Why don't we use a linear scale for f-stops, rather than one based on square roots? A linear scale will be easy to remember (in fact, there will be nothing to remember) and to calculate how much brighter one stop is over another. That is, f/6 will have half the brightness as f/3, which will have two-thirds the brightness of f/2.

This will still have the advantage of the current f-stop scale, which is that we can compare them independent of focal length / angle of view.

What am I missing? Or is the only answer to my question that it's an accident of history?

  • \$\begingroup\$ f/3 is actually 4 times brighter than f/6 \$\endgroup\$
    – phuclv
    May 31, 2014 at 13:34
  • \$\begingroup\$ @LưuVĩnhPhúc Yes, that's basically the question, but it's confused by reusing the notation in the examples for the proposed alternate system. \$\endgroup\$
    – mattdm
    May 31, 2014 at 13:42
  • \$\begingroup\$ I had the same thought as well, it would be much more intuitive to use f number squared. This would not be linear though, it would be 1, 2, 4, 16, 32, 64, etc (instead of 1, 1.4, 2, 2.8, etc). But it would be intuitive and match the way you calculate exposure for Shutter and ISO (doubling/halving the value will double/halve the exposure). \$\endgroup\$
    – mike
    Jan 31, 2018 at 20:43

3 Answers 3


The f number is the inverse square root of the light collection efficiency. Why such a crazy relationship?

First - let me define "light collection efficiency". You build an image by collecting photons (light particles) with your lens, and focusing them onto the film / sensor. If you double the area of your lens, you collect twice as many photons - so you can increase your shutter speed by 2x and still collect the same number of photons as before.

Now the f number is the ratio of the focal length of a lens to its diameter : that's a nice easy thing to measure. A 25 mm diameter lens with a focal length of 50 mm has a theoretical f number of 2.0. In practice it will be less, because you lose some light on the surfaces of the lens.

If you double the focal length, the image you create is bigger. In fact, the area of the sensor (for the same field of view) would have to be 4x greater to see the same image. Consequently, each "pixel" on your film/sensor sees 1/4 of the photons it saw before.

Similarly, if you double the diameter of the lens, you increase the area of the lens by 4x. If you do both - double the lens diameter, and double the focal length - then you collect 4x as much light and have to distribute over 4x as much area, so light per unit area stays the same: you have the same "light collection efficiency".

If you just want a factor 2x change in light collection efficiency, then you need to change the f stop by the square root of 2. And this is why you have the series that we tend to use: 1.4, 2.0, 2.8, 4.0, 5.6, 8, 11, 16, 22, ...

The answer to your question, then, consists it two components:

1) Historically, the light gathering power of a lens was most easily described by the ratio of focal length to lens diameter - it is something that is easy to measure. That f-number became the standard, and several attempts to dislodge it have been unsuccessful.
2)And the reason that this (generally accepted) scale is not linear is explained above.

So why don't we use the square of the ratio? Sometimes that would make sense - at other times it would not. The math can actually be easier this way. For example if you have a flash with a guide number of 45 at an ISO of 100, you can compute the f stop needed as (guide number / distance). To shoot a subject at 12 meters you would need a f stop of 4. At 4 meters you would need 11. On the other hand if you don't use a flash then there is no advantage on the math: a shutter speed of 1/125 and an f number of 8 gives the same exposure as 1/250 with f 5.6 - in other words shutter speed times f number squared is constant. I agree that is not helpful.

So back to the original answer. "Just because history".

  • \$\begingroup\$ Thanks for the explanation -- it helped me understand the physics better, and get a better intuitive idea. But it doesn't really answer my question. \$\endgroup\$ Jun 1, 2014 at 4:09
  • \$\begingroup\$ I added a post script which I hope provides a more specific answer to your question... \$\endgroup\$
    – Floris
    Jun 1, 2014 at 4:25

It's not linear because it's based on area rather than lines.

Or, to put it another way, the scale isn't based on square roots. It's based on the actual math that represents the physical properties of the system, and it happens that exposure doubles when aperture size opens by a factor of the square root of 2.

If you remember back to your elementary-school math, the area of a circle is πr². For a given aperture area A, how do we change r to get it to double? Well, we need to increase that r by a number which, when squared, is 2 — there's our √2.

We could certainly renumber those stops to give a linear scale, but since that would be an extra layer of abstraction from the physical system, that actually makes it more complicated rather than easier. This is more than theoretical, as the numbers are used in other places like flash guide numbers.

It's not really hard to understand and memorize the basic scale (since the practical range is only a handful of numbers anyway). In the 19th century, there were a number of competing systems using linear numbering, but by the beginning of the last century, the f-number system had won out.

  • \$\begingroup\$ Thanks, mattdm. I understand that it's not BASED on square roots. The wikipedia link you gave says, "these systems all involved some arbitrary constant, as opposed to the simple ratio of focal length and diameter". But this is not true -- one can just use the square of the ratio between the focal length and the entrance pupil. Yes, it would be an extra layer of abstraction over the physical system, but wouldn't it make it easier to compare and convert between apertures? In other words, it's a good abstraction (one that makes life easier for us) rather than a bad one. \$\endgroup\$ Jun 1, 2014 at 3:56
  • \$\begingroup\$ Thanks also for the tip about how to easily remember the f-stop scale. I will keep that in mind. Though, to go back to my original question, it's still not as easy as a linear scale, where there's nothing to remember. As for flash guide numbers, I did read the link you gave, but I'm afraid I'm not qualified to talk about it :) \$\endgroup\$ Jun 1, 2014 at 4:02
  • \$\begingroup\$ I would add to @KartickVaddadi comment that the f number is the focal ratio - focal length divided by diameter of the lens. When you do double or halve the lens area, the effective lens diameter changes as described and the new ratio of focal length to diameter is the f number. \$\endgroup\$ Jun 2, 2014 at 6:11
  • \$\begingroup\$ "since that would be an extra layer of abstraction from the physical system, that actually makes it more complicated rather than easier." IMO, using "f squared" for aperture (1, 2, 4, 8, etc) would be more intuitive since that's the scale used for shutter and ISO. \$\endgroup\$
    – mike
    Jan 31, 2018 at 20:45
  • \$\begingroup\$ @mike Maybe. Too late, though. \$\endgroup\$
    – mattdm
    Jan 31, 2018 at 20:49

Because an f-number is not an arbitrary number, it is the ratio between the focal length of the lens and the diameter of the entrance pupil (the effective aperture, if you will). To double the area of the entrance pupil at any given focal length one must increase the diameter of the entrance pupil by multiplying by the square root of two.

  • \$\begingroup\$ Thanks. My question was: instead of using the ratio between the focal length and the entrance pupil, why don't we use the square of this ratio? \$\endgroup\$ Jun 1, 2014 at 3:51
  • \$\begingroup\$ Because then the same number would mean different exposure values at different focal lengths. \$\endgroup\$
    – Michael C
    Jun 1, 2014 at 18:22
  • \$\begingroup\$ Needless to say, the formula to calculate exposure value in terms of F-number would be adjusted -- it would no longer square the F-number, since it's already squared. \$\endgroup\$ Jun 2, 2014 at 12:32
  • \$\begingroup\$ In case that was not clear, I am referring to the formula from en.wikipedia.org/wiki/Exposure_value#Formal_definition : exposure value = log base 2 (N^2 / t) where N is the f-number, and t is the exposure time. Now, with my proposed f-number, which is the square of the conventional f-number, the formula would change to log base 2 ( N / t). So the end result would be the same. \$\endgroup\$ Jun 4, 2014 at 4:33
  • 1
    \$\begingroup\$ @KartickVaddadi "Needless to say, the formula to calculate exposure value in terms of F-number would be adjusted -- it would no longer square the F-number, since it's already squared." No it wouldn't with regard to flash guide numbers. \$\endgroup\$
    – Michael C
    Jun 4, 2014 at 11:57

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