The size of the iris inside a lens (or the opening that the iris can create) is referred to as F/Stops. The smaller the number, the bigger the hole that lets the light through. Every consecutive F/Stop is exactly half the amount of light (F/5.6 is half of F/4, F/8 is half F/5.6) A smaller aperture (higher number) gives a bigger DOF (and less light) than a larger aperture (Smaller number)

A good explanation is given here by Mathew Cole:
“F stop sequence: 1.4    2.0    2.8    4    5.6    8    11    16    22

On the face of it, going from f/4 to f/5.6 doesn’t sound like halving the amount of light. What’s more, 5.6 is a larger number and sounds like it ought to be more light, not less. Neither does f/4 to f/2.8 sound like doubling the amount of light. In fact, each of the numbers in this sequence is a halving/doubling of the amount of light from its immediate neighbours, just like the shutter speed settings are. Not only that, but it makes sense, as I shall show below.

The reason that both the halving and doubling and the smaller numbers mean more light things make sense is that the f/stop is a ratio. The ratio is between the diameter of the aperture in the lens and the focal length of the lens. The focal length is generally measured in millimeters, so we’ll stick with those as our unit of measure. On a 50mm lens, f/2 is saying that the diameter of the aperture is 25mm. The ratio is this 50/25 = 2. A good question might be, what is the area of that aperture? Well, the aperture is usually a set of five to fifteen blades which form a roughly circular hole, so we’ll use the formula for the area of a circle, which as you all remember from fifth grade math is π * radius2. For π I’ll use 3.14159265. On our 50mm lens, the aperture at f/2 has a diameter of 25mm which is a radius of 12.5mm. The area of the aperture is thus π X 12.52, or 3.14159265 X 156.25, or 490.9 square millimetres.

This fact by itself isn’t all that useful. It is useful in relation to the adjacent f/stops. What is the area of the aperture at f/2.8? Well, because the f/stop is a ratio of the focal length to diameter, our 50mm lens at f/2.8 would have a diameter of 50/2.8 = 17.86mm. The area of the circle thus formed would be π X (17.86/2)2, or 250.5 square mm. That’s about 250 sq. mm at f/2.8 and 500 at f/2, a double/half relationship. Aha! So that’s it! The area of the hole doubles and halves, it’s just represented by a ratio on the lens! No wonder it’s so darn confusing. ”

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