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What does focusing mean? What does it have to do with the formula

1 / u + 1 / v = 1 / f

u is the object distance,

v is the image distance,

f is the focal length.

Does focusing mean that both sides of the formula are equal?

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  • When I look through the viewfinder of my camera I do not do mathematical equations, I turn the focusing ring until what I have chosen or what I feel is the subject is in focus. For me it has nothing to do with that formula. Perhaps there’s a mathematical equation to explain what I feel but I do not care to know what that is.
    – Alaska Man
    Mar 30 '20 at 8:31
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Both sides of the formula are always equal (for theoretical, thin lenses; for real lenses, the formula is only an approximation). What it says is where you should place the image plane (the film or the sensor) to have a sharp image.

In practice

  • focal length f is fixed (because you have a prime lens, or because you have choosen the desired focal length of your zoom lens)
  • the object distance u is also fixed (almost: it changes a bit when moving the lens for changing focus)
  • then it follows there is one specific image distance v

Focusing means changing the distance between lens and film/sensor so that it matches v.

Again, only approximately because real lenses are not theoretical thin lenses; also focusing generally moves the lens so that object distance u changes a bit.

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    If u + v is fixed, you can still move the lens back and forth: that increases u and decreases v by the same amount, or vice versa. What do you mean by 'the depth of field is infinite'? It absolutely isn't. Mar 30 '20 at 12:41
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    This formula calculates what u or v need to be for a perfectly sharp image. DOF calculations show how far u and v can be off their ideal position while still having an acceptably sharp image.
    – Orbit
    Mar 30 '20 at 13:12
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    For focus at infinity, 1/u = 0, so 1/v = 1/f and v = f. In other words, the sensor/film should be exactly at the focal point of the lens. Note that focus at infinity is not the same as infinite depth of field. Is focus at infinity what you're after? Or rather focusing at the hyperfocal distance, which comes closest to infinite depth of field? Mar 30 '20 at 14:04
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    @enbinzheng: other factors come into play (circle of confusion) when talking about depth of field and hyperfocal, thigs I'm not very knowledgeable about and certainly don't fit in a comment. Maybe best put it into a new question. Mar 30 '20 at 18:24
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    @enbinzheng: That's an old question on a different subject, and already has an accepted answer. I'm perfectly happy to answer a question now and then if I feel I have anything valuable to add, but I don't have the time to go on wild tangents, sorry. Mar 31 '20 at 8:12
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Yes, it means changing either f or v (u+v being constant) so that a point on the subject is just a point on the sensor/film (ie, v is the distance from the lens to the sensor).

In other words, the formula is always true, but if v isn't the distance between the lens and the sensor the subject is out of focus. Focusing is moving the lens (or changing its focal length) to that v is the distance between lens and sensor.

In practice, real camera lenses are not the theoretical "thin lenses" on which your formula is valid...

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  • Does focusing mean that both sides of the formula are equal?
    – enbin
    Mar 30 '20 at 11:09
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    The formula is always true. But if v isn't the distance between the lens and the sensor the subject is out of focus. Focusing is moving the lens (or changing its focal length) to that `v| is the distance between lens and sensor.
    – xenoid
    Mar 30 '20 at 11:43
  • If u + v is fixed and f is also fixed, then when v is what, the depth of field is infinite.
    – enbin
    Mar 30 '20 at 11:59
  • If u+v is fixed, you can still move the lens longitudinally: (u-Δ)+(v+Δ)=u+v
    – xenoid
    Mar 30 '20 at 12:05
  • So what v will have infinite depth of field?
    – enbin
    Mar 30 '20 at 12:39

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