f-stop system supprise (to me) New to math site so forgive if worded poorly:
As you know, cameras and other optical applications frequently use the f-stop system. The focal ratio (f/#) set in common usage is:
1 – 1.4 – 2 – 2.8 – 4 – 5.6 – 8 – 11 – 16 -22 -32 -45 -64
The f-number set is grounded on the geometry of circles. The f/# is obtained by dividing the focal length by the working diameter of the aperture. The resulting ratio is universally used to compare one lens to another with regards to its light transmission potential. The above set has a delta of 2X. Each increment going right decreases light transmission 50%. Each increment going left increases light transmission 100%. The basis is; Multiply or divide the diameter of any circle by the square root of 2 computes a revised circle diameter with an area that is twice or half, in other words such adjustment of the diameter of the aperture stop of a camera, doubles or halves the exposing light energy. For this set, each number going right is its neighbor on the left multiplied by 1.414. Each number going left is its neighbor on the right divided by 1.414.
Now modern cameras often need to adjust using a finer increment. Industry practice is to use a number set in 1/2 f-stop increments. The number set is:    1 – 1.2 – 1.4 – 1.7 – 2 – 2.4 – 3.5 –  4 – 4.5 – 5.6 – 6.7 – 8 etc.
The multiplier/divisor is the 4 root of 2 = 1.1892
Often camera lenses require 1/3 f-stop changes. The number set is:
1 – 1.1 – 1.3 – 1.4 – 1.6 – 1.8 – 2 – 2.2 – 2.5 – 2.8 – 3.2 -3.6 – 4 etc.
The multiplying/dividing factor is the 6th root of 2 = 1.1225
One would expect that if you constructed a series of circles the delta for the 1/2 f-stop set  would be a 50% area change. It is not, it is 41%. Why?
One would expect that if you constructed a series of circles the delta for the 1/3 f-stop set would be a 33% area change. It is not, it is 26%. Why?
I computed revised circle areas using the multiplying/dividing factors, then a percent change. I am surprised by fact that what I thought was 1/2 f-stop increments only yield a delta of 41% and the 1/3 f-stop set yields a delta of only 26%. Why? Why? How am I going astray? If I am correct, then maybe the industry thinks these sets are more elegant?  Or am I overflowing with gobbledygook?  
 A: You want the area to double when you change $1$ full f-stop.  If you want to make two steps with the same fractional increase, they should each multiply the area by $\sqrt 2$.  In that case, each one is $\sqrt 2-1 \approx 41\%$ larger than the previous one.  Similarly if you want to make three steps, each one should multiply the area by $\sqrt[3]2 \approx 1.26$ so each should be $26\%$ larger than the previous one.
A: Here is an example from financial mathematics.
Say you have 100 dollars, and you deposit it into a bank account that earns interest compounded annually at a rate of $i$.  In other words, whatever the amount $P$ in the account at the beginning of the year, the bank adds $Pi$ at the end of the year to your balance, for a total $P(1+i)$ in your account at the end of the year.  So if you deposited $100$ on January $1^{\rm st}$, by December $31^{\rm th}$ of the same year you would have $100(1+i)$.
What would the value of $i$ need to be in order for you to have doubled your money--i.e., to have $200$--at the end of the second year?  Of course, we would observe that at the end of the first year, you would have $100(1+i)$, which then gets compounded at the end of the second year to yield a total balance of $100(1+i)(1+i) = 100(1+i)^2$.  Then if we set this to equal $200$, we get $(1+i)^2 = 2$, or $i = \sqrt{2} - 1 \approx 0.41$.  Note this is not $i = 0.50$, because in a sense, the interest you earned on the initial deposit also has a chance to ean interest in the second year.  Had we had $i = 0.5$, you would have at the end of two years $100(1.5)^2 = 225$, not $200$.
Similarly, if we were to ask what the interest rate would need to be in order to double your money at the end of three years, you would solve $(1+i)^3 = 2$, or $i = 2^{1/3} - 1 \approx 0.26$, not $1/3$ as you might think.
So, if we go back to $f$-numbers, if you were to make a half-stop scale so that each number is $50\%$ larger than the previous, the scale would look like this:  $$1, 1.5, 2.25, 3.375, 5.0625, \ldots.$$  This of course doesn't work.
