Does the angle of refraction always have $n_2$ as its refractive index? As far as I know, $$\frac{\sin\theta_i}{\sin\theta_r} = \frac{n_2}{n_1}.$$ So I got this question here:
Calculate the critical angle for a flint glass-air ($n=1.58$) boundary.
Since the critical angle has an angle of refraction of 90° that is $\theta_r= 90^\circ$, but I am confused on what values should I substitute for $n_2$ and $n_1$. So does the refractive index of the angle of incidence always equal $n_1$?
In physics, where possible, don't jump into applying a formula until you have a pretty good picture of the physical effect you are studying. In this example, the picture is easy to form: it is light rays going from one medium to another. Draw some example diagrams, sticking to the rule that in each case the angle between the ray and the normal from the interface is larger in the medium with lower index. In this way you can answer your own question about which refractive index refers to which medium in the formula you quoted.
(I have deliberately not told you the answer because I think if you try the method I advocate, you will learn more. I hope you will find this helpful rather than frustrating.)
To avoid confusion, here is a different way of writing Snell’s law. $$ n_1\times\sin(i) = n_2\times\sin(r) $$ Here, $n_1$ is the refractive index of the medium in which the incident ray exists $n_2$ is the refractive index of the medium in which the refracted ray exists.
So for this question as the light goes from denser to rarer medium, $n_1$ would be the refractive index of flint and $n_2$ would be that of air.
