When light moves from a medium with a high refraction index to a medium with a low refractory index (water to vacuum), that there exists a "critical angle" at which no more refraction occurs. What is happening conceptually (chemically, physically, thermodynamically, electromagnetically, etc) at the critical angle that is stopping the light from refracting (transmitting) through the medium?

Guesses: Does the light not have enough energy at that particular angle to cross the barrier? Does it reflect too quickly before it can be transmitted? If so, why is that)?

Why is there no critical angle when light moves from low index to high index (other than the Snells law explanation)?

Guesses: Light is moving faster (more energy) so it can easily pierce the barrier regardless of angle.

  • $\begingroup$ Don't guess. Do research. It's really not that hard to find info on Snell's law. $\endgroup$ Nov 25, 2015 at 12:46

2 Answers 2


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.

  • $\begingroup$ This is a (good!) answer of the "it has to be this way because otherwise some condition is violated" variety. It would be really nice to have a cause-and-effect explanation as well. $\endgroup$
    – DanielSank
    Nov 24, 2015 at 23:44

The ideal perfectly smooth flat surface has translational invariance symmetry. That means that there is no mechanism for the scattering of a wave in the horizontal direction, and no mechanism for the change in the component of wave vector parallel to the boundary. That is, the horizontal component of wave vector is conserved.

For light incident at angles smaller than critical angles, a direction (in the low-index medium) can be found for which $$\frac{2\pi}{k} = \frac{c}{f}$$ and $k_{||}$ is the same on both sides of the interface.

For light incident at angles greater than the critical angle, the horizontal component of the wave vector is larger than any wave vector that could exist in free space: $$\frac{2\pi}{k_{||}} > \frac{c}{f}$$

so there's no possible direction for the light in the low-index medium. There can be no wave there.

If you're looking for a more "mechanistic" explanation, I suspect that the Ewald-Oseen extinction theorem should be in play. According to that theorem, the incident light would continue unchanged across the boundary, but the oscillating dipoles in the high-index medium would generate a wave identical except for being 180 degrees out of phase, thus perfectly canceling the other wave. Result: no wave at all in the low-index medium.

Another point of view: each oscillating dipole in the high-index medium radiates in all directions, and interferes with the radiation from every other dipole. For incident angles greater than the critical angle, the net result of all that interference is complete destruction of the wave in the low-index medium. The Ewald-Oseen thm is an interpretation of the mathematics behind all that interference.


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