (or "T" as you keenly insist; thank you, I am familiar with it) in every single practical situation of photography, no matter how much you try to balance the needle on its tip.
    - ghshephard 15y ago
      hackermom, I mean this with earnest sincerity - so please take it in that nature; In fifteen years of photography, photography classes, photography magazines, and photography books - I have never, ever, had anyone suggest that ƒ is synonymous with light transmission. It is possible, that I, and every photographer, photography magazine, photography book, photography instructor, technical reference, manufacturer, and Wiki-Page, has been using the term incorrectly all this time - but could you stop for a second and possibly consider that perhaps you might be the one using it somewhat differently than everyone else? At the very least - look at a camera, and note that when you set the "F-Stop" to, or read it to be, say, F/3.5, it does precisely the same thing to the camera if you are in a completely pitch-dark room or outside on a very sunny day. That alone should suggest that F-Stop is not a measure of light, but instead of something else, and, perhaps also consider that it is, as I have suggested numerous times on this thread, the ratio of the focal length to aperture diameter.
- ghshephard 15y ago
  You can't adjust your F-Stop with dark lenses. F-Stop is not a measure of light coming in, it's a measure of the focal length to the aperture. Light conditions prior to hitting the aperture (Either by flashes, sunlight, or dark lens) are entirely separate from the concept of F-Stop. For a given focal length on a subject in focus at a reasonable distance (not at infinity), increasing the F-Stop will always increase the DOF.

barrkel 15y ago

Can you explain why the ratio of aperture to focal length would change simply by using dark lenses or grey filters? The implied meaning of "ƒ only tells you how much light is passed to the film plane" is that ƒ would change merely by adding or removing a filter, rather than adjusting aperture or focal length.

hackermom 15y ago
ƒ is in every practical use an indicator of how much light comes out of the lens, nothing else. Two lenses of identical diameter and identical focal length can at wide-open aperture have two completely different ƒ values (due to, among other things, the brightness of the glass used in the groups of lenses inside), yet produce identical DoF. This is why ƒ tells you not the state of the aperture, which is primarily what affects your DoF, but rather how much light reaches the film plane.
- ghshephard 15y ago
  F is not an indicator of light coming out of the lens. It is the ratio of the Focal Length to your aperture diameter. Nothing else. It says nothing about the amount of light coming out of your lens. And, indeed, cannot. Two Lenses of identical diameter [edit: aperture size] and identical focal length, by definition of the concept "F-Stop", will always have the same F Values. Period.
- andrewjshults 15y ago
  This explanation is going to confuse non-photographers because for modern 35mm format lenses the difference between the maximum f-stop (which is purely the ratio between the focal length and the diameter of the aperture) and the maximum t-stop (the actual amount of light transmitted through a lens) are minimal (no where near full t-stops of light loss). Two lenses with the same focal length at the same f-stop with the same distance to the subject are going to produce the same DoF[1] and same exposure. [1] To really understand how to calculate DoF for the format you are shooting with and the size you are viewing at, you also need to understand Circles of Confusion (CoC). The same (effective) focal length lens at the same aperture/focal distance and printed at the same size is going to give dramatically different DoF if one camera is 35mm format and the other is large format (8x10).

lutorm 15y ago

I don't get it either. ƒ relates to DoF only in that it's relative to the maximum aperture of a specific lens. No, the f/ratio is f/D, which determines the divergence angle of the ray bundles that end up on the detector plane, and this is what determines the depth of field. You can not produce two exposures that have the same focal length and f/ratio that have different depths of field. Adding an ND filter will just make one image darker, it will not change the depth of field. You can't change the aperture without changing the f/value, because the f/value determines the aperture.

dagw 15y ago
You can not produce two exposures that have the same focal length and f/ratio that have different depths of field Just to nitpick, you can if you change the size of the film/sensor. f2 on a medium format camera looks very different than f2 on a digital P&S sensor. But other than that you are correct.
hackermom 15y ago
"You can not produce two exposures that have the same focal length and f/ratio that have different depths of field." I am not sure why we are both establishing to one another that DoF is determined by the dispersion of the light hitting the film plane when we already know this. My Mamiya Sekor 50mm ƒ/6.3 has a DoF not even tangibly different from my Nikkor 50mm ƒ/1.4, unless I stop the latter down to ƒ/6.3 which produces two identical exposures at same ƒ/ratio but with very different DoF.
- ajkessler 15y ago
  This is simply not true. If you stop your Nikkor down to 6.3, it will have the identical depth of field as your Mamiya does at 6.3.
  - eftpotrm 15y ago
    No; the larger sensor size of the Mamiya as a medium format camera will give less depth of field for a given aperture than the 35mm Nikon. Otherwise you're right, so in the odd circumstance the poster has a Nikkor 50/1.4 medium format or Mamiya small format 50/6.3 then their depth of field characteristics will be as you said. I'm also pretty sure that even on 6x7 50mm @ f/6.3 wouldn't give DoF characteristics anywhere near 50mm @ f/1.4 on a DX sensor (as the extreme examples to artificially push them closer together), and they'd have a radically different field of view.