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So I'm looking at the wikipedia article on Aperture, and I'm a bit confused. I understand aperture to be the diameter of the hole that lets light in. In the article, someone states that

The amount of light captured by a lens is proportional to the area of the aperture, equal to:

Area = pi ( focal length /(2 * f number) )^2

But in the f number article, they define f number as

f = focal length / aperture

It then seems trivial to substitute in:

Area = pi ( focal length / (2 * focal length / aperture) )^2

Area = pi ( focal length * aperture / (2 * focal length) )^2

Area = pi ( aperture / 2)^2

Area = pi ( radius )^2

So, my question is: Why did they bother putting something so blatantly obvious into the article? Didn't we already know that the area of the aperture was pi * r^2? Is this just me reading too much into a wikipedia article?

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The focal length and f-number are photographically relevant quantities, so the formula is expressed in terms of those convenient variables. Photographers don't generally know the radius of their lens aperture for every f-stop.

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To enlarge on "photographically relevant", we are interested in relative doubling or halving exposure more so than specific areal metrics, and the f-stops are typically chosen so every two stops represents a factor of two change in exposure (for constant shutter time). So, go down 2 stops (i.e. make the aperture larger), halve shutter time to get back to the same exposure.

I should mention that the concept of "stops" seems these days to only apply to DSLR lenses with manual f-stop rings, which are typically made to give a tactile click feel as you rotate them, so they can be operated whilst looking through the viewfinder. I just checked out how my little p&s (Fuji F500EXR) deals with f-stops, and it is very coarse indeed.

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  • \$\begingroup\$ On aperture stops and point-and-shoots, see Why do compact digital cameras not have the aperture range of DSLRs?. \$\endgroup\$
    – mattdm
    Apr 7, 2012 at 12:46
  • \$\begingroup\$ Regarding the comment that "we are interested in relative doubling or halving exposure", is there a downside if the aperture were specified as 1 sq mm, 2 sq mm, etc? You'd still know that 10 sq mm is double the exposure of 5 sq mm. Is there any downside to this as compared to the f-stop scale? In fact, this scale seems easier, since it's linear rather than sqrt(2). \$\endgroup\$
    – Kartick Vaddadi
    Dec 30, 2013 at 16:36

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