geodesicsopticsrefraction
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?
"Is it possible for a medium (here calcite) to have two refractive indices?"
Yes. Birefringence refers to the property that the refractive index of a material depends upon the polarization or direction of light through the material.
The answer is YES. See diagram:
The key here is that you can write down the expressions for the angles $a_2, a_3, a_4$ in terms of the angles of the prism $\alpha, \beta, \gamma$ and the first angle $a_1$. Now $\alpha + \beta + \gamma = 180$ and you know you must have total internal reflection with the first three angles, but not with $a_4$.
$$a_2 = \gamma - a_1\\ a_3 = \beta - a_2\\ a_4 = \alpha - a_3$$
So expressing $a_4$ in terms of $a_1$ we get
$$\begin{align}\\ a_4 &= \alpha - (beta - (\gamma - a_1))\\ &= \alpha - \beta + \gamma - a_1\\ &= \alpha + \beta + \gamma - (a_1 + 2\beta)\\ &= 180 - (a_1 + 2\beta)\end{align}$$
Playing around a bit with these equations, I find that the optimal solution is obtained with
$$\alpha = 36\\ \beta = 72\\ \gamma = 72\\ a_1 = 36$$
In this case, you have $a_2 = 36, a_3 = 36, a_4 = 0$. In other words, as long as you can get total internal reflection with an angle of incidence of 36 degrees, you can do it - and as a bonus, you go "straight in" and come "straight out" - see below (approximately accurate):
This requires a refractive index just greater than 1.7 ($1/\sin(36)$ - definitely possible.
If you are willing to have the exit direction ($a_4$) be something other than zero, then you can improve on the above solution (make it possible with lower refractive index).
Best Answer
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.