Critical Angle Formula

Question

The formula is given by $n=\frac{1}{\sin(C)}$, $n$ is the refractive index of the denser medium, C is the critical angle. From this formula, it seems to be that we are substituting the angle of refraction as the angle of incidence, therefore $\frac{\sin(90)}{\sin(C)}$, but why can we do this? Why not $\frac{\sin(C)}{\sin(90)}$. Or this is just how it is when we derive the formula using Snell's Law...

Answer 1

From Snell's law, we have $n_1\sin(\theta_i) = n_2 \sin(\theta_r)$, where $n_1$ is the index of refraction on the incident side, $\theta_i$ is the incident angle, $n_2$ is the index of refraction on the refracted side and $\theta_r$ is the angle of refraction. Here we are assuming that the incident side is the denser medium side, and the refracted side is the less dense medium. That generally means that $n_1$ is bigger than $n_2$, and in fact you seem to assume that the refracted side is air, so we can take $ n_2 = 1$, a reasonable approximation. In this case, the refracted angle is greater than the incident angle. That then gives us $n_1\sin(\theta_i) = \sin(\theta_r)$. The largest that the sine on the RHS can be is 1 (which is obtained when the angle of refraction is $90^\circ$). That's the criticality condition. That then gives us $$ n_1 = \frac{1}{\sin(\theta_C)} $$ That makes sense, since $n_1 > 1$ because of the dense medium, which allows us to find a real angle that solves this equation. If the less dense medium side was the incident side, there would be no way to make the refracted angle $90^\circ$, because it would always be less than the incident angle.

Answer 2

Yes, $n=1/ \sin C$ is derived from Snell's Law. We can't write $n=\sin C/ \sin90^\circ$ because $n>1$ is refractive index of denser medium (in which light ray is incident) w.r.t. rarer medium (in which light ray refracted). Mathematically also, $C=\sin^{-1}(n)$ becomes undefined therefore $n\ne \sin C$

Derivation: Let $\angle i$ be the angle of incidence in denser medium with refractive index $n_1$ & $\angle r$ be the angle of refraction in rarer medium with refractive index $n_2$. Using Snall's Law $$n_1\sin\angle i=n_2\sin \angle r$$

But, for internal reflection $\angle i=C $ & $\angle r=90^\circ $, $$\therefore n_1\sin C=n_2\sin 90^\circ$$
$$\frac{n_1}{n_2}=\frac{\sin 90^\circ}{\sin C}=1/ \sin C$$ $\frac{n_1}{n_2}=n$ is the refractive index of denser medium w.r.t. rarer medium then $$n=1/\sin C$$