5
$\begingroup$

In the special case of the illumination problem where one uses only one light ray (instead of illuminating in every direction), is possible to illuminate a rectangle everywhere densely? Does this hold for almost every angle and starting position?

$\endgroup$
  • $\begingroup$ I assume it will for any angle with finite, non-rational tangent value, which is almost all angles, and any starting point. But I have no proof. $\endgroup$ – Arthur Nov 5 '12 at 11:46
3
$\begingroup$

The only way for a ray not to illuminate the rectangle densely is if it eventually hits its starting point exactly and then continues cyclically.

This is a bit easier to see if instead of a rectangle with mirrored ways we imagine a rectangle where opposing walls have been identified to form a flat torus -- which corresponds to adding to the original rectangle its reflection about, say the top and right sides. The "flat torus" model is a quadruple cover of the mirrored rectangle, so if the torus, so if the torus is illuminated densely, then so is the mirrored rectangle.

Now consider the distance (along $S^1$ in some preferred direction) between two successive crossings of the horizontal edge. If that is a rational multiple $p/q$ of the width of the room, then after $q$ crossings the ray will hit its starting point exactly. However, if it is an irrational multiple, then the horizontal edge will be hit in densely many places (it can't ever hit the same place on the edge, so by compactness there will be two points arbitrarily close to each other separated by say $k$ windings, and then the hits at every $kn$ windings ($n\in\mathbb N$) henceforth will eventually cover the edge with that small spacing). Once we see that the horizontal edge is hit densely, the pieces of ray emerging from those dense points will illuminate very level of the flat torus equally densely.

Now, there are only countably many angles that achieve a cyclic ray. (Imagine the infinite lattice the rectangle, with mirrored walls, looks like from inside -- in order to get a cyclic ray you'll have to aim for one of the countably many images of the starting points). Therefore almost every direction the initial ray could take will lead to dense illumination.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ How do you KNOW that this is the only way? I think that's the difficult part. How do you KNOW that it doesn't skip an area in the corners, or in the center of the rectangle, for instance? $\endgroup$ – Arthur Nov 5 '12 at 11:48
  • $\begingroup$ @Arthur: Better now? $\endgroup$ – hmakholm left over Monica Nov 5 '12 at 11:57
  • $\begingroup$ Yes. What I couldn't get myself was the argument going from a compactness (which only says that the beams hit densely around some point) to proving that they hit densely everywhere. $\endgroup$ – Arthur Nov 5 '12 at 12:02
0
$\begingroup$

In case of the rectangle I believe an important question to ask would be what happens in the corners? The trajectory line cannot be perpendicular to the walls as it would result in a cyclic path. On the other hand if the line is not perpendicular, then what should happen to the direction on rebound? The only reasonable choice would be that the incident and exiting lines are parallel, thus creating an "endpoint" in the trajectory. As the two lines would be identical, the trajectory starting from the corner would be the inverted version of the original one (that continues further after reaching the original starting point). As a corner rebound creates an endpoint, and incident and outbound trajectories from this point are symmetric the trajectory can only have 2 such points, thus it cannot cover all 4 corners of the rectangle. For this reason the domain including all 4 corner points cannot be uniformly illuminated by a single ray.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ But maybe some choice of starting point and angle would get points arbitrarily near the corners. All one needs for the question is to cover a dense set of points in the rectangle. $\endgroup$ – coffeemath Nov 5 '12 at 15:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.