1976 is the year this lens, a Nikkor 50mm f/1.4, was made. This is the time when lenses were just metal and glass, with no electronics or motors. The lower ring had apertures engraved on it, going from 1.4 (its maximum) to 16 (its smallest).
The aperture ring had quite noticeable clicks at those f-values, but no intermediate clicks at 1/3 stops or anything. You went from 5.6 to 8 with nowhere to stop in between (though, with everything being analog and continuous, you could turn the ring to any position, even if it didn’t click in place).
(Side rant: I wish modern lenses and cameras were like that. Who needs 1/3 stop increments? I will take full-stop increments and fractional exposure compensation any day of the year. End of rant.)
Anyway, the reason for this post is not to let you long for a time when cameras were much simpler, but to explain why those pesky f-numbers are the way they are and not something else. In order to do this, we need math (brrrrrr!)
Ready? OK! Take 8 consecutive numbers, starting at 2 and doubling each time:
2, 4, 8, 16, 32, 64, 128, 256
Now take the square roots of those numbers, approximating to only keep 2 digits:
1.4, 2, 2.8, 4, 5.6, 8, 11, 16
OK, so now you know f-numbers are just the square roots of consecutive powers of 2. Big deal, but why?
The reason lies in the way the f-number corresponding to a given aperture hole size is defined:
f-number = focal length / diameter of the aperture
Or, solving for the diameter:
diameter of the aperture = focal length / f-number
So, for this 50mm lens, an f-number of 4 gives a diameter of 12.5mm (50/4) and an f-number of 5.6 gives a diameter of 8.9mm (50/5.6).
“Er… what hole are you talking about?”
The one that is inside the lens and that I can make larger or smaller by turning the aperture ring, as you can see in the second and third images. The one you cannot see in your modern lenses that don’t have an aperture ring that mechanically controls the aperture blades. I feel bad for you 😉
Sorry for the digression, let’s get back to our math. We still haven’t explained why we use the square roots. The answer is that the light that enters a hole is proportional to the area of the hole, not to its diameter.
In fact, if we approximate the aperture hole as a circle (it is close enough, for practical purposes) its area is given by
A = π r*r
Where r is the radius, or half the diameter. Solving for the diameters we have computed earlier for our 4 and 5.6 f-numbers, we obtain:
A(4) = 122.72 sq. mm
A(5.6) = 61.36 sq. mm
As you can see, the aperture area corresponding to f/5.6 is exactly half the aperture area at f/4 and, consequently, the amount of light entering the camera at f/5.6 is exactly half the amount of light entering the camera at f/4 (1).
Do you remember that a stop of difference in exposure is defined as half or double the amount of light? Well, now we have found out that the amount of light at f/5.6 is half that at f/4, so this is why there is one stop of difference between those two f-numbers or between any pair of consecutive f-numbers on that scale.
(1) To obtain the exact values, I used the value of sqrt(32) that is not 5.6, but more like 5.656854249492381. You wouldn’t want that engraved on a lens, for sure!