A = PI*R*R, where A is the area of exit pupil, R is the radius, taking first order of derivative gives

dA = 2*PI*R*dR, and this gives

dA/A = (2*PI*R*dR / PI*R*R), cancelling out the PI and one R gives

dA/A = 2*dR/R ==> dR/R = 0.5*dA/A (1)

Note, the term dA/A is THE stop difference, ie, when people say one lens is dA/A stops faster, and dR/R is the relative change in radius (diameter).

**This relevation shows how elegant and simple to estimate dR/R for a given (small) stop difference without using calculator.**

The formual (1) is true for infinitely small dA, however, it is still close enough to estimate dR/R, take the following:

1/10 stop --> 0.05 or 5%

1/5 stop --> 0.1 or 10%

1/4 stop --> 0.125 or 12.5%

1/3 stop --> 0.167 or 16.7%

Numbers corresponding to 1/4 and 1/3 is probably high, maybe 2nd order derivative should be factored in. I have left school so long, so I tend to use numerical analysis to analyze anything higher than 2nd order. As matter of fact, it is not difficult to write a Python program to produce ACTUAL result. But that loses the elegance of estimating dR/R quickly in mind.

**The underlying assumptions are: everything else about the two lenses are identical**Or maybe the same lens under different stops.