f-stops explained

This is a subject that for a long time has caused a lot of confusion in my head. Also, I have the feeling that very few people actually know what they're talking about when refering to f-stops or apertures in photography. Because I'm a geek I decided to go a little deeper and finally cleared all my doubts. Well at least almost all of them. I'll try to explain here what I've learned in the most simple way in the hope of making someone happier, and also to test my own knowledge. First things first...


What's the aperture?

I assume you have a slight clue on this, but anyway, simply put, the aperture is the actual hole that sits in your lens and lets light go through it until it reaches the sensor (or film plane).  The size of the hole can be controlled by the photographer if the lens (on older systems) or the camera allows it. The f-stop scale that we see in most lenses has this kind of progression:

f/1.4 - f2 - f2.8 - f4 - f.5.6 - f8 - f11 - f16 - f22 - f32

Most lenses nowadays have a slightly longer scale because they offer some steps in between.

How does it really work?

This is the question I have asked myself for a long time. Obviously there is a logical reason behind it, but it's not that easy to get there just by looking. You probably heard that each f-stop represents exactly half the light that you'd get in relation the previous f-stop (considering that every other variable remains the same - speed and ISO). So, f/2.8 is half the light of f/2. Why didn't they make it f/4 then? It would be easier to have a linear scale... But let's go a little deeper. First of all, it's important to know what the f-stop really means. Every f-stop number represents a ratio of the focal distance of the lens (take 50mm for example) to the actual diameter of the diaphragm opening.

The diaphragm is the device that controls effective diameter of the lens opening. It's usually formed by a series of blades that form an almost circular shape.

So, f/2 is equal to the Focal Distance / 2. Taking our 50mm lens as an example, that's 50 / 2 = 25mm (diameter of the aperture). Still following? Ok, so for a 50mm lens at f/2 the diaphragm has an opening with a diameter of 25mm. If you take a 100mm and do the same math, you can easily see that the same f/2 aperture results in a 50mm diameter (twice the focal length, twice the aperture diameter). You must asking "well if the aperture's bigger, how come the f-stop is the same (f/2)?". It's a great question that haunted me for a long time! And if you go and shoot a photo with both lenses at f/2 you will conclude that the amount of light that makes up your picture is the same. More on that later. At this point we know how to calculate the aperture size (diameter) and we know that it's directly related to the focal distance. What you may not know is that to determine the actual amount of light reaching the sensor (or the film plane) we need to consider the area of the aperture (and not the diameter). Remember how to calculate the area of a circle? Ok, I'll give you a little help: π * radius² (that's pi * square radius). Back to our two lenses, the 50mm (25mm diameter, 12,5mm radius) and the 100mm (50mm diameter, 25mm radius):

  • The area of our 50mm lens @ f/2 is = 3,14 * 12,5² = 490,6 mm²
  • The area of our 100mm lens @ f/2 is = 3,14 * 25² = 1962,5 mm²

Notice that an f/2 aperture in the 100mm lens is actually 4 times bigger than the same aperture in the 50mm lens (1962,5 / 490,6 = 4).


Why did we double the focal length and the aperture area is 4 times bigger?

It's math! There's a rule that says that the area varies as the square of the radius. So, if you double the radius that's the same as saying which is 4. By setting both lenses @ f/2 the 100mm gets double the aperture radius to that of the 50mm. 


If the opening is 4 times bigger how can it be the same amount of light (f/2) on both situations?

The answer has to do with the focal distance, once more. And also some physics: the inverse square law. Wikipedia says:

In physics, an inverse-square law is any physical law stating that some physical quantity or strength is inversely proportional to the square of the distance from the source of that physical quantity.

Read the rest of the article here.


Why the odd progression of the f-stop scale?

  • The amount of light is related to the area of lens opening not the diameter
  • The f-numbers represent the diameter of the opening (focal length / f-number)
  • If you double the area of the aperture, you double the amount of light
  • The area varies as the square of the diameter and consequently:
  • The diameter varies as the squared root of the area

So if each f-stop (moving backwards from right to left in the scale) represents double the light as the previous number that means its opening has double the area as well (for the same subject  in the same light conditions, if you double the area you double the light).  What happens to the diameter? It varies as the squared root of the area. Because we doubled the area, the diameter will be 1,414 times as great (which is √2 - square root of 2 - double). And that's how the scale is built, by using this multiplication factor:

  • f/2 * 1,414 = f/2.8
  • f2.8 * 1,414 = f/4
  • f/4 * 1,414 = f/5.6 and so on...

Well if you are still reading this then you reached the end! Thank you for the patience and I really hope I have made it easier for you to understand what you're doing next time you pick up the camera :-) Feel free to comment and leave me your feedback or your doubts if you still have them (comment here or send me a tweet).

 

5 comments to f-stops explained

  • Henrik

    This was great. Short and easy and made perfect sense. Thank you.

    I will come back to this blog the next time i need something explained, hoping you have made a path.

  • Jack

    Well, let me add to above article as it appears to simplify things and distort actuality. The formula you are referring for aperture calculation uses the diameter of “entrance pupil” which is most of the time NEVER equal to actual opening of the diaphragm. Overall yo have a good grasp of the problem, except the above carries importance for a “geeky” individual. There are some good reference books on optics which would explain the “pupil” issue vs. actual opening of the iris, or you can simply google it. Bottom line being, the pupil may be smaller or larger than the physical iris opening and depends on lens construction.

  • Tim Smith

    “The f-numbers represent the diameter of the opening (focal length / f-number)”

    Should this not read (focal length / diameter) ?

  • Vanessa

    Thanks…….!!!

  • Vanessa

    Thanks……..much clearer to me now!

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