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You have a setup like in the image above. But it seems like detector A does some weird things. You should better check it with detector B. What is the minimum number of mirrors you have to move (translate and rotate arbitrarily) to bring the laser beam onto detector B?

  • Please leave the mirrors which are used to direct the laser to detector A as they are, because if it turns out detector A is not broken you want to switch back to it. I mean it's quite well shielded from stray light compared to detector B.
  • The thick black lines are walls. Please don't burn any holes in them.
  • The mirrors all look the same. They have only one side (blue) with reflective coating.
  • The grey grid is only for orientation.
  • The detectors work for any angle of incidence.
  • Make sure that the beam actually hits the detector, not only stray light.
  • And don't look into the laser beam.
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  • $\begingroup$ If you move the detector, then you can complete the task in zero 'moves'. $\endgroup$ – Strawberry Jan 3 at 16:44
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Assuming only 45 degree mirrors,

We are going to need two moves, and there are several ways to do it.

But since the detectors are omnidirectional, we can do better:

First, take a look at the two mirrors nearest detector B. Mark the position B' as the location of the detector B as seen through the nearest mirror. Then, mark position B'', which is B' as seen through the second mirror. From that point, draw a straight line to the center of the third mirror in the rightmost column.

Notice that the line moves 1 square sideways in 6 vertical squares. It also passes through another mirror in between. Pick up this mirror, we'll use it when we figure out where to place it.

Now, reflecting at the mirror, continue the line 6 squares to the left, during which time it moves 1 square up, because of the reflected angle. What joy, we are at a spot through which the original beam passes, so we can move the mirror we picked up here. We must place the mirror at an angle that is exactly half of the desired beam deflection (which is $\text{arctan}(1/6) \approx 9.5°$), or about 4.7 degrees, in order to hit the detector.

Like so:

enter image description here

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  • $\begingroup$ Nice explanation! rot13(Gurer vf ng yrnfg bar zber fvzvyne fbyhgvba, gurersber V jvyy jnvg fbzr gvzr orsber V npprcg gur nafjre jvgu gur uvturfg fpber.) $\endgroup$ – A. P. Jan 2 at 0:15
  • $\begingroup$ This would be my intended answer, but yours is equally good. $\endgroup$ – A. P. Jan 2 at 5:22
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I think you can do it with two moves: one translation and one rotation:

enter image description here

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    $\begingroup$ I prefer this answer because we can keep mirrors at 45 degree diagonals and it has the same number of moves as Bass's answer. And in real life, hitting the laser sensor at an angle means you might not be transmitting 100% of the laser beam, which we don't have to worry about here. $\endgroup$ – person27 Jan 2 at 14:49
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    $\begingroup$ @person27 if you read through the question carefully, you'll notice that this answer uses two moves, while my answer uses one. Also, the question clearly states that the detectors work at any angle of incidence. All that being said, I think this is nicer than any of the 2-move solutions I could find, so +1 in any case. $\endgroup$ – Bass Jan 3 at 15:14
  • $\begingroup$ @Bass I skimmed your answer and thought the mirror that was moved was the one to the left of its original position, so I thought you had two moves. Whoops, fair point. $\endgroup$ – person27 Jan 5 at 2:24
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Bass' answer is perfectly valid, but I just wanted to share another answer that works in the same way:

One can solve it moving only one mirror.

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  • $\begingroup$ That's really interesting. I didn't catch anything in the question saying no arbitary rotations, etc. $\endgroup$ – Adrian Zhang Jan 2 at 5:24
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    $\begingroup$ For a general solution, you could replace detector B with a powerful light bulb, and then pick up the mirrors one by one. Wherever a spot gets illuminated by both the laser and the bulb, that's a possible spot for the picked up mirror. $\endgroup$ – Bass Jan 2 at 14:43
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A lazy laser physicist would balk at the extra calculations they'd have to do if any of the mirrors aren't oriented exactly at 45 degrees (to horizontal). They are not actually allowed to adjust the newly positioned mirrors on a trial and error basis (to lazily escape the calculations) because (a) they'll burn a hole or more in the walls and (b) perhaps someone will accidentally look into an unstable laser beam.
The solutions (including the one by the OP) which use mirrors at angles other than 45 degrees are smart (respect!!!) but not really suitable for use by lazy people.
Since it appears that two is the minimum number of mirrors to move, our lazy scientist would prefer one of the easiest solutions (both on their hands and brains) which should be the ones involving simple translation (with no rotation) assuming the setup allows them to do it easily enough.

One such solution would be:

enter image description here The arrow marks show the translation to be done; the green line is the new laser beam.
Please excuse the shabby and not-to-scale drawing, if any.

With the assumption of easy reversibility of translations (so that we can go back to the original configuration easily), we can take it a step further and use just one movement:

enter image description here The translation can be reversed to reach detector A. In effect this movement can be toggled to use detector A or B as required.

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  • $\begingroup$ While you're right that moving the mirror in your second solution is the easiest solution, it violates the instructions on the puzzle, requiring that you leave the mirrors that reflect to A alone. $\endgroup$ – Andrew Jan 3 at 14:46
  • $\begingroup$ For a lazy physicist the beam path calculations are basically for free ($k_B T \ln 2$ for each non-unitarily processed bit), which is a lot less than moving his arm to grab the second mirror^^ $\endgroup$ – A. P. Jan 4 at 0:53
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Possibly you could

do it with one mirror

if you

can position the mirror pointed to by the blue arrow as show below in blue in front of the mirror used by detector A and reflect directly to the detector (Thanks Dr Xorile)

Shown here

enter image description here

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  • $\begingroup$ But here you changed one of the mirrors that are used to direct the beam onto detector A. $\endgroup$ – A. P. Jan 1 at 21:53
  • $\begingroup$ sorry @A.P. .i'll clear up what i meant.... rot13(chg n qvssrerag zveebe va sebag bs gur bar hfrq sbe qrgrpgbe N) $\endgroup$ – SteveV Jan 1 at 21:58
  • $\begingroup$ Ok that would solve the aforementioned problem. But what I don't like about this solution is that you only direct stray light into detector B, not the whole beam. I will make this point clear in the question. $\endgroup$ – A. P. Jan 1 at 22:03
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    $\begingroup$ @A.P. I'm not sure what you mean. Why would it be stray light? if they are perfect mirrors, you will get the whole beam. What i'm not sure about is if there is a gap big enough to hit the second mirror by detector B at the right angle, but thats what I'm trying! i suggest using a low power laser to check :) $\endgroup$ – SteveV Jan 1 at 22:06
  • $\begingroup$ The "oblique reflection" made me think that you mean stray light. Can you show your solution with less wrinkeled beams and a real-size mirror? Because I actually tried my best to avoid exactly this solution. $\endgroup$ – A. P. Jan 1 at 22:10
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With some minor fidgeting, I'm pretty confident a setup like this can work well. Just move any unused mirror into the blue spot as shown:

idk

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    $\begingroup$ Creative idea, but unfortunately mirrors are not that creative. Keeping the reflected angle the same as the incident angle, the beam has the same direction after being reflected from two parallel or orthogonal mirrors. So the downwards tendency after the inserted mirror should be the same two mirrors later. $\endgroup$ – A. P. Jan 1 at 22:31
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I found a solution with 2 mirrors:

enter image description here

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  • $\begingroup$ As a lazy physicist who hates aligning mirrors, I'd remove the 2 mirrors vertically below B, and shift the one mirror from below A to the right. $\endgroup$ – The Photon Jan 2 at 19:57

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