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 Double Reflection (Posted on 2016-03-24)
Two flat mirrors are placed to form a wedge with the reflective sides facing inward. A laser is shined in from the end of one mirror. It hits each mirror once and leaves by the end of the other mirror. How large can the angle between the mirrors be?

Crude diagram: the laser starts at point 1 (at the end of one mirror), travels through 2, reflects off of 3, travels trough 4, reflects off of 5, travels through 6 and ends at point 7 (at the end of the other mirror).
```     /\
/  \
/    \
/3 4  5\
/  6  2  \
7          1
```
What if I want the laser to bounce off of each side n times?

 No Solution Yet Submitted by Brian Smith No Rating

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 solution | Comment 1 of 2
Part 1:

As in other light reflection problems, we can model the broken (reflected) ray by a straight ray if we reflect any secondary mirror(s) in the primary one.  In this case there are only two mirrors, and so only one secondary, except as the primary one becomes reflected in the second, etc.

Whatever the angle between the two mirrors, the reflection of one in the other will be another segment attached to the first two at the same point where they meet, and the angle will be the same as the first angle. For the presented situation the final picture will have four equal segments with one end point each coinciding at the one point, making three equal angles. The free endpoints of the outermost segments are connected by the straight ray.

The three equal angles must add up to less than 180°, so the angle beteen the two mirrors in the setup must be under 60°.

Part 2:

Each two additional bounces (increasing n by 1) adds two additional segments with the accompanying two additional angles. So the number of angles is 2n+1, and the angle between the actual mirrors must be less than 180°/(2n+1). The part 1 case involved n=1 (one reflection on each of two sides).

 Posted by Charlie on 2016-03-24 10:23:45

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