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:
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:
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.
Firstly, your point about total-internal-reflection (look into the tag wiki for further info about the phenomenon) resulting in 100% transmission of light energy and hence 100% intensity of incident light being preserved during reflection is conventionally considered true, but is not entirely correct.
Let me explain:
Reflection of any type, even if it is total internal reflection, cannot reflect 100% of the incident light energy. Total Internal Reflection nearly 100% (99.9999% maybe, but not 100%) efficiency compared to conventional reflection from a surface.
An interface between 2 optical media has the property of a "critical angle", the angle of incidence beyond which if light is incident on the interface from the optically denser medium into the optically rarer medium, it will be reflected back entirely into the denser medium, with no refraction into the rarer medium. Normally, at an interface between 2 transparent media of different optical densities, the light wave will be partially reflected into the medium from which it was incident and partially refracted into the medium into which it was incident, until the angle of incidence goes beyond this critical angle. This phenomenon of total internal reflection is usually observed for light, but is also applicable for other types of waves, like sound and waves on a string. This phenomenon is explained by Huygen's Wave Theory of Light in a classical sense. It occurs at the interface between 2 media of different densities in which the wave travels.
Hence, the intensity of incident light is preserved nearly entirely in total internal reflection, except for that minor fraction which may be absorbed by the denser medium itself, which is low due to the medium being transparent. There is also some energy loss due to photon tunneling across the interface, but again, it is not conventionally significant (total about $10^{-3} \%$). There are also evanescent-waves across the interface, but they do not result in net energy transfer across the interface. Refer Wikipedia here.
In conventional reflection, 2 surfaces are involved:
1) A transparent unsilvered surface
2) An opaque silvered surface
For light transmission through the unsilvered surface, similar energy losses are applicable as in case of total internal reflection.
However, for light reflection at the silvered surface, the surface being opaque, will absorb a significant portion (say about 1%-2%) of the incident light energy, while it will reflect most of the light energy due to it being silvered and reflective.
This 2nd significant energy loss of incident light at the silvered surface is not applicable for total internal reflection, hence we commonly say that total internal reflection forms images at 100% the brightness (intensity) of the incident light.
Best Answer
Heading over to PhET's Bending Light Simulator, you can try out this experiment yourself. As it turns out, you cannot get the light to be infinitely contained in the glass if you shine light from the outside. But it is possible if the light source is within the glass, as is shown below.
But why is this true?
To answer this question, let us first try to find out the necessary conditions for the light to have an angle of incidence equal to the critical angle at its second incidence (when it goes from glass to air).
[I’ll be using “the first incidence of light” to refer to when the light goes from the air to the glass, and “the second incidence of light” to refer to when it goes from the glass to the air.]
Have a look at the following image.
First you must understand that the normals drawn at points B and C have the radii of the circle as their part, i.e., if extended, they intersect at the centre always. Why is this true? Because the normals are by definition perpendicular to the tangents to the circular glass drawn at those points — B and C. But it is known that the radius of a circle is the line segment that is perpendicular to a tangent drawn at any point on the circumference. Therefore, the normals must be extensions of the radii of the circle.
If you didn’t get the above part, you may ignore it; just know that OB and OC are radii of the circle, hence they are equal. This would imply that triangle BOC is isosceles, and so ∠OBC = ∠OCB.
Now we know by Snell’s law that $\sin{i} \propto \sin{r}$. In other words, more the angle of incidence, more the angle of refraction, and vice versa, because the sine function increases from $0°$ to $90°$, and those are the only angles we are considering here. Thus, if we look at the first incidence, to maximise $r_{1}$ (∠OBC), we need to maximise $i_{1}$ (∠EBA).
But $i_{1}$ cannot be any bigger than $90°$, because if it were greater, the light would basically have its source within the glass, and we know that the light does get infinitely contained if its source is in the glass.
Now what is the angle of refraction $r_{1}$ corresponding to a $90°$ $i_{1}$? Let’s find that out using Snell’s law! Assuming $1.5$ to be the refractive index ($\mu$) of glass,
\begin{align} & \frac{sin(i)}{sin(r)} = \mu \\ & \implies \frac{sin(90°)}{sin(r_{1})} = 1.5 \\ & \implies \frac{1}{sin(r_{1})} = 1.5 \\ & \implies r_{1} = sin^{-1}(1/1.5) \\ & \implies r_{1} \approx 41.8103149° \end{align}
[Which is exactly equal to the critical angle for glass and air, in accordance with the principle of reversibility of light.]
$\therefore ∠OBC = ∠OCB \approx 41.81°$
But this wasn’t our true objective was it? We wanted the light to get totally internally reflected, but for that, $r_{2}$ must be greater than $90°$. Hence $i_{2}$ must be greater than $41.81°$ (because $i \propto r$, by Snell's law), which implies that $r_{1}$ must be greater than $41.81°$ (because $r_{1}$ = $r_{2}$), which means that $i_{1}$ must be greater than $90°$ (again, because $i \propto r$). That is impossible, unless the source be in the glass itself, thus proving our observations.