Above, Michael Belsley says it best and Thomas Walther also gives a good answer -- the key is the imaginary component to the propagation vector in the lower index medium. This means that beyond a short distance into the lower index medium, there is negligible E-field. The Poynting vector into the lower index material is zero. Why does this occur. Belsley and Walther explain: boundary conditions at the interface demand continuity of the tangential components of the E (and H) fields. But what does this look like? Well, in the lower index material, there is a layer within a (small) skin depth where there is an exponentially decaying (in direction perpendicular to interface) electric field due to the incident wave from the higher index side. If you could put an E-field meter into this layer, you would sense power. (Aside: That is how TIR fingerprint readers work, your fingerprint ridges enable some of this evanescent power from this thin region on the low index side to tunnel out, going into your finger, while the valleys of your fingerprint are too far from the thin region of E-field to couple any power -- the difference between fingerprint ridges and valleys shows up as measurable contrast. Practical stuff that you can try at home by looking down into a clear glass of water at your fingers holding the glass below the water line.) So picture a region on the low index side as an exponentially decaying standing wave. It's real but it doesn't propagate, hence Belsley's explanation using an imaginary propagation constant. And therefore, no energy enters the low index material so that ALL the energy bounces off the interface back into the high index material. Just conservation of energy. Hence the 'total' in TIR. That the angle of incidence equals the angle of reflection is also a consequence of boundary conditions but this is no different from the general case, say, before the critical angle is reached. The coolest thing, in my opinion, about TIR is that evanescent field. As an exercise that you can work out on paper, imagine a high index material, say glass, with a very thin air gap separating it from another slab of glass. And imagine light from within the glass incident at the glass/air interface at an angle exceeding the critical angle. If the air gap is much greater than the decay depth in the air, the evanescent field in the air never sees the other slab of glass. But as you narrow the air gap to the point at which the second slab of glass enters the evanescent field, some of the incident radiation actual tunnels into the second slab of glass, escapes into the second slab and destroys the TIR condition. Measuring this behavior can tell you all about the evanescent field. So, after a lot of words -- Belsley and Walther got it right. I hope my enthusiasm for the topic added some insight.

With TIR which occurs for angles of incidence exceeding the critical angle, the reflected wavefront has the same velocity because it stays within the medium. For the same reason that for specular or Fresnel reflection whre the law of reflection holds, it is thus true for TIR also except that the reflectance is close to 100%.

Total internal reflection only occurs when the light from the optically denser medium hits an interface. According to Snell's Law you can work out the angles, but the sine of an angle can never be larger than unity. If this 'critical angle' is reached, the refracted ray cannot emerge on the side of the optically thinner medium.

@Thomas Walther

Yes, of course, that is the explanation you see everywhere. "The sine cannot be larger than unity (1), so it cannot leave the medium (2), and because of this it reflects back at an identical angle (3)".

For me, statement 1 does explain statement 2, not not statement 3. What I'm missing is a reason why it reflects back in the identical angle. Why does it not reflect in 90°, or scatter isotropically, or just vanish into phonons?

The rule reflection angle = incidence angle holds for particles as well as light and really is due to momentum and energy conversation. For light, the in-plane momentum component is given by p_perp=h/(2*pi)*k_perp, and so the in-plane component of k-vector (due to momentum conversation) and its total length (due to energy conservation) must be retained - this directly gives the above.

it depends on

Snell's law basically enforces that the propagation wave-vectors parallel to an interface are continuous. As many of the above answers state, when the incident angle is above the critical angle Snell's law predicts that the refracted ray will propagate such that the sine of the angle it makes to the normal is greater than one. This can be analytically satisfied by letting the angle become complex. Using the continuity of the wave-vector parallel to the interface this implies that the wave-vector of the wave attempting to enter the more dense medium becomes complex, leading to a decaying evanescent wave. It does not become the reflected wave in the less dense medium as your question seems to imply. The reflected wave is just the normal reflected component, whose amplitude becomes equal to that of the incident wave for angles of incidence above the critical angle.

Above, Michael Belsley says it best and Thomas Walther also gives a good answer -- the key is the imaginary component to the propagation vector in the lower index medium. This means that beyond a short distance into the lower index medium, there is negligible E-field. The Poynting vector into the lower index material is zero. Why does this occur. Belsley and Walther explain: boundary conditions at the interface demand continuity of the tangential components of the E (and H) fields. But what does this look like? Well, in the lower index material, there is a layer within a (small) skin depth where there is an exponentially decaying (in direction perpendicular to interface) electric field due to the incident wave from the higher index side. If you could put an E-field meter into this layer, you would sense power. (Aside: That is how TIR fingerprint readers work, your fingerprint ridges enable some of this evanescent power from this thin region on the low index side to tunnel out, going into your finger, while the valleys of your fingerprint are too far from the thin region of E-field to couple any power -- the difference between fingerprint ridges and valleys shows up as measurable contrast. Practical stuff that you can try at home by looking down into a clear glass of water at your fingers holding the glass below the water line.) So picture a region on the low index side as an exponentially decaying standing wave. It's real but it doesn't propagate, hence Belsley's explanation using an imaginary propagation constant. And therefore, no energy enters the low index material so that ALL the energy bounces off the interface back into the high index material. Just conservation of energy. Hence the 'total' in TIR. That the angle of incidence equals the angle of reflection is also a consequence of boundary conditions but this is no different from the general case, say, before the critical angle is reached. The coolest thing, in my opinion, about TIR is that evanescent field. As an exercise that you can work out on paper, imagine a high index material, say glass, with a very thin air gap separating it from another slab of glass. And imagine light from within the glass incident at the glass/air interface at an angle exceeding the critical angle. If the air gap is much greater than the decay depth in the air, the evanescent field in the air never sees the other slab of glass. But as you narrow the air gap to the point at which the second slab of glass enters the evanescent field, some of the incident radiation actual tunnels into the second slab of glass, escapes into the second slab and destroys the TIR condition. Measuring this behavior can tell you all about the evanescent field. So, after a lot of words -- Belsley and Walther got it right. I hope my enthusiasm for the topic added some insight.

I would like to thank everyone for answering this question in depth, as I was unsatisfied how many textbooks just handle this fascinating phenomenon with a plain "because it has to be that way".

I knew how an evanescent field appears "on the other side", which is why the cladding of a waveguide must be transparent, although it doesn't actually carry light, and the basis for tunable beam splitters built from two prisms with a sub-wavelength gap, but the basics of the phenomenon were not clear to me. Now I have a much better intuitive understanding, and will look for some material which models it with a complex angle to support this understanding with solid math.

I am also not satisfied with the answers. Of course we all know that the result can be found through Maxwell's equations, which gives a result in complex manner. But it is very strange to obtain an electric field for a wave which is travelling in half circles to see if it finds where to transfer its energy, if not, goes back to a total reflection. Moreover, if it finds a less dense medium, and decides not to transfer energy. And I could not find an answer through a model of dipole reemission, as I can for refraction and reflection. At the same time, I had intense discutions with seismologist, who consider the Snell law being valid for seismic waves, but at the TIR the wave may travel paralel to the interface, and can exit in a symmetrical way before. See one of the explanations attached, just go to the final part.

The original question was: 'why is the total internal reflection angle equal to the angle of incidence'. Te answer is rather simple: field continuity conditions at the interface show that the angle of reflection is always equal to the angle of incidence (law of reflection), and the angle of refraction follows Snell's law. This can be understood as constructive interference in the forward and backward directions, of the fields emitted by the dipoles in the media, for TIR conditions, there is still the angle of reflection governed by the law of reflection, and a non-propagating condition for the refracted part. So it is easy to understand that the energy that would flow in the refracted field, is transferred to the reflected field. I agree with Porf. Lunazzi that is difficult to think of the evanescent field in the less optically dense medium in terms of a sum of the dipolar contributions

Dear Patrick Bollgruen,

the equality of incidence and reflection angles can be derived from Huygens' principle of elementary waves and composite waves.

You can assume a system of parallel wavefronts going perpendicular to the incident ray. Each wavefront, upon hitting the interface, creates spherical elementary waves around each point of incidence. These elementary waves combine (add/interfere) to produce another system of parallel wavefronts.

Based on a simple geometric construction, you can see that the reflected wavefronts (the envelope of the new circles) are associated with a ray whose angle is equal to the incidence angle.

By the way, you can use Huygens' principle to construct refraction, too.

Kind regards,

Sascha Grusche

I must add the solution to the problem of it happening, in a simple way. If Snell's law indicates sinTETA = 1, that means that light must exit across the interface, as the sismologists indicates, but we can apply the principle of reciprocity: In geometrical optics, if you have one path for a light ray, the inverse path is also valid (like seeing the movie backwards in time). So that this indicates that light taking its way along the surface must go back in perfect reflection. It is the coincidence of the refracted and the reflected light. You may think in dipoles or Huyguens secondary waves. The fact is that we are considering our ray as being the direction of an infinite plane wave. If you have a small diameter beam, then your construction is not the same, you have lateral side effects due to no light outside the beam. In that case, there is a very interesting displacement of the reflected beam, the Goos-Hanchen shift. Although a few wavelenghts in value, it was measured in 1947 (I do not guess which intuiton carried to think on that experiment). There is also another effect, the Imbert-Fedorov, but it is enough for today, Is'nt?

you can read a possible answer at https://arxiv.org/pdf/1403.4468