if $\frac{\sin(I)}{n}=\sin(r)$ where $n$ is the refractive index, $I$ is angle of incidence and $r$ is angle of refraction (these are relative to the normal).
How come blue light refracts at a higher angle than red light if red travels faster in glass than blue, surely $n$ for blue is larger than $n$ for red making $\sin(r)$ for red larger than that for blue, and the larger $\sin(r)$, the greater the angle (to a point)?
You have it backward, faster speed of light in a material corresponds to less refraction, not more. In the limit that the speed of light in a material was the same as the speed of light in vacuum, there would be no refraction at all.
It can be shown that in a material the index of refraction is the speed of light in vacuum, $c$ divided by the speed of light in the material $c_m$.
$ n ={ c \over c_m} $
So, slower speed in a material corresponds to a larger index of refraction ,and higher speed to a lower index of refraction. The index of refraction is always greater than or equal to 1, because $c$, the speed of light in vacuum, is always greater than the speed in a material.
So, as you have stated, red light has a lower index of refraction than blue light since it also has a shorter wavelength, so lower index of refraction corresponds to higher speed in a material.
Now Snell's Law is stated
$${n_1 \over n_2} = {\sin \theta_2\over \sin \theta_1}$$
where the geometry is as shown:
So if $n_1=1.0$ and $\theta_1 = 20$ degrees we have
$n_2 =1.51$ for red light and $\theta_2 = 13.09$ degrees,
$n_2 = 1.53$ for blue light, and $\theta_2 = 12.91$ degrees.
As you can see from the diagram, a smaller value of $\theta_2$ corresponds to a larger amount of refraction because it is a larger change in angle and thus a larger change in direction of the incident light.
If the question is about why red and blue light refract at different angles, this is because the refractive index, $n$ for many materials depend upon the frequency of light and they are referred to as dispersive. More often, the variation is given in terms of the wavelength ($\lambda$) and in the optics industry people quote the refractive index for the Helium d line ($\lambda = 587.6 nm$), and for the Mercury e line ($\lambda = 546.1 nm$), and a few other properties at other wavelengths, see Kaye and Laby.
The reason why the refractive index depends on frequency is related to the atomic structure of matter.
I think I understood you. You are correct in all that you said, but very likely you are drawing the rays in the incorrect way. Yes, the red angle of refraction will be bigger if you apply Snell's law. Since refractive index for red is smaller it should bend LESS, but the angle you get for the red is BIGGER. And you are stuck here because this LESS and BIGGER seems to contradict each other. But they don't. It's all fine.
I guess you must be drawing in one of those 2 ways (which are both incorrect):
Applying Snell's Law on the above images makes no sense:
$$n_1 \cdot sin(\gamma) = n_2 \cdot sin(\theta) $$
since $\large \;n_{2,red} < n_{2,blue}\; \quad$ and $\quad n_1=1 \; (air)\;\;$ ,
$$red \Rightarrow\quad \uparrow sin(\theta) = \frac{sin(\gamma)}{n_2 \downarrow} $$ $$blue \Rightarrow\quad \downarrow sin(\theta) = \frac{sin(\gamma)}{n_2 \uparrow} $$
$$\therefore \quad \theta_{red} > \theta_{blue}$$
But that is not what we see on the above images and that is what might be confusing you.
Image on the right: it's incorrect because the Normal inside the prism is not perpendicular to the surface, but only a continuation of the incoming outside ray.
Image on the left: it's incorrect because the deflected rays inside the prism deflect on the other side of the Normal.
The correct drawing you should be looking along with Snell's Law is presented bellow:
You can see that now it makes sense, as the blue angle is smaller and at the same time it is deflected farther away from it's incoming direction.
