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Suppose we have a medium with varying refractive index and a source of light inside that medium emitting rays. Is it possible to bend the ray into any closed loop?

As the medium has varying refractive index, is it possible? And if possible, how will it look like if anyone stand in the path of the ray?

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Let's try to design an axisymmetric medium in which a concentric circle of radius $R$ is a possible light ray. The index of refraction is $n(r)$. In polar coordinates, light rays close to the desired circle are described by $r(\theta) = R + \delta r(\theta)$.

The arc length $ds$ of a small segment of such a ray is the hypotenuse of a right triangle whose tangential leg is $d\theta\, r(\theta)$ and whose radial leg is $d\theta\, r'(\theta)$. Fermat's principle says that a valid ray must extremize the travel time $$cT = \int ds\, n = \int_0^{2\pi} d\theta\, \sqrt{r(\theta)^2 + r'(\theta)^2}\, n\bigl(r(\theta)\bigr).$$ Expanding to first order in $\delta r$ gives $$cT = 2\pi R\, n(R) + \int_0^{2\pi} d\theta\, \bigl(\delta r(\theta)\, n(R) + R\, n'(R)\, \delta r(\theta)\bigr).$$ (Note that $r'(\theta)^2$ does not contribute here because it is second-order in $\delta r$.)

So the condition for an extremum is $$n'(R) = -\frac{n(R)}{R}.$$ The medium has a refractive index gradient such that the index is higher in the inner part and lower in the outer part. This is consistent with the observation that in a nonuniform medium, such as the atmosphere, light rays bend toward the higher index.

As a further extension, for all concentric circles to be valid light rays, we need $n'(r) = -n(r)/r$ for all $r$, giving $n(r) \propto 1/r$. Realistically, this cannot be extended beyond a finite range of $r$.

For an observer in this medium, a circular path will appear straight. A preexisting loop of light will be disrupted and absorbed by the observer. A source behind the observer will be visible after its light travels all the way around (appearing to be in front of the observer). The apparent shape and size of a non-point source depends on the dynamics of other rays that are not concentric circles, which have not been addressed here.

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  • $\begingroup$ It would be amazing to see a ray-traced simulation of such an environment. Non-linear refractive media could be an interesting video game mechanic. $\endgroup$
    – Quantum7
    Sep 29 at 13:47
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I think so, and not just only loops but also knots. Any optic cable tied into a loop or a knot will probably do. It's easy to imagine theoretically, but probably much more difficult to set up practically as there is the problem of getting the light into the loop into the first place.

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    $\begingroup$ An optical cable was my first thought too, but they work with total reflection, right? But the question is if it is also possible to achieve via refraction. @The Space Guy correct me if I'm wrong. $\endgroup$
    – tomtom1-4
    Sep 26 at 8:38
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    $\begingroup$ Yup, I want to know if it is possible through refraction!? $\endgroup$ Sep 26 at 8:43
  • $\begingroup$ @tom-tom: Good point, I just checked and the total reflection is due refraction bwtween two internal layers. Nevertheless, it's possible to do what space guy is asking for via just refraction. Though building something that works might not be so easy. $\endgroup$ Sep 26 at 8:44
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    $\begingroup$ It is possible to construct an optical fiber that guides light through a graded index. Graded-index fiber. $\endgroup$
    – mmesser314
    Sep 26 at 17:24
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    $\begingroup$ @TheSpaceGuy You should have mentioned in your question that you want to do this with pure refraction, not total internal reflection (which is reflection, even though it occurs at the interface between media of different refractive indices). But it's too late to edit that info into your question now, since that would invalidate several of the existing answers. $\endgroup$
    – PM 2Ring
    Sep 27 at 10:34
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Ok so here are my thoughts. I wanted to construct something like a clock where each circular sector is made of a different material. The incoming angle of the light ray with respect to the boundary of the first to the second layer is $\alpha$ (see image). To make things easy I wanted to choose the second material and circular section in such a way, that the incoming angle in the second refraction is still $\alpha$. If the refractive index of the second material is bigger than the index of the first material, this means that the circular sector of the second material must be smaller $\alpha - \epsilon$. Since the sum of the angles in a triangle is $\pi$ we find that the outgoing angle $\beta$ must be \begin{equation} \beta + \pi/2 + \pi/2 - \alpha + \alpha - \epsilon = \pi \quad \Rightarrow \quad \beta = \epsilon. \end{equation} Hence we must choose two materials with \begin{equation} \frac{\sin \alpha}{\sin \beta} = \frac{\sin \alpha}{\sin \epsilon} = \frac{n_2}{n_1} \end{equation} For the next circular sector, we proceed in the exact same way. The sector has the angle $\alpha - 2 \epsilon$ and the incoming angle is still $\alpha$, thus the ratio $\frac{n_3}{n_2}$ is equal to $\frac{n_2}{n_1}$. This process can be repeated until we got around the entire circle. Important is that we have to choose $\alpha$ and $\epsilon$ in a way that we get around the entire circle, i.e. \begin{equation} \sum_n (\alpha - n \epsilon) \ge 2\pi. \end{equation} For instance choose $\alpha = 30^\circ$ and $\epsilon = 1^\circ$.

This does not entirely answer your question though, as the loop is not closed. The distance to the center of the circle gets larger and larger with every refraction. In the ideal case $\epsilon = 0$ you would have a closed-loop but for that, you would need a material with an infinite refractive index. So for a setup like this, we have shown that it is not possible to construct a closed loop. And even if I did not show it I predict that it is not possible at all (but I am happy to be convinced of the opposite).

Also notice, that even if you are happy with a loop that is not closed, you can not build such a device in real life as you run out of materials with a high enough refractive index. For $\alpha = 30^\circ$ and $\epsilon = 1^\circ$ you would already require $n_2/n_1 = 29$ which is impossible to achieve in real life (as far as I know).

enter image description here

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OP seems to be asking about the case that uses a smoothly varying isotropic refractive index $n({\bf r})$ with no jumps and with no reflections/mirrors.

Let us assume that OP's given curve $\gamma:[a,b]\to \mathbb{R}^3$ is smooth and has no self-intersections, i.e. it is a simple curve. Also let us assume that $\gamma$ has finite length.

Then we can sketch a construction as follows:

In a tubular neighborhood $\Omega\subseteq\mathbb{R}^3$ of the curve $\gamma$, we can pick a function $W:\Omega\to\mathbb{R}$, such that the gradient $\nabla W$ restricted to $\gamma$ is non-vanishing and parallel to $\gamma$.

(E.g. for each point $p$ along the curve $\gamma$, there is a unique hyperplane $\Pi_p$ perpendicular to $\gamma$. We could choose the function $W$ to have the hyperplanes as level sets.)

($W$ plays the role of the eikonal in geometric optics and is related to Hamilton's characteristic function in Hamilton-Jacobi theory, cf. Ref. 1.)

We next choose the refractive index $n:=|\nabla W|$. The surfaces of constant $W$ becomes the surfaces of constant optical phase and thus define the wave fronts. The curve $\gamma$ becomes a geodesic/ray trajectory in this geometry.

(If the curve $\gamma$ is closed, then $W$ becomes multi-valued, but since we are just interested in the length of the gradient $|\nabla W|$, we just have to make sure that it doesn't depend on the branch of $W$.)

We could accentuate the construction by letting the refractive index profile $n$ become gradually smaller away from $\gamma$, but it is in principle not necessary. (This is the idea behind an optical fiber with a smooth gradual fusing of the cladding, cf. many of the other answers.)

Finally extend $n$ in a smooth way from the tubular neighborhood $\Omega$ to all of space $\mathbb{R}^3$.

References:

  1. H. Goldstein, Classical Mechanics, 2nd edition; Section 10.8.
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Fiber-optic gyroscope
Yes, this is routinely achieved in Fiber-optic gyroscopes and other similar interferometric devices. Moreover the light makes multiple loops (there can be hundreds or even thousands of them), although it never returns to the same point - for simple topological reasons (there should be point where the light enters and exits). The optical fiber is, of course, a way of confining light using the difference in refraction indices inside the fiber and outside of it.

What is a ray?
The ambiguity in the question is what we mean by a ray. Simple picture is a plane wave in an empty space, which can be more or less adapted to the ray passing into a medium with a different refractive index. In practice, the light has a mode structure determined by the resonator that confines it, i.e., by the boundary conditions. One can (and does) fabricate a toroidal resonator or a closed loop of optical fiber and excite propagating electromagnetic modes in it - but can we call it a bent ray?

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If a glass prism in air counts as "a medium with varying refractive index", you just need to arrange enough prisms in a circle to create a closed path.

The challenging part is to make sure that the sum of the deviation angles in all prisms is exactly 360º, because otherwise a light ray would spiral outwards or inwards in every turn. That part can be solved by making prisms with slightly concave faces, so that light bending is smaller in the inner part of the circle than in the outer, making outer rays to spiral inwards ans inner rays to spiral outwards to an stable loop.

When building this set in 3D, a related problem is that any deviation from parallelism in the axis direction can transform the circular path in an helix, making light leave the prisms after some turns, but that can likely be corrected by appropriate bending of faces.

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  • $\begingroup$ Same idea I had; also, since refraction is different for different wave lengths, I suspect that only one single colour will be part of the loop, while the other colours will be deviated outside the loop. $\endgroup$
    – Berello
    Sep 28 at 13:06
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    $\begingroup$ If the idea of prisms with slightly concave faces actually works, different wavelengths will have loops of slightly different radius. $\endgroup$
    – Pere
    Sep 28 at 17:46
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Two Eaton lenses would form parallel closed optical path loops. Standing in between the two Eaton lenses (much larger than your head) and looking at one Eaton lens you would be able to see the back of your head.

An "inside-out Eaton lens" also has closed optical path loops.

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If you allow reflection in addition to refraction, that's what a fiber optic cable does. If you want to do it just with refraction, it's more complicated, but possible. For a change in medium to change the direction of a light beam, the interface (the surface between the mediums) must be oblique (that is, not perpendicular) to the beam. If you go from a medium with a high index of refraction to a medium of low index of refraction, the beam will bend towards the interface, and vice versa. So we can alternate media of high and low indices of refraction, and alternate the direction that the interface slants.

This is how a prism works. Suppose light comes into a prism from the left and exits the right side. When it enters, the edge of the prism is slanting up, and the index of refraction is increasing, so the beam refracts away from the interface, which is down. When it exits the prism, it now refracts towards the interface, but the interface is sloping down, so the beam again refracts down. If you arrange several prisms in a circle, the light beam will follow.

If the light beam is made up of different wavelengths, however, the index of refraction will be different for the different wavelengths, causing dispersion. This can be counteracts with several different materials, rather than just two.

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Simple, practical example:

Take a fiber-optic cable, bend it into a loop. Connect the ends.

Getting light into the cable in the first place is left as an exercise for the questioner. (And note that no medium is perfectly transparent, in practice the light vanishes in timeframe that a human would perceive as instantaneous.)

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  • $\begingroup$ I think the OP wants refraction to be the only thing taking care of the bending of the light. In this case the the light ray itself won't bend, but the channel through which it shall travel will. $\endgroup$
    – mnuizhre
    Oct 2 at 7:05
  • $\begingroup$ @mnuizhre And what other than refraction bends the light ray in a fiber optic cable?! $\endgroup$ Oct 3 at 3:36
  • $\begingroup$ Check it out here. $\endgroup$
    – mnuizhre
    Oct 3 at 6:49
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On a larger scale I’m sure light is bent into a closed loop somewhere in the universe. Like a photon sphere around a black hole or light bending around multiple black holes until it returns or almost returns to where it started.

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