# Real light increments using aperture scale in fractional stops

In the case of using f-number with full stops, the +1EV step means 2*light because the Area (A = πr^2) in the diaphragm is doubled (i.e. radius*√2).

However, I don't know why in fractional stops such as +1/2EV; +1/3EV; +2/3EV, the light is not increased 1/2*light, 1/3*light, 2/3*light, respectively. I mean, the radius is multiplied by √2^(1/2), √2^(1/3), √2^(2/3), which means an increment in the Area by 1.41, 1.26, 1.59 (resolving the Area equation), respectively.

Why are not taken the real Area/light increments of 1.5, 1.33, 1.66 using the radius √(3/2), √(4/3), √(5/3) as the real fractional stops of +1/2EV; +1/3EV; +2/3EV?

This comes down to understanding how logarithms work. Specifically in photography, the log base-2 of a doubling ratio equates to a difference of 1 stop.

We use logarithms to reduce repeated multiplication and division by growth factors to simpler linear addition or subtraction of stop values. If you are familiar with the decibel scale, this is the same process, except that in the dB scale, the logarithms are base-10 (and being SI values, decibels are tenths of a Bel—literally deci Bel).

So increasing by ⅓, ½, and ⅔ of a stop (i.e., adding EV) is the same as multiplying a light amount by the ratios 2 (≈ 1.26), 2½ (≈ 1.41), and 2 (≈ 1.59).

In aperture terms, as you note, doubling the light means doubling the area. And also you also note, since area is proportional to the square of the radius (Ar²), then aperture ƒ-numbers increase (are multiplied) by a factor of (√2), etc.

• @RobinHood - Exposure Value is a logarithm, they are not fractions of light received, they are fractions of stops. – Mattman944 Nov 21 '19 at 3:01
• @RobinHood Because +1/3 EV is an increase of 1/3 stop of light. The term stop specifically means "a doubling of a quantity" in photography. 1/3 of something means that 3 iterations of the 1/3 results in 1 (i.e., 1/3 + 1/3 + 1/3 = 1). So when it comes to iterated multiplication, what number when multiplied by itself 3 times results in +1 stop (that is, a total multiple of 2)? Well, that's 2^(1/3) because 2^(1/3) * 2^(1/3) * 2^(1/3) = 2^(1/3 + 1/3 + 1/3) = 2^1 = 2. When we say n stops or n EV (because the "units" of EV is stops), we are saying a ratio of 2^n. – scottbb Nov 21 '19 at 3:24
• @scottbb - A tip of the hat from Alan Marcus – Alan Marcus Nov 21 '19 at 5:04
• @RobinHood For instance, in your original question, you ask that for when we say +½ stop, why does that not correspond to increasing the area by a ratio of √(3/2). But if it were that way, then applying +½ stop twice should be a total of 1 stop increase, but that would mean the area increased by a ratio of 3/2. So expecting a doubling of exposure (1 stop) wouldn't be achieved because area only increased by 50%, rather than doubling as desired. – scottbb Nov 21 '19 at 22:09
• @RobinHood The U.S. system (Uniform System, from England, 1880) basically did that, with aperture values 1, 2, 4, 8, 16, representing actual multiples of exposure increase (see en.wikipedia.org/wiki/F-number#Aperture_numbering_systems ). But it was obsoleted more than 100 years ago by the f/stop system, which had major advantage of being the same exposure (f/8 is always f/8 exposure) on any lens of any size (including focal length, which otherwise would affect exposure), which was greatly more useful to photographers, and has served us very well. – WayneF Nov 21 '19 at 22:16
Why are not taken the real Area increments of 1.5, 1.33, 1.66 using the radius √(3/2), √(4/3), √(5/3) as the real fractional stops of +1/2EV; +1/3EV; +2/3EV?

Surely, you would like adding +1/2EV twice to produce +1EV which means 2*light.

Now, if +1/2EV was 1.5*light, then +1EV would be 2.25*light.

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%.

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%.

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 is this? Asked by me on Stack Exchange Mathematics