Timeline for How does this trick with mirrors work?
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Mar 3, 2014 at 0:04 | comment | added | DumpsterDoofus | @MasonWheeler: Retroreflectors make you see yourself because $\sigma_x\sigma_y\sigma_z\mathbf{v}=-\mathbf{v}$ for any vector $\mathbf{v}$, where the $\sigma_\alpha$ are reflection operators. | |
| Apr 17, 2013 at 11:14 | comment | added | Mason Wheeler | All right, so it's a retroreflector. But, as far as I can see at least, neither this answer nor the linked Wikipedia article actually answers the question: When I look into a 90 degree retroreflector, no matter where I'm looking into it from, why do I always see a reflection of myself at the point where the two mirrors touch? | |
| Apr 17, 2013 at 4:47 | comment | added | Brandon Enright | @MasonWheeler okay what you describe is exactly the 1 right-angle mirror the image. | |
| Apr 17, 2013 at 4:21 | comment | added | Mason Wheeler | Do you have a medicine cabinet in your bathroom, with multiple mirrored doors? Open one of them out to 90 degrees and stand somewhere inside the 90 degree arc. No matter where you move, you will always see half of your face on either side of the line where the two mirrors intersect, in addition to the "straight" reflections you'd see from looking into the mirror directly. That's not a diagram, but it's a very simple way to understand it. | |
| Apr 17, 2013 at 4:06 | history | answered | Brandon Enright | CC BY-SA 3.0 |