How Aperture Relates to the other legs of the Triangle

The other aspect of aperture that is complicated is how it relates, numerically, to the other values in the exposure triangle.
Hang with me here! Or take my word for it and skip ahead.

Let’s say we have that 3 inch simple lens that is 1 inch across. What if we doubled the size of the lens to 2 inches, keeping the same focal length (somehow)? How much more light is transmitted through this lens? Since the light goes through the entire lens (the left side and the right side, and also the near side and the far side), the size of the lens has doubled in two directions! So the amount of light has increased by a factor of 4.

Another way of looking at this is to compare the area of the lenses, as you look down at it from the top. The area of a circle is

A=πr^2=1/4 π d^2

For the first lens:

A=1/4 π(1)^2

A=0.7854 〖inch〗^2

For the second lens:

A=1/4 π(2)^2

A=3.14159 〖inch〗^2

The ratio of these two areas is 4.0.
So, doubling the value of the aperture, changes the amount of light that goes through the lens by a factor of 4.

To double the amount of light that goes through a lens, you increase the aperture by a factor of the square-root of 2 (√2 = 1.414).

These values for the aperture are twice as big as the preceding one:

…, f/1, f/1.4, f/2, f/2.8, f/4, f/5.6, f/8, f/11, f/16, f/22, f/32, …

Each of these steps is “one stop” of exposure.

Next: Putting it all together: Exposure