If you drop a candle into an infinite well made of perfect mirrors, would you see it forever? Part 1: 
If you drop a candle into an infinite well made of perfect mirrors, would you see it forever?
Part 2:
So when you drop a candle into an infinite well not made of perfect mirrors, the reason you don't see it after some time is because too few photons are hitting your eyes? (Most of the photons have been absorbed by the walls and due to the equation of the volume of a sphere, the intensity at your eye is small). Thank you!
 A: I'll post this as an answer though it's not conclusive. I think it's bit tricky.
I'm imagining there is a spherical emitter falling through the hole. The emitter emits light energy at a rate $\Gamma$ with dimensions of Energy/Time. The hole is square and the walls are rectangular. The emitter has some trajectory $z(t)$. If it is free-falling under gravity then $z(t) = -gt^2$. First I'll point in this geometry that half of the light goes up out the hole and half the light continues down the hole.
Consider the emitter fixed at $1m$ into the hole and suppose that at $t=0$ it is turned on for $1s$. Some of the light will be emitted directly out the hole but some of it will be directed towards the walls at some angle. The light which is directed towards the walls will need to reflect off the walls and thus take a longer path to get to the opening. In fact, some of the light will be emitted almost directly sideways towards the walls. This light will need to bounce many many times and take a very long time to exit the hole. This means if you put an infinitely sensitive detector at the opening of the hole you would initially see a large pulse of light (lasting about $1s$ and starting after the speed of light delay) whose integrated energy would be almost $E = \frac{\Gamma}{2} \times 1s$ but in fact for ALL time out to infinitely you would still see small small amounts of light from the light which was directed at very steep angles.
This is the first insight into the problem. The point is that a pulse of light emitted at a certain depth results in a pulse of light on the detector which has a DIFFERENT shape. In fact the pulse of light on the detector is "spread out" in time compared to the pulse which was emitted deep in the well.
Now, it should be clear that no matter how deep the emitter is if you turn it on and leave it in the same location then eventually all of the light will "integrate up" until the total intensity leaving the opening is again up to the $\frac{\Gamma}{2} level.
However, if the emitter is moving very quickly then the intensity which is received at a given time from the emitter at a certain depth might be decreasing too quickly for the total intensity coming out of the hole to increase..
This is where my argument gets less logical. But I think that depending on the particular trajectory $z(t)$ you can get two limits. In one limit (moving slow or with a small scaling, say linear or polynomial motion) the intensity on the detector will increase asymptotically towards $\frac{\Gamma}{2}$. In the other limit (moving fast or with fast scaling like exponential or something) the intensity will increase to a maximum and then decrease towards $0$.
I don't know what the answer would be for falling under gravity.
Now imagine instead that all of the light from the emitter was emitted directly upwards (say the emitter was a laser pointing upwards). In that case I think the intensity coming out of the hole would always be $\frac{\Gamma}{2}$.
edit: To relate this to the question a bit more I assume a threshold intensity $I_0$ below which the human eye cannot detect. Thus depending on which asymptotic  situation we are in, in one case you would be able to see the light forever, but in the other case the integrated intensity would fall below the detectable level so you would not be able to see it.
The question then is where are the photons if you cannot see them? The answer is that there are many many photons bouncing around inside of the hole. There must be for energy conservation. If you could see all of them then you could see the candle but they're slowly bouncing their way out in such a way that only a few of them actually make it out per second. 
A: Fiber optics show us that light can travel very far with little attenuation if it bounces on mirrors. Therefore with perfect mirrors it could go infinitely far  without attenuation (neglecting the air absorbtion).
So yes you'd see something forever but I would expect blur to increase progressively as some rays (aligned with well) travel faster than others. I assume that the candle wouldnt reach relativistic speeds
