The Math of the F/Stop Progression

As a quick post, I’m going to mention something that everyone seems to have difficulty with.  This is a little math-heavy and I’ll try to simplify it.

F/stop progressions.  Why do I have to double my shutter speed when going from F/2.0 to F/2.8?  Wouldn’t it make more sense to double my shutter speed when I go from F/2.0 to F/4.0?

The F/stop is related to the diameter of the aperture of the lens, or the width of the circle of light that shines on the sensor or film.  The key word here is ‘circle’.

Everyone remembers that the area of a circle is area = pi * r^2.  The number in the F/stop is related to the Diameter, which is 2 * radius.  If you want to cut the area of the circle in half, you need to divide the diameter (and thus the radius) by the squareroot of 2.  The squareroot of 2 is 1.4142136 … but for our purposes, 1.4 is good enough.

This is why the F/stop progression is 2.0, 2.8, 4.0, 5.6, etc.  Each of these numbers are about 1.4 apart (2 * 1.4 = 2.8, 2.8 * 1.4 = 2 * (1.4 * 1.4) = 2 * 2 = 4).  Each stop is the same as increasing the diameter/radius by 1.4, and doubling the size of the opening.

Now you may be thinking to yourself, “self, this makes it look like the numbers are in reverse order”.  The piece to understanding the order is in how the F/stop is normally stylized, F/2.8, F/4.0, etc.  The / in math means divided by.  It means the aperture is set to the focal length divided by the number represented in the F/stop.  For example, on a 120mm lens, at F/4.0, the aperture is 120mm/4.0 or 30mm.  If we stop down to F/8.0, the aperture is 15mm.  A circle with a 30mm diameter has an area 4 times the size of a circle with a 15mm diameter, thus, F/8.0 lets through one quarter the light of F/4.0, and is a change of ‘2 stops’.

I hope this helps understand everyone understand one of the less obvious parts of photography.

-Brad

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