# Physical explanation for why total internal reflection occurs

I have been trying to understand total internal reflection (and have read several posts on this site already). Mathematically, I feel that I understand how the evanescent wave decays exponentially as it crosses a boundary surface and why there is no transmission. However, I have no intuition as to why it happens. Is there a deeper physical explanation for total internal reflection that explains what the atoms are doing?

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It's not about atoms, it's about the facts that atoms arranged in different ways (i.e. different materials) make light move at different speeds. Because of the different speeds there is a deflection (see Snell's law) and for certain angles the deflection is so strong that actually light is reflected in the starting medium. The thing can be made more detailed and formal, of course. – Bzazz May 4 '13 at 20:58
@Bzazz; I agree with you. Again, what are the atoms doing as light is internally reflected? – Carlos May 4 '13 at 21:57

I don't think you need a microscopic explanation for the phenomenon. All you need is the fact that the phase of the wave must be continuous across the water-air boundary.

If you take any two points at the interface, spaced a distance $w$ apart, then the length of the line segment, $l$, I've labelled d$\phi$ is just $w \sin i$, and the phase difference is just $2\pi l/\lambda$, so the phase difference between the light beams at the two points is is just:

$$d\phi = w \sin i \frac{2\pi}{\lambda} = w \sin i \frac{2\pi f n_i}{c}$$

Now we appeal to Huygen's principle and say that we can regard our two points as sources for the light on the other side of the interface, and because the phase is constant across the interface the phase difference for the emitted rays must be the same $d\phi$ that we calculated above:

$$d\phi_r = d\phi_i$$

or:

$$w \sin r \frac{2\pi f n_r}{c} = w \sin i \frac{2\pi f n_i}{c}$$

A quick rearrangement gives Snell's law, and because $\sin r$ can't be bigger than unity we conclude there can't be any transmission for $i$ greater than some critical angle i.e. we get total internal reflection.

So the physical origin is the continuity of the electromagnetic field that makes up the light and you don't need to think about what the atoms are doing. I suppose you could say that the electron density around the atoms at the interface will be oscillating in phase with the incoming wave and also in phase with the refracted wave (which is why the phase across the interface must be constant) and this wouldn't be possible if any light were transmitted above the critical angle.

Response to comment:

If you consider the atoms/molecules at the interface, these will have electrons in outer orbitals and/or molecular orbitals that are polarisable so the light will induce an oscillating dipole.

So you have a layer of oscillating dipoles at the interface, and these will themselves radiate an EM wave. The direction(s) the radiation is emitted depend on the phase changes as you move within this layer because the dipoles will interfere with each other, constructively in some directions and destructively in others. For example if all the dipoles are in phase the light will be emitted normally. If the phase changes linearly with distance as you move in the layer the light will be emitted at an angle. In the case of total internal reflection there is no direction in the low refractive index side where all the dipoles can interfere constructively, but there is such a direction in the high refractive index side. Therefore all radiation from the dipoles will be emitted into the high $n$ side i.e. you have total internal reflection.

But be cautious about taking this as anything more than an analogy as you can't treat the light and the dipoles separately. I'm not implying that the light is absorbed by the dielectric then re-emitted.

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I believe I follow your argument by applying boundary conditions to make the wave continuous across the water-air boundary. I can also derive Snell’s laws quickly using Maxwell’s equations (as shown in Griffiths Electrodynamics book (3rd ed), see page 388) and immediately get the same result. What interests me most is your last paragraph. Can you please expand on this more? – Carlos May 6 '13 at 11:21
@Carlos: I've expanded on my answer. I hope this makes things clearer and not even more confused! – John Rennie May 7 '13 at 6:18
@ John: I love your answer and it has helped me to somewhat visualize what’s happening at the boundary layer. I’d say you answered the question I asked. Thanks. – Carlos May 11 '13 at 11:06

Imagine the bars of a jail cell. Tennis balls shot directly at them will only bounce back if they happen to hit a bar and at the right angle. This is not completely comparable to reflection, but wait for the rest of the analogy. Now imagine shooting tennis balls at the bars (horizontally) at an angle of 5-10 degrees from parallel to the "wall" of bars. The odds of the ball hitting and bouncing back off of one of the bars is 100% ~ sorta like total internal reflection. The balls aren't being "reflected" but the idea is similar. Bars are comparable to electrons and tennis balls to light waves.

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I believe that I was able to improve on your analogy. If you throw a tennis ball at the bars with a small angle relate to the normal, the tennis ball “sees” a larger opening or effective cross sectional area in order to pass through. As the angle is increased, the effective cross sectional area decreases to the point that no tennis ball will be able to pass through. Hence, all tennis balls experience total internal reflection. Thanks for the analogy, it works very well. – Carlos May 6 '13 at 11:51
Very interesting. Could you extend the analogy to throwing balls from the other side? Lower refractive index? Say from inside the jail ? :) – roadrunner66 May 9 '13 at 4:32