A sequence is a set of numbers that can be constructed using a formula known as a recurrence relation. The most obvious of these is the set of natural numbers (integers). How? Start with any integer (positive or negative). To get the next integer, simply add 1!

Some sequences are very obvious to decipher, while others require more mathematical manipulation, such as the Fibonacci sequence. Sequences often show up in pure mathematics, number theory, and computer science. [More about sequences].

One of the most widely used sequences is the f-number or f-stop (f/stop) series of numbers in photography. No matter where you stand as a photographer, you will be faced with these numbers. Often times, you will have to rely on your calculations to determine how many stops there are between two stop numbers. You will have to memorize them or just rely on your camera - unless you know the mathematics behind these numbers.

In this article, I will explain the method I use to remember the f-stop sequence. All that is needed is to remember the first two numbers. I will first quickly present the method so that you don't have to read this entire article. I will then present the mathematical formalism for the way f-stop numbers are constructed.

How to Remember the f/stop Numbers

I will start with the most common f-stop numbers. These are given by the following set of numbers

$\bg_white \120dpi {\color{red} 1\quad1.4\quad2\quad2.8\quad4\quad5.6\quad8\quad11\quad16\quad22\quad32\quad45\quad64\quad90\quad128}$

Let us write these in the following way

$\bg_white \120dpi \large \begin{matrix} 1 \overset{ }{\longrightarrow} 2\overset{ }{\longrightarrow} 4\overset{ }{\longrightarrow} 8\overset{ }{\longrightarrow} 16\overset{ }{\longrightarrow} 32\overset{ }{\longrightarrow} 64\overset{ }{\longrightarrow} 128\\ \;\quad 1.4\overset{ }{\longrightarrow} 2.8\overset{ }{\longrightarrow} 5.6\overset{ }{\longrightarrow} 11\overset{ }{\longrightarrow} 22\overset{ }{\longrightarrow} 45\overset{ }{\longrightarrow} 90 \end{matrix}$

Looking at each row separately, you will quickly notice that these form what is called a geometric sequence. A geometric sequence is a set of ordered numbers, in which any number is obtained by multiplying the previous number by a constant value. This constant value is known as the common ratio. Guess what the common ratio (in our case) is? It is 2 (see proof below). Here's another graphic for that

$\bg_white \120dpi \large \begin{matrix} 1 \overset{\times 2}{\longrightarrow} 2\overset{\times 2}{\longrightarrow} 4\overset{\times 2}{\longrightarrow} 8\overset{\times 2}{\longrightarrow} 16\overset{\times 2}{\longrightarrow} 32\overset{\times 2}{\longrightarrow} 64\overset{\times 2}{\longrightarrow} 128\\ \;\quad 1.4\overset{\times 2}{\longrightarrow} 2.8\overset{\times 2}{\longrightarrow} 5.6\overset{\times 2}{\longrightarrow} 11\overset{\times 2}{\longrightarrow} 22\overset{\times 2}{\longrightarrow} 45\overset{\times 2}{\longrightarrow} 90 \end{matrix}$

As you can see, you only need to remember the first two f-numbers, i.e. 1 and 1.4. Separate them into two sets: the even set (first row) and the odd set (bottom row). Then construct the entire f-stop range just by multiplying by 2.

Remarks:

Note that each set presents jumps in two stops, not one stop. f/1.4 lets in four times more light than f/2.8. Being multiplied by two should emphasize that fact - for f/1.4, the lens diameter is twice as much as that for f/2.8. (As will be explained below, to go one full stop at a time, you'd have to multiply by Sqrt(2)~1.4, e.g. 1.4*1.4 ~ 2 and so on...)
Looking at the odd set (bottom row), you can notice that 5.6x2 = 11.2, not 11. So why do we choose 11? To the best of my knowledge, it is just a convention to keep the numbers easy to remember. The actual f/stop used by the lens is 11.312).

As will be shown below, if we start by f/1.0 as the smallest possible f/stop, the next full stop is 1.0xSqrt(2) = 1.414. This is the first f/stop that corresponds to the bottom row. Now
1.414x2 = 2.828
2.828x2 = 5.686
5.686x2 = 11.312
11.312x2 = 22.624
22.624x2 = 45.248
and so on...

Mathematical Analysis

For those of you who are mathematically inclined, the analysis that follows provides the rationale behind the construction of the stop number.

In photography, the lens aperture is that opening in the lens (or on the camera body) that determines the amount of light that is to be admitted to the light sensitive medium (film or CCD ...). The surface area of this opening can be adjusted by the use of a diaphragm. The action of closing or opening the diaphragm is called stopping down the lens (whether full or half). We define the f-stop number as

$\bg_white \120dpi \large S=\frac{f}{D}=\frac{\text{Focal Length}}{\text{Aperture Diameter}}$

In science & engineering, S is referred to as a dimensionless number, meaning that it does not have any units associated with it. In our case, since the focal length and the diameter both describe a length (m, cm, mm...), their ratio is dimensionless because the units can be simplified (just like simplification of numbers).

The advantage of using a dimensionless quantity is that any results drawn form an experiment on a specific device (lens in this case) will equally apply to any other device with the same dimensionless number. For example, you've probably heard of the Mach number "M". The Mach number is a dimensionless quantity used in aerodynamics and describes how fast an object is moving in a medium (air) compared to the speed of sound in that medium. It is probably the most popular dimensionless number on the planet! (I think the f-stop number should be added to the list of dimensionless number). Now any experiment carried out on a model jet with M = 3 for example, will illustrate exactly what happens when the real jet is flying at M = 3 in the atmosphere (shockwave structure, pressure and temperature distributions ...).

Here's an example in photography. The amount of light REACHES THE SENSOR using a 28 mm lens with S = 2 is exactly the same as that of an 80 mm lens with S = 2, although the diameters of both apertures are different (if all factors that affect image brightness are held constant). (If you can't prove this for yourself, let me know and I'll write up my proof).

This dimensionless number is a very useful tool for determining properties of lenses (and therefore the light coming through) without referring to diameters or any other data.

In cameras, when we set the f/stop number, we are essentially setting the Diameter of the lens aperture. This can be computed by knowledge of the focal length. Therefore, for a given stop S, the diameter opening of your lens is

$\bg_white \120dpi \bg_white \120dpi \large {\rm D} = \frac{f}{S}=\frac{\text{Focal Length}}{\text{Stop Number}}$

For example, a lens set at a focal length of 70mm and a stop number of 5.6 has an aperture diameter of

$\bg_white \120dpi \large {\rm D} = \frac{70}{5.6} = 12.5 \;{\rm mm}$

Note that S is inversely proportional to D which explains why as the stop number increases, less light enters the camera since D decreases (f/32 lets in less light than f/16).

Let us now compute the stop number (S) required to let in Twice as much light, for a given lens set at a fixed focal length.

We start by computing the diameter that will let in twice the amount of light. This is equivalent to saying that the aperture surface area has to be twice as much to let in twice as much light. For example, at the same focal length, an aperture with a surface area of 10 mm^2 will let in twice as much light as an aperture with a surface area of 5 mm^2. This is how it looks mathematically

We now set

$\bg_white \120dpi \large {\rm A}_0 = 2 {\rm A}_1$

$\bg_white \120dpi \large \pi \frac{{\rm D}_0^2}{4} = 2 \pi \frac{{\rm D}_1^2}{4}$

finally

$\bg_white \120dpi \large {\rm D}_0 = \sqrt{2}\,{\rm D}_1 \approx {\rm D}_1 +0.414\; {\rm D}_1$

In other words, for an aperture to let in twice as much light, its diameter must increase by approximately 41%.

Now that we have a relation between the diameters of both apertures, we can use the f/stop equation to deduce the recurence relation between the stop numbers. This is done as follows:

$\bg_white \120dpi \large \left. \begin{matrix} {\rm D}_0 = \sqrt{2}\;{\rm D}_1\\[16pt] {\rm D}_0 = \dfrac{f}{S_0} \\[16pt] {\rm D}_1 = \dfrac{f}{S_1} \end{matrix}\right\} \frac{f}{S_0} = \sqrt{2}\frac{f}{S_1}\Leftrightarrow S_1 = \sqrt{2}S_0$

Thus, to let in half as much light, we multiply the previous stop number by the square root of two ~ 1.4. Alternatively, to let in twice as much light, we multiply by the reciprocal of the square root of two ~ 0.7. In general, for a stop number Sn, we have

$\bg_white \120dpi \large \begin{cases} S_{n+1}=\sqrt{2}\;S_n \\ S_{n-1}=\frac{1}{\sqrt{2}}S_n \end{cases}$

where S{n+1} is the stop number that lets in HALF as much light as Sn, while S{n-1} is the stop number that lets in TWICE as much light as Sn. For example, if we are a stop number of Sn = 5.6, we have

$\bg_white \120dpi \large S_n = 5.6: \begin{cases} S_{n+1}=\sqrt{2}\;S_n \approx 7.92 \quad \text{(or 8 on cameras)} \\ S_{n-1}=\frac{1}{\sqrt{2}}S_n \approx 3.96 \quad \text{(or 4 on cameras)} \end{cases}$

In its present form, our recurrence formula depends on a fixed stop number. It would be useful if we can write our recurrence formula based on some initial reference stop number S0. For this, we do the following

$\bg_white \120dpi \large \begin{align*} &S_0 & & \text{A given initial stop number} \\ &S_1 = \sqrt{2} S_0 & & \text{1 stop from}S_0 \\ & S_1 = \sqrt{2} S_1 = \sqrt{2}\times\sqrt{2}S_0 & & \text{2 stops from} S_0\\ & S_3 = \sqrt{2} S_2 = \sqrt{2}\times\sqrt{2}\times\sqrt{2}S_0 & & \text{3 stops from} S_0 \\ & \vdots & & \vdots\\ & S_n = \sqrt{2} S_{n-1} = \left(\sqrt{2} \right )^n S_0 & & \text{n stops from} S_0 \end{align*}$

With this general recurrence formula, we can calculate the stop number at any given number of stops from a starting number S0. For example, if our lens is set at S0 = 5.6 and we want to calculate the stop number corresponding to 3 stops (i.e. letting in 3 times as much LESS light), we have

$\bg_white \120dpi \large \left.\begin{align*} & S_0 = 5.6 \\ & 3 \; \text{stops}\implies n = 3 \end{align*}\right\} S_3 = \left(\sqrt{2} \right )^3 S_0=15.83\approx 16$

This means that n is in fact the number of stops from S0. It is simply a counter of stops.

Smallest Stop Number

The question now arises as to what is the smallest stop number that a lens can achieve and how difficult is it to manufacture such lenses. I do not have any experience with lens manufacturing (although I have been contemplating learning that skill lately), but the mathematics could give us a hint. Looking at the equation for the aperture diameter

$\bg_white \120dpi \bg_white \120dpi \large {\rm D} = \frac{f}{S}=\frac{\text{Focal Length}}{\text{Stop Number}}$

one could argue that the longer the focal lens of a camera, the more difficult it is to achieve small stop number. Here is why. Let's say that you have a 50mm lens, for a stop number S=1, this means that the aperture diameter is equal to the focal length, i.e. 50 mm. Threfore, the actual diameter of your lens MUST be at least 50mm! If you add the barrel and the internal mechanisms, the actual diameter will be even larger than 50mm.

Look at the Nikon 50mm F1.4D lens for example. Its maximum aperture diameter is D = 50mm/1.4 = 35.7mm while the lens' actual diameter is a whopping 64.5mm!

Let's take another extreme. For a lens with a focal length of 500mm, a stop number of 1 means that the aperture diameter is 500mm! Imagine carrying a lens that's half a meter in diameter (of 1.65 ft!). These become impractical (unless you're dealing with a telescope). For example, the maximum aperture diameter for the Nikon 600mm f4.0 is D = 600/2 = 150mm. The lens has a diameter of 165mm and weight of about 5kg!

Bottom line is that the longer the focal length of a lens, the more difficult it is to manufacture it with wide apertures. That is also why fast lenses are very expensive! And that's why fast TELEPHOTO lenses are even more expensive! There are important design considerations to take into account in that case...

It is however convenient to choose S = 1 as the smallest stop number and start from there. In this case, our recurrence formula becomes

$\bg_white \120dpi \large S_n = \left(\sqrt{2} \right )^n = 2^{\frac{n}{2}}$

and here's how the famous f/stop numbers are generated:

$\bg_white \120dpi \large 1 \overset{\times \sqrt{2}}{\longrightarrow} 1.4 \overset{\times \sqrt{2}}{\longrightarrow} 2 \overset{\times \sqrt{2}}{\longrightarrow} 2.8 \overset{\times \sqrt{2}}{\longrightarrow} 4 \overset{\times \sqrt{2}}{\longrightarrow} 5.6 \cdots$

Remember, each full stop lets in twice as much or half as less light. The above equation is for reducing the aperture size, i.e. letting in less light. (the converse recurrence relation can be easily derived).

Intermediate Stops

Now what about half stops, one-third stops or one-fourth stops? How are these numbers constructed? If a full stop lets in half as much light, does a half stop let in 75% light? Let's look at that.

Keeping in mind that the f-stop sequence is a geometric sequence (multiplicative), any value sought within an interval has to obey the rules of a geometric sequence. Let us insert a "partial stop" in the middle of a full stop interval

$\bg_white \120dpi \large \begin{matrix} S_{n-1} \xrightarrow[\text{1 stop}]{\times \sqrt{2}} S_n \\[12pt] S_n = \sqrt{2}S_{n-1} \\[12pt] S_{n-1} \xrightarrow[\text{half stop}]{\times P}\phi \xrightarrow[\text{half stop}]{\times P} S_n \end{matrix}$

$\bg_white \120dpi \large \left. \begin{matrix} \phi = S_{n-1}\times P \\[12pt] S_{n}=\phi \times P \end{matrix} \right \} S_n = P^2 S_{n-1}$

but, we know that

$\bg_white \120dpi \large S_n = \sqrt{2}\,S_{n-1}$

form which we can computer P, the common ratio for the half-stop sequence

$\bg_white \120dpi \large P^2 = \sqrt{2};\quad \text{or} \quad P = \sqrt{\sqrt{2}}=2^{\frac{1}{4}}$

The same principles applies for deriving one-third, one-fourth,... one-mth stops (divide the interval into m sub intervals). In general, for a one-mth stop increment, we will have

$\bg_white \120dpi \large P^m = \sqrt{2};\quad \text{or} \quad P = 2^{\frac{1}{2m}}$

For example, for a 1/3 stop increment from S0=1, we have

$\bg_white \120dpi \large \left. \begin{align*} &S_0 = 1 \\ &\frac{1}{3} \quad \text{stop }\implies m = 3 \end{align*} \right\} S_{\frac{1}{3}} = 2^{\frac{1}{6}}=1.122\approx1.1$

There is an even easier way to derive non-integer stop formulas. Using

$\bg_white \120dpi \large S_n = \left(\sqrt{2} \right )^n = 2^{\frac{n}{2}}$

we notice that there is no restriction on "n" being an integer. As discussed previously, n is simply a stop counter. So, for example, if you want 1/3 stops, you simply substitute n = 1/3. If you want 2.4 stops, use n = 2.4 etc...

Number of Stops Between Two Stop Numbers

One can derive an equation for the number of stops between two stop numbers using the formula

$\bg_white \120dpi \large \bg_white \120dpi \large S_n = \left(\sqrt{2} \right )^n S_0$

By taking the natural logarithm of both sides of the equation, we get

$\bg_white \120dpi \large n = \frac{\log\left(\frac{S_n}{S_0} \right )}{\log\left(\sqrt{2} \right )} = \log_{\sqrt{2}}\left(\frac{S_n}{S_0} \right )$

For example, to calculate the number of stops between f/22 and f/1.4, we set

$\bg_white \120dpi \large n = \log_{\sqrt{2}}\left(\frac{22}{1.4} \right )=7.98 \approx 8 \; \text{stops}$

so that f/1.4 lets is 8 stops away from f/22 and lets in ~ 256 times more light than f/22! (at the same focal length. At other focal lengthes, the amount of light that reaches the sensor is ~ 256 times more).

Voila!

Cite as:
Saad, T. "The Mathematics of f/stop Aperture Numbers". Weblog entry from Please Make A Note. https://pleasemakeanote.blogspot.com/2010/10/mathematics-of-fstop-aperture-numbers.html?m=0

15 comments:

AnonymousDecember 14, 2010 at 12:00 AM
Thanks, I think I will do better on my photography finals because of this information!
PhilFebruary 4, 2012 at 3:49 PM
Thanks for writing this piece!

Can you edit this sentence, though?

>>
"f/2.8 lets in twice more light than f/1.4."
>>

Whereas, f/2.8 actually lets in 1/4 the amount of light that f/1.4 does.
PhilFebruary 4, 2012 at 3:52 PM
Thanks for writing this piece!

Can you edit this sentence, though?

>>
"f/2.8 lets in twice more light than f/1.4..."
>>

Whereas, f/2.8 actually lets in 1/4 the amount of light as does f/1.4.
yNotFebruary 4, 2012 at 6:44 PM
Yes thank you for catching that!
Hailey JonesSeptember 8, 2012 at 10:40 PM
Thanks for posting this, so fascinating!

Can you clarify something for me though? You mention in the beginning that:

"the next full stop is 1.0xSqrt(2) = 1.414"

How did you get 1.414 out of the square root of 1? I thought it would always come out to be 1.

Thanks so much,
-- Hailey Rose

Hailey JonesSeptember 8, 2012 at 10:43 PM
Nevermind! I just found my error, you're multiplying 1 by the square root of 2, which is 1.414.

Thanks though,
-- Hailey Rose
glJanuary 16, 2013 at 3:28 AM
Thanks - but that sentence near the end is wrong: "so that f/1.4 lets in 8 *times* more light as f/22 (at the same focal length"

It should read "8 stops" (which is 256 times more light).
UnknownAugust 17, 2013 at 10:29 PM
Very nicely conceived article and useful. Thank you yNot.

A couple of suggestions:

Scrolling down about 45%, I see that your formula line for "two stops from S zero" should lead with a two, not a one.

In the paragraph just below that, your reference to three stops (5.6 to 16) being three times as much less light should read 1/8 as much light. Two to the third power is eight, and its inverse is one eighth.

Thanks again. Bookmarked this page for reference.

Henry in Atlanta

Kurt DCOctober 3, 2013 at 8:41 PM
Very Insightful article. First time I've seen this explained with much clarity and made easier to understand. Thank you!
UnknownJanuary 2, 2014 at 3:31 PM
Thanks a lot, had a great time reading.
Just a minor error when writing "D = 600/2 = 150mm" :)
Callie DanielsMarch 18, 2015 at 9:45 AM
Thanks for the information. Can you recommend a textbook that I may refer to to gain similar information regarding the mathematics of photography?
yNotMarch 18, 2015 at 5:01 PM
I'm afraid that I do not know of a textbook that covers these things.
UnknownSeptember 20, 2015 at 4:46 PM
The links to the graphics is broken. Other than that, an interesting article.
remOctober 16, 2015 at 10:41 AM
mathematical odyssey of a photogrpher

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Monday, October 4, 2010

The Mathematics of f/stop Aperture Numbers