Let there be light!!!

It has been a long time since I have posted anything on my blog. Well, I guess I am just out of focus now a days. Anyways, today’s post is not about a puzzle but it is rather about a couple of open problems. I apologize for not giving any reference for these problems.

Problem 1 :

There are finite number of  non-intersecting line segments in a 2D plane. Each of these line segments act as a mirror from both sides. That is, any ray incident at an angle i is reflected back at an angle r where i=r. It is conjectured that for any point p in the 2D plane. If a light source is put at p then there will be at least one light ray which will escape to infinity ( or in other words, the path of the light ray can not be bounded by a closed curve on the 2D plane ). No other physical phenomena like refraction, deflection etc can be used for the argument.

This innocuous looking problem is notoriously hard to prove or disprove.

Problem 2 : Let there by any polygon in a 2D plane. It has been conjectured that if a light source is put at any point in the polygon, it will illuminate the entire polygon ( for each point p in the polygon, there exist a ray from light source which reaches at point p observing only the law of reflection as described in the Problem 1 ).

This is again an open problem and so far it is invincible. Any takers???

–Saurabh Joshi

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4 Responses to “Let there be light!!!”

  1. Boris Yakubchik Says:

    Hey, thanks for the fun problem. I spent over an hour thinking about problem 2 and figured out some things (not very deep although). I can be sure that the following approach can’t work: proving that you can reach any point within some radius of your light-source _from any direction_. Of course if that was possible – the problem is solved. Here is a reason: there exists at least one polygon such that there exists at least one point such that there exists a light-ray that never comes back to this point. While not so eloquently put – I hope you can follow and agree. Notice sending a light at slope sqrt(2) from the origin makes it so you never hit integer coordinates ever.

    I also noticed that any ray that enters some opening (say your polygon is concave) it always leaves (thus it does it infinitely many times – though perhaps using the same path that it entered through) … but I don’t think this helps much.

    What’s important is that we have continuity: if you imagine the light-source being a laser that spins 360 degrees, whatever surface it hits before first-bounce – it covers fully. And you can take the two end-points where the ray stops hitting the surface – reflect them, and whatever they hit, anything between (so long as it’s along the same wall) it will be fully ‘covered’

    I wish I was better at linear algebra – perhaps I could have had a better-than-trivial insight. But I’d love to see other people’s ideas here – any other observations.

  2. Saurabh Joshi Says:

    Well, I could partially understand what you want to say. Anyways, let me share a comment with you.

    For any convex polygon the claim in problem 2 is trivially true. Now, Any concave polygon can be made by tiling up a few convex polygons. If we can show that for all convex polygons, each point on the surface of the polygon receives light from all directions then we are done. The reason is these points on surface receiving lights from entire 180 degree will act as a light source on the other side of the surface.

    In fact, only one point in each edge of a convex polygon should exist such that it receives light from entire 180 degree. But alas, none of this happens. Let’s take a regular triangle ( all sides equal ) and a light source somewhere inside it. Draw three mirror images of this triangle with each side acting as a mirror. You can fill up entire 2D plane this way. Now, there are countably infinite light source in the plane. You draw a straight line from any of these light source to a point on the edge of the original triangle. This line will tell you the way light will reflect in the from the light source of original triangle to reach this point. Thus, it is easy to see that receiving light from _all direction_ is not possible. But we can have dense ness, in a sense that given any angle \theta and $\latex \theta ‘$ we will have at least one ray which hits the point with angle lying between this range.

    For non-regular convex polygon even this property is not easy to prove because while taking mirror images and trying to fill entire 2D plane, the images overlap which invalidates any ray path from a light source in a 2D plane to a point on the edge of the original polygon to be a physical possibility.

    I hope you are able to visualize what I want to say. I guess a couple of figure would explain it better.

  3. Boris Yakubchik Says:

    I think I understood most if not all. Thanks for the comment.

    I talked to my friend who had an interesting idea. Imagine the point light-source sends out a certain angle of light instead (say 1 degree) in some direction. Each time upon reflection, the length of some wall that is illuminated grows. Since the segment gets reflected infinitely many times, the segment grows to infinity.

    What is not clear of course is whether it will hit all parts, or be somehow stuck within some part of the polygon. I wonder if there is a way to “trap” in a polygon so that the 1 degree light-beam never reaches some wall of a polygon. Perhaps prove by method of contradiction? … I really don’t know … I’ll think some more – this is a fun problem 🙂

  4. Saurabh Joshi Says:

    Sorry for the late reply.

    Well, this light beam does not necessarily increase the illuminated segment of some wall. It depends on the angle at which it strikes the wall and polygon geometry.

    Apart from that, what can we say about behaviour of light ray directly hitting a vertex of the polygon? Because you can not define incident angle properly. If you assume that all vertices act as a black hole, that is a light ray hitting exactly any vertex does not come back. In this scenario someone has given a proof that there can exist dark regions inside the polygon as it effectively renders all line segments as open mirrors.

    Other possibility is that the light ray gets reflected about the angle bisector at that vertex. Well, no proof or contradiction says anything about this case so far.

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