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A lot of textbooks and exam boards claim that light incident at exactly the critical angle is transmitted along the media boundary (i.e. at right-angles to the normal), but this seems to violate the principle of reversibility in classical physics. How would a photon or ray travelling in the reverse direction "know" when to enter the higher refracting medium? It can't know, so I conclude that such light is simply reflected?

Is this correct?

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This is a non-problem. Light would never be traveling along the interface. –  André Neves Sep 17 at 14:05

2 Answers 2

When one is exactly at the critical angle, the light behaves in a way that may be interpreted as "something in between" refraction and reflection: it continues in a direction that is tangent to the boundary of the mediums.

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When the angle is smaller than the critical angle, we get refraction. At the critical angle, $\theta_2$ of the refraction becomes 90 degrees, so we get the tangent propagation. At angles larger than the critical ones, there is a discontinuity: the equation for $\theta_2$ (arcsine of something) has no solutions which is why we get a total internal reflection.

There is nothing about these facts that would contradict reversibility or time-reversal symmetry of the laws of physics. If we time reverse the behavior at the critical angle, it indeed looks like the light must "randomly pick" a moment at which it enters the medium with the higher refraction index and there isn't any unique way to pick the preferred moment.

But that's not a problem because the probability that the direction of light is "exactly" tangent to the boundary is zero. In a real-world situation, the light will be a superposition of beams with angles $\theta_2=\pi/2-\epsilon$ for various small values of $\epsilon$, and for any nonzero $\epsilon$, the light will know very well when it hits the boundary. So your problem only occurs at a negligible, "measure zero" portion of the situation, so it is at most a "measure zero" problem. When one adds the appropriate degree of realism and specifies the precise angles and deviations from the "idealized model", the problem goes away.

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I agree this is not a real world problem, since it has measure zero, rather it is a question about whether textbooks and exam boards are correct. However it does seem from your answer (which I agree with) that we should remove the middle diagram from the illustration and just say that light relects internally at angles greater than or equal to the critical angle, and refract at smaller angles. It simplifies the answer and is (more) correct. –  Michael C Price Jan 30 '13 at 12:28
    
It is not more correct. Note that reciprocity holds. Suppose you have an infinite medium with an interface as pictured above, then a plane wave (which rays represent) propagating exactly in the $-x$ direction would have an electric field vector in the $y$ direction, exciting the medium, and resulting in the a refracted ray. At the photon level, the electric field would be realized by the interaction of the photons with the medium... –  daaxix Jan 30 '13 at 21:48
    
Also, the above diagrams assume a lossless (i.e. purely real index of refraction) medium, which doesn't exist so far. See this interesting discussion on this issue : osa-opn.org/home/articles/volume_21/issue_1/features/… –  daaxix Jan 30 '13 at 22:49
    
@Daaxix, the assumption of a lossless medium is not a problem for a gendanken experiment, which strip away such irrelevant details to expose the physical principles underneath. –  Michael C Price Jan 31 '13 at 14:36
    
Yes. Like I said, in the gendanken experiment, there isn't a problem either and reciprocity holds at the critical angle. –  daaxix Jan 31 '13 at 14:42

It's more correct to say that light rays split into reflected and refracted subcomponents at boundaries, rather than reflect or refract. Figuring this out is done by rigorously looking at the boundary conditions of the electric and magnetic fields at the interface. The result ends up being polarization-dependent, and is known as Fresnel's equations.

Per Fresnel's equations, as expected, some of the light is transmitted at the boundary, and some is reflected at the boundary, even below the critical angle. At the point that you hit the critical angle and the light ray "goes horizontal", the transmission coefficient goes down to zero. So, while there is a "solution" that moves horizontally, exactly 0% of the incident light ray will actually be doing this. The reflection coefficient hits 100% exactly at the critical angle.

More rigorously, consider that an analysis of the boundary conditions at the interface requires that the two polarizations of light have reflectances:

$$R_{s} = \left(\frac{n_{1}\cos\theta_{i} -n_{2}\cos\theta_{t}}{n_{1}\cos\theta_{i}+n_{2}\cos\theta_{t}}\right)^{2}$$

$$R_{p} = \left(\frac{n_{1}\cos\theta_{t} -n_{2}\cos\theta_{i}}{n_{1}\cos\theta_{t}+n_{2}\cos\theta_{i}}\right)^{2}$$

Then, consider that we're at the critical angle, which tells us that $\theta_{t} = \pi/2$ and $n_{2} = n_{1}\sin\theta_{i}$ (I henceforth remove the $i$ subscript). Then, we have:

$$R_{s} = \left(\frac{n_{1}\cos\theta-0}{n_{1}\cos\theta+0}\right)^{2}=1$$

and

$$R_{p} = \left(\frac{0-n_{2}\cos\theta}{0+n_{2}\cos\theta}\right)^{2}=1$$

By conservation of energy, this means that the transmission coefficients are therefore both zero, and no light is refracted at this angle.

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