Why is it brighter under a rainbow? Part 4

 Doing the maths on the two simultaneous equations produces a curve like this. The deviation angle cannot be bigger than 180 degrees but has a definite minimum. Notice that there is a range of other angles of incidence close to the minimum which will have almost exactly the same deviation - they will come out along the same path.

Here's an attempt to show it. There are 3 rays that come out close together but notice that the outliers A and B both end up ABOVE those 3 rays. That's a consequence of it being a minimum.

This grouping of those 3 rays together coming out along the same path is apparently called the RAINBOW RAY - a reinforcement. Now the rainbow ray is wavelength specific - it is different for each colour because each wavelength has a different refractive index and thus is bent by a different amount. 

The final bit that it took me some time to comprehend is the idea that raindrops fall. If my eye is lined up with the direction of the rainbow ray, I see a colour. The drop has to be in a particular place. If it is above that point, notice that no light comes out of the drop below the rainbow ray so I won't see it - it looks dark.

However, a drop below the rainbow ray will send light to my eye. It will be the rays I've labelled A and B. You get these rays whatever the wavelength, so there is a mixing and it appears white. That's why it looks brighter underneath a rainbow. I got there in the end thank to https://plus.maths.org/content/rainbows