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 Category: photography photography
 Published: 30 September 2014 30 September 2014

This is a short article explaining how to calculate the number of fstops between two apertures.


Calculating fstops
There is considerable confusion in internet forums about the topic of fstops, more precisely, of how many stops are there between two given lens apertures. For example, how much faster is a f/1.2 lens versus a f/1.4 lens? I decided to write this short article to hopefully clarify this topic and offer a simple formula that anyone with a scientific calculator can use.
The formula
I've written before in this site about the logic behind the fnumber sequence we see in lenses (1, 1.4, 2, 2.8, 4 ...). The key point is that every element in this sequence of numbers equals the preceding element times the square root of 2 (see the article for more details). To put this mathematically, if 'b' and 'a' are two apertures separated by 'n' stops then the relation in Figure 1 must hold. With that in mind, it's simply a matter of applying logarithms to boths sides and solving for 'n' to get the number of fstops as a function of 'a' and 'b':
Figure 1  Formula and Derivation
Notice that in Figure 1, I used natural logarithms (usually marked as 'ln' in most scientific calculators), but you can also use base 10 logarithms (usually 'log' in a calculator) and the formula will still work. In fact, provided you use the same type of logarithm in the numerator and denominator the formula is still valid. Let's now look at some examples.
Examples
Example 1: how much faster is a f/5.6 lens versus a f/2 lens?
An experienced photographer will tell you immediatelly this is a 3stop difference. But let's look at is using the formula above:
ln(5.6/2) = 1.0296
ln(sqrt(2)) = 0.3466
1.0296/0.3466 = 2.9705
Which is close enough to 3stops as we predicted. Why the small difference? Well the reality that many people don't realize is that the fstop numbers manufacturers assign are rounded to one decimal point. This is done for practical reasons of course (how useful would it be to have three or four digit precision anyway?). Figure 2 shows the fstop calculated with two digit precision and you can tell that, except for the powers of 2, the other numbers are actually rounded. If you repeat the calculations above with more precise fnumbers, you will get something closer to 3. As a practical matter tough, I usually don't bother doing this and just round the number Let's look at a few more examples.
Figure 2  More precise fstops
Example 2: how much faster is a f/1.2 lens versus a f/1.8 lens?
At first glance, 1.2 and 1.8 doesn't seem like much of a difference. However, once you do the calculation, you realize it is one stop difference! This is twice the light (and, theoretically, twice as fast the shutterspeed for the same ISO and lighting conditions).
ln(1.8/1.2) = 0.4055
ln(sqrt(2)) = 0.3466
0.4055/0.3466 = 1.16 => 1 stop
(Again, the error is due to rounding).
The fact of the matter is that, for smaller fstop numbers (faster lenses), the difference between consecutive stops is much smaller than it is for larger fstop numbers. This is just a characteristic of logarithmic functions (or exponential functions, depending of how you look at it). Figure 3 illustrates this last point. The scale is linear and there's a dot for each of the fnumbers separated by 1stop (1, 1.4, 2, 2.8, 4, 5.6, ...). Notice how the values 'clump' together towards the left end of the scale (nearer f/1) and grow further and further apart towards the right.
Figure 3  Fstop distribution
Hopefully this article sheds some light (no pun intended) on how fstops are calculated and why these values seem to "shrink" for faster lenses.
Comments, questions, suggestions? You can reach me at: contact (at sign) paulorenato (dot) com