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Suppose that I have a block of some transparent material, glass, for instance, with a certain index of refraction. Suppose that the transparent material is placed in air or in any other transparent medium with a smaller index of refraction (smaller index because we want to have total internal reflection, see second paragraph).

I shine a light ray into the block at some angle. Depending on the angle, there might be one or more total internal reflections.

Question

My question is, is it possible to have a well-defined form/shape of block with a well-defined index of refraction, and have a special angle at which if we send a light ray into the block, the light ray will undergo total internal reflections forever inside the block? (striking boundaries forever at angles larger than the critical angle; not just many times, but infinitely)

Approaches

Of course, there are trivial solutions such as a rectangular prism of infinite length (like an infinitely long wire) ... But I am looking for nicer, more interesting cases such as blocks of finite volume. Maybe it's simpler to start by thinking about 2D polygons. Perhaps a combination of flat and curved boundaries in a 2D or 3D object could also work.

EDIT

I want to consider this question only on a theoretical, almost mathematical, basis. (No attenuation, no scattering, etc.)

Ideas?

Thanks

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    $\begingroup$ "rectangular prism of infinite length".. I do not understand that. I think you meant infinitely long wire like those used in fibre optics. Did you? $\endgroup$ – Ari Sep 22 '15 at 4:08
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    $\begingroup$ Yes, wires that are infinitely long. In such a wire, if the initial angle is large enough, the light could always reflect off the boundaries forever (since the boundaries are infinite). $\endgroup$ – Andrew Trubatchev Sep 22 '15 at 4:10
  • $\begingroup$ There is always attenuation, so, no. $\endgroup$ – CuriousOne Sep 22 '15 at 4:28
  • $\begingroup$ If you are interested in "light traps" for geometric rays, try Geometry and Billiards. $\endgroup$ – Keith McClary Sep 22 '15 at 14:34
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A physicist's answer to this is that the second law of thermodynamics forbids such a construction. You are describing a perfect black body, and the indefinite input of light in the way you propose will inevitably heat any finite cavity with the properties you propose. If your input light comes through a perfect waveguide from a black body at some temperature $T$, then the second law forbids the device's rising to a temperature above that of the source. So some light has to eventually leave the device.

However, what of trapping a short pulse of light (where the heating effect would not be a problem)?

There are contrived mathematical solutions to similar problems. For a 2D example, consider a perfect circular mirror with an infinitely thin slit in it for a ray to pass through (here we strike another difficulty of applying ray optics to this problem: real light cannot pass through an infinitely thin slit and indeed diffracts strongly on the far side if (1) the slit is much less than a wavelength width and (2) the wall thickness is thin enough for transmission through the slit. So already we must begin to consider full EM theory rather than rays. But let's state the ray solution for completeness:

Cavity

Theangle $\Delta\theta$ between the angular positions of successive bounces is constant. This angle is a continuous function of the incidence angle, and equal to $\pi$ when the incidence angle is nought. Clearly all values of $\Delta \theta$ in some neighborhood $(\pi-\epsilon,\,\pi+\epsilon)$ of nought are reachable by adjusting the incidence angle. So we choose a $\Delta\theta$ that is an irrational multiple of $2\pi$. The ray hits the slit again (and thus escapes) after $n$ circulations, where $n\,\Delta\theta=2\,\pi,\,m$, for integers $n$ and $m$. But this is impossible if $\Delta\theta$ is an irrational multiple of $2\,\pi$, whence the device "swallows" such a ray permanently.

Take heed how infinite precision is needed for this argument: a nonzero width slit in this device will always lead to an eventual escape. To understand that there must be an eventual escape of the ray in this case, either $\Delta\theta$ is a rational multiple of $2\,\pi$, in which case it hits the slit precisely after some finite number of bounces, or it is an irrational multiple of $2\,\pi$. If the latter case, it can be shown that the set of intersection points where the ray bounces from the mirror is dense in the circle, therefore, any nonzero angular interval (and thus nonzero width slit) contains at least one of these intersections, so the ray escapes in this case too.

We can make a more realistic ray solution escape proof. A strip of horizontal light rays will be trapped by a Cassegrain like structure with two arcs of parabolas with common focus and vertical directrices will do it:

black hole with a gap

See this MathOverflow Thread "Symmetric Black Hole Curves" for more information. But this solution also has the catch that the incoming ray must be perfectly horizontal for trapping to happen. Since diffraction is roughly tantamount to a nonzero angular spread of rays, this means that real light will eventually escape such a structure.

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  • $\begingroup$ But what if you only introduce a couple photons? No heat problem, tho' the guarantee of ultimate escape is still true. $\endgroup$ – Carl Witthoft Sep 22 '15 at 11:59
  • $\begingroup$ @CarlWitthoft Yes, we need a full field analysis for this. See updated answer. $\endgroup$ – WetSavannaAnimal Sep 22 '15 at 12:36
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A counter-argument to all the naysayers :-) .

First of all, the evanescent wave has nothing to do with light escaping. For example, single-mode fiber optics are often designed such that there's a significant evanescent wave, but since there's no high-index material outside the core, no energy escapes (other than quantum probabilistic long-range tunneling).

Next, and this is the unobtanium part, suppose you shoot a particle and its antiparticle into a, say, octagonal right prism, such that the photons generated when they destruct are emitted in the prism plane and at an angle equal to the face angle, which is also less than the critical angle for the prism material. Those photons will bounce from face to face forever.

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    $\begingroup$ An optical fiber of finite length is always leaky. My argument above was that no solution of the wave equation can be purely evanescent outside a finite-size medium. $\endgroup$ – higgsss Sep 22 '15 at 12:41
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    $\begingroup$ Inventive! Also, note that if you allow solutions of this kind, there is always at least one trapped ray in any surprisingly general, as this MathOverflow post by the late, great Bill Thurston (RIP) shows $\endgroup$ – WetSavannaAnimal Sep 22 '15 at 13:00
  • $\begingroup$ @WetSavannaAnimalakaRodVance THanks-- that's a very interesting bit of math there. $\endgroup$ – Carl Witthoft Sep 22 '15 at 15:00
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    $\begingroup$ You wouldn't need to anihilate exotic particles if the prism could be made from a fluorescent material that was perfectly transparent at the emitted wavelength. $\endgroup$ – Solomon Slow Sep 22 '15 at 16:01
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The answers here is beautiful. But, i’ll give another simple example. Just take a glass square. The critical angle of glass is 42 degree. So pass a light ray through a very small slit , such that it strikes the side of the square glass slab, at 45 degrees. It will reflect and will ALWAYS strike the faces of the slab at angles greater than 45 degrees. And you’re in business!

enter image description here

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In the language of wave optics, I think that your question boils down to the following:

Given that dissipation is negligible, can a dielectric medium of finite size support a mode that is propagating only within the medium and evanescent outside?

The answer to this question is certainly "No."

Regardless of the form of the electric field inside the medium, it should be matched to the field outside the medium satisfying the transverse wave equation: \begin{equation} \nabla\times(\nabla\times\textbf{E}) + \frac{1}{c^{2}}\frac{\partial^{2}\textbf{E}}{\partial t^{2}} = 0. \end{equation} For a propagating mode only within a finite-size medium, the field outside must be evanescent in all three dimensions. But the above equation does not have a solution with such a property. Therefore, any mode that appears to be localized should in fact be matched to a propagating solution outside the medium, making the mode leaky.

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    $\begingroup$ This wouldn't be true for a reflecting medium, though. $\endgroup$ – CuriousOne Sep 22 '15 at 6:03
  • $\begingroup$ @CuriousOne What do you mean by a "reflecting medium"? $\endgroup$ – higgsss Sep 22 '15 at 6:04
  • $\begingroup$ A perfect conductor will not let any waves escape. The latter is, of course, impossible. $\endgroup$ – CuriousOne Sep 22 '15 at 6:27
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    $\begingroup$ An ordinary dielectric sphere in three dimensions can indeed support fully bound modes. To see this you just need to look at the Helmholtz equation in spherical coordinates: the bound solutions can be represented in terms of spherical harmonics and the evanescent decay outside is described by a "Modified Spherical Bessel Function of the Second Kind". Of course, there is no way to excite them using incident external plane waves. $\endgroup$ – Nanite Sep 22 '15 at 8:10
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    $\begingroup$ @higgsss I am thinking more like real frequency, imaginary wavevector in vacuum, modes like this in particular: en.wikipedia.org/wiki/File:Whispering_gallery_modes_sphere.png . However I just realized that the Helmholtz equation might not be fully applicable due to polarization. In the example I just linked, there is actually some escape of light even in the ideal case; it is not fully bound. So my previous comment is probably wrong. $\endgroup$ – Nanite Sep 22 '15 at 8:28

protected by Qmechanic Sep 22 '15 at 7:29

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