# Critical Angle Formula

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...

• If your textbook doesn't state explicitly that this is the critical angle in vacuum, you might want to get a better book. It is an important point. May 20, 2020 at 14:59

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.
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 $$n=1/\sin C$$