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…
[Physics] Critical Angle Formula
refractionvisible-light
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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: $$ n_x = {\sin(r) \over \sin(i)}$$
Imagine the speed of light to be $1$ meter per second and the speed of light in the medium with a high refractive index to be $\frac{1}{2}$ meters per second.
If you have a single peak of a wave in the slower medium, that peak must move forwards at speed $\frac{1}{2}$, no matter what angle it's facing. In the faster medium, that peak must move forwards at speed $1$, no matter what angle it's facing.
The critical angle comes into play when you consider where the peak of the wave is on the boundary between two mediums. If $\theta$ is the angle between the wave direction and the surface normal and $v$ is the speed of the wave, this point travels at a speed $\csc(\theta) v$. This makes sense: if $\theta=\pi/2$, the point at the boundary is just the wave speed. If $\theta=0$, the wave passes instantly and so the question isn't really defined (because there is no point where the peak of the wave is on the boundary).
We demand $\csc(\theta_1) v_1=\csc(\theta_2) v_2$. That is, there should be a single point on the boundary where the peaks of both waves meet. The velocity of the point can be expressed in two ways, and both must be equal.
Unfortunately for the faster medium, if you have a full wave, the point on the boundary can never move slower than $v_2$, in this case, $1$ meter per second. But we're trying to send in a wave whose boundary point can move as slow as $\frac{1}{2}$ meter per second. There is absolutely no way any wave in the faster medium can satisfy that.
The result of this is an evanescent wave, where something is "transmitted" but decays exponentially (so that nothing is truly transmitted over long distances). You can't see that very well in optical light, but you can in microwaves. Take for example this Sixtysymbols video. Around three minutes in, two microwave prisms get pushed together. The reading starts to increase slowly before the two prisms are mushed up right next to each other because there is an evanescent wave "escaping" the prism but transmitting nothing over long distance. If the evanescent wave hits another prism, there is some actual transmission.
Best Answer
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.