Rainbows are the result of a property of matter called dispersion. What dispersion means is that, when light passes through a material, the refractive index of the light will depend upon its wavelength (or frequency, if you prefer). If you recall Snell's law, you'll recognize that the angle that light is refracted through a material depends on the refractive index. Therefore, if the refractive index itself depends on wavelength, this will cause different colours to deflect at different angles and split apart. The splitting that we see gives rise to rainbows.
The dispersion relation for water is fairly complicated. I will be using the relation found here, where I've taken a temperature of 293.15K (20C), and density of $1000 kg/m^3$. Solving this for n, we get a relation that looks like this:
$n = \sqrt{\big(\frac{1+2C(\lambda)}{1-C(\lambda)}\big)}$
Where $C(\lambda)$ is the numerical relation listed in the reference. Graphically, the dispersion relation looks like this, which probably tells us all we really need to know
The refractive index doesn't change by much, but it changes enough for it to make a big difference, as we'll see.
Let's now calculate the refraction of light through a spherical droplet. You can see the setup that I'll be using in the figure.
I've labelled the incident angle $a$, the refracted angle $b$. The normals are all in blue, and I've shown the paths of a red and green ray to illustrate the differences this will likely make. Note that, by definition, the outgoing ray a will be the same input and output angles $a$ and $b$, respectively. We're interested in the angle $q$, which is the change between the entry ray and the exit ray. Note that we don't know any of the angles at this point. We don't actually need them at this point, because the angles are all related.
The angles $a$ and $b$ are related by Snell's Law. That is,
$\sin a = n(\lambda) \sin b$
I've set the refractive index of water to 1. I note here that $n$ is dependent on $\lambda$ as above. I won't include it beyond this point, however, for brevity's sake. Now, you'll note that all of the normal line converge in the middle, by definition of the way circles work. The angles $c$ form isoceles triangles with the angles b. In a total circle the angles add up to 360 degrees, hence.
$360 = 2a + 2c + q$
But $b$ and $c$ are related as well.
$360 = 2a + 2(180 - 2b) + q$
This simplifies to a relation between q, b, and a.
$q = 4b - 2a$
Let's take the derivative of Snell's law and the derivative of the equation above, with respect to $b$.
$\frac{dq}{db} = 4 - 2\frac{da}{db}$
$\cos a \frac{da}{db} = n\cos b$
$\cos a \frac{da}{db} = n\cos b$
Combining these two equations, we can form a third relationship between the angles.
$\frac{dq}{db} = 4 - 2n\frac{\cos b}{\cos a}$
Let's set the derivative $\frac{dq}{db} = 0$, which will maximize/minimize the angle $q$ in terms of $a$ and $b$.
$4\cos a = 2n\cos b$
Let's multiple Snell's law by 4, square it, and add it to the square of this equation.
$16 \cos^2 a + 16\sin^2 a = 16n^2\sin^2 b + 4n^2\cos^2 b$
Noting that $\sin^2 b = 1 - \cos^2 b$ and $\sin^2 a + \cos^2a = 1$, we can solve for $b$.
$cos^2 b = \frac{4}{3}\big(1-\frac{1}{n^2}\big)$
We can then go back and find $a$.
$cos^2 a = \frac{n^2 - 1}{3}$
Now we can go back and find $q$. Unfortunately, there doesn't seem to be a particularly elegant way to write this out; it's a bit of a mess. So rather than writing the whole thing out, I'll just plot it for you here, including the dispersion relation for n.
You can see that we observe a deflection of about 3 degrees across the visible light, and that at the 600 nm, we get the "rainbow angle" of 42 degrees. At a rough estimate, if the rainbow appears to be 1 km away, then the red and blue light would be separated by 26m, more than enough to create the effect that we call a rainbow. Note that the example I'm doing here is a first-order rainbow--we can also make rainbows be counting further reflections within the droplet, and we'll see rainbows showing up at various other points. The method of calculation is pretty much the same.
And that is how we get a rainbow.
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