0

Today I woke up wondering, after I had thought about reflectors the day before, whether or not it is possible to trap a beam of light inside a box where the beam fails to escape it due to total internal reflection? I’m guessing the answer is no, but I’m wondering whether I can understand why not with high school physics.

This other post is on the same question but there is only one answer which says ”quantum tunneling” with a wikipedia link: With a box that has perfect mirrors on the inside would it be possible to trap light? However, I would like to know if that is the only way to explain it by reading up on quantum physics or whether or not there is a more down-to-earth explanation of the same phenomenon. Could classical physics not explain this phenomenon?

That same question was also marked as a duplicate of another one but I checked it out quickly and couldn’t find a satisfactory answer to this question.

Any help and links to further reading would be appreciated. Thank you in advance.

Max123456789
  • 51
  • 1
    You may be interested in looking into Optical Resonators and Cavity Resonators; I'm not sure if this addresses what you mean by trapping the light, but it certainly holds the EM energy in a confined space. – xXx_69_SWAG_69_xXx Feb 16 '21 at 19:14

1 Answers1

1

Yes, it is possible. Optical fibers confine light via total internal reflection, with very low losses. Still, the losses are enough that the light intensity decays substantially after a few tens of milliseconds.

S. McGrew
  • 24,774
  • Thank you for your answer, are the losses due to the fact that not all reflections are ’total internal reflections’ or more so that there really are no ideal ’total internal reflections’ to begin with? Meaning that what we call total internal reflection doesn’t actually mean that ALL light is reflected but rather that 99.9999% is reflected? – Max123456789 Feb 16 '21 at 19:20
  • This answer is an oxymoron. The right answer is: 'no, it isn't possible' because reflectivity is always below one, no matter how close to one it actually is. – Gert Feb 16 '21 at 19:35
  • Losses are typically due to scattering from nonuniformities and impurities in the fiber or its cladding. TIR is TIR if the surfaces and materials are perfect; but perfection is very rare in reality. – S. McGrew Feb 16 '21 at 19:41
  • Also, no material is perfectly transparent. I don't know what the attenuation length of the best optical fibres is, though. – PM 2Ring Feb 16 '21 at 20:21