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Reasonably accurate. add.: it gives a good foundation regarding teaching exposure itself, but, apart from not explaining (nor visualizing) that focal length an

Reasonably accurate. add.: it gives a good foundation regarding teaching exposure itself, but, apart from not explaining (nor visualizing) that focal length and perspective are two separate attributes of optics, it also misrepresents the meaning and function of the ƒ value in a very common way often repeated even by seasoned photographers, in that it suggests (and indirectly claims) that ƒ only has to do with the aperture of the lens, and that ƒ is an indicator of what depth of field you will have. In reality, ƒ only tells you how much light is passed to the film plane; it does not tell you the state of the lens' aperture, nor does it say anything about the depth of field. This misleading explanation regarding ƒ is widespread even in literature.

kenjackson 15y ago
Could you expand on the 'f' value thing?
- hackermom 15y ago
  I hope the above two posts are detailed enough. If not, I'll gladly explain if you have a specific question.
  - kenjackson 15y ago
    It's very good. Thanks alot.
sasha61 15y ago
Actually this is not quite correct. F is ratio of aperture to focal length. As such, it does not tell you the absolute values of either, but it definitely does relate to DOF (although this will be affected by the focal length value - e.g., for the same F, you could have DOF of a few mm at short focal length, and a few dozen metres with long zoom lenses).
- hackermom 15y ago
  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.

kenjackson 15y ago

Could you expand on the 'f' value thing?

hackermom 15y ago
I hope the above two posts are detailed enough. If not, I'll gladly explain if you have a specific question.
- kenjackson 15y ago
  It's very good. Thanks alot.

sasha61 15y ago

Actually this is not quite correct. F is ratio of aperture to focal length. As such, it does not tell you the absolute values of either, but it definitely does relate to DOF (although this will be affected by the focal length value - e.g., for the same F, you could have DOF of a few mm at short focal length, and a few dozen metres with long zoom lenses).

hackermom 15y ago
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.