opticsrefraction
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)?
Snell's law still applies to the curved surface, but you have to measure the angles of incidence and refraction relative to the surface where the light hits.
The image is my attempt to show parallel rays of light falling on a curved surface. Even though the rays are parallel, the angle of incidence is different for the two rays because it has to be measured relative to the normal at the point the light strikes the surface. Hence the angle $i$ is not the same as the angle $i'$.
Response to comment:
It has become clear from the comments that the problem is that the value of $n$ depends on whether the light is passing from the air to glass or from glass to air. To be precise the two refractive indices are reciprocals of each other i.e.
$$ n_{air-glass} = \frac{1}{n_{glass-air}} $$
The refraction of the light ray happens because the speed of light, and therefore the wavelength, changes when the light enters and leaves the glass. The refractive index when a light ray passes from a medium 1 to a medium 2 is:
$$ n_{1-2} = \frac{v_1}{v_2} $$
where $v_1$ is the speed of light in medium 1 and $v_2$ is the speed of light in medium 2. So in our example the refractive index when passing from medium 2 to medium 1 is:
$$ n_{2-1} = \frac{v_2}{v_1} $$
i.e.
$$ n_{1-2} = \frac{1}{n_{2-1}} $$
It is important to note that the equation you mention gives the index of refraction of one medium with respect to another. If light travels from one medium, with refractive index $n_i$ and incident angle $i$, to another medium, with refractive index $n_r$ and refraction angle $r$, then the relationship is described by Snell's Law as such: $$ n_i\sin(i)=n_r\sin(r)$$ which can be rewritten like this: $$ {\sin(i) \over \sin(r)} = {n_r \over n_i}$$ If we say that $n_i < n_r$, meaning the light propagates from a rarer to a denser medium, then the above equation gives the index of refraction of the denser medium in relation to the index of refraction of the rarer medium.
If instead we wanted to consider the case where light travels from a denser to rarer medium, then the only change would be that now $n_i > n_r$, in which case the above equation would yield the index of refraction of the rarer medium in relation to the denser medium. Notice that, in this second case, if we still desire the index of the denser medium with respect to the index of the rarer medium, we must rearrange the equation like this: $$ {\sin(r) \over \sin(i)} = {n_i \over n_r}$$ But this is simply a consequence of which ratio we are looking for. For example, say we consider the propagation of light from air to some unknown denser material. In this case, $n_i \approx 1$ is the index of refraction of air, and $n_r = n_x$ is the index of refraction of the unknown material. We would then say that, from the first relationship defined, the index of refraction of the unknown material is: $$ n_x = {\sin(i) \over \sin(r)}$$ If instead we said this light traveled from the denser unknown medium to the air, then $n_i= n_x$ and $n_r \approx 1$, in which case we would find the index of the unknown material by using the second relationship: $$ n_x = {\sin(r) \over \sin(i)}$$
Best Answer
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.