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I have heard there are none right now...but i saw something that said something about 100% reflection. Forgotten completely. sorry.

I want it, i think it'd be very amazing to save sunlight in it from day and let little out in night.

But i have heard it's theoretically not possible to have surface that's 100% reflective. I don't believe that. Would it be possible to reflect light without absorbing or normal reflection like using some technique to bend it 180degrees.

explain please...

Qmechanic
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Muhammad Umer
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3 Answers3

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Well there is one well known situation where 100% reflectance does occur; and that is the case of Total Internal Reflection (TIR). It occurs when light in a medium with refractive index (N1) , impinges on a boundary with a second medium of refractive index (N2) where N2 < N1 and the angle of incidence on the boundary is greater than arcsin(N2/N1) , which is called the critical angle. It derives naturally from Snell's law of refraction: N1Sin(I1) = N2sin(I2) and Sin (I2) cannot exceed (1).

It is employed often in optical systems such as the prisms in binoculars.

  • Not to forget Fiber Optic cables too – mcodesmart Oct 11 '13 at 22:22
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    But even total internal reflection is not perfect because the materials are always absorbing and not ideally lossless. I think this idealized case is what the question refers to. – Andreas H. Oct 12 '13 at 00:35
  • """"".....is there 100% reflective mirror…i just want one.....""""" Well that is exactly what the question refers to. Says not a word about other unrelated loss mechanisms. TIR IS Total reflection; 100% –  Oct 12 '13 at 00:46
  • Besides, the entire optical system could be completely inside the higher index medium; including radiation sources and detectors. –  Oct 12 '13 at 00:57
  • Well, if absorption does not count than any metal mirror would be ideal, wouldnt it? Same thing with TIR, which will not be ideally 100% reflective if the outer material is lossy (even given the case that the higher index material is completly lossless) (due to evanescent fields). To me no difference to a normal mirror where the reflecting material (metal) is lossy. – Andreas H. Oct 12 '13 at 03:19
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    Remember the question is about how to store light from the day for the night! Or could a cavity based on TIR in a hypothetical case store light for hours, if yes in which? – Andreas H. Oct 12 '13 at 03:25
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    I quoted verbatim, EXACTLY what the OP's question is; what you just stated is NOT a question, but the OP's musings. I'm not going to waste time on troll nitpicking. –  Oct 12 '13 at 19:26
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In Misha Lukin’s group at Harvard, they developed a technique called atomic array. Generalizing it to 2D, another PI at Harvard Susan yellin proposed that using this 2D array there could be a perfect mirror.

RoderickLee
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Classically you have been given references in the other answer. After all the black body simulation as a cavity filled with photons assumes total reflection.

Mirrors and reflections though, when one is talking of limits, as is the case of total reflection, are quantum mechanical objects.

Quantum mechanics means

a) that the Heisenberg uncertainty principle holds .This means that there is always a width of acceptable angles

b)that reflections are the quantum mechanical solution of the electromagnetic interaction with the electrons on the shells of the mirror. By nature, when one has an interaction at the quantum mechanical level there is always a partial solution giving other results than total reflection. These will be with very low probability, but considering you are thinking of trapping light for at least 24 hours this means there are zillions of interactions and then the absorption probability adds up.

So it all depends on probabilities and time constants. My calculation tools are rusty but I guess that even with a clever focusing device the photons would be absorbed within nanoseconds, raising the temperature of the walls of your cavity.

anna v
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  • Anna, The cavity derivation of "blackbody" radiation spectrum; REQUIRES that the cavity walls and the cavity radiation are in thermal equilibrium. That means that they are at the same isothermal Temperature. It is required that the walls be 100% reflective at ALL frequencies or wavelengths, and also that the walls have zero thermal conductivity (or energy would be lost from the walls). No real material has those properties; even at any one single frequency. The alternative black body description requires a body with zero reflectance, and infinite thermal conductivity. None exists –  Oct 14 '13 at 06:32
  • @GeorgeE.Smith right. It is a classical electrodynamics and thermodynamics approach, and that is why it broke down at the ultraviolet, and photons had to be postulated. I am just stressing that when one goes to limits like 100% reflectivity quantum mechanics enters the game. – anna v Oct 14 '13 at 07:55
  • My point Anna, is that so-called "black body radiation", is a purely hypothetical concept, in that NO physical object has the properties required for the existence of such a device; yet it plays such an important role in modern physics. It is questionable as to just how "quantized" Planck's derivation is, since the allowable photon energies are in no way constrained to specific values. Yes, you can only have an integral number of photons, but any frequency and photon energy is allowed, even extremely large photon energies; but few of them. –  Oct 15 '13 at 20:35
  • what if you could bend light into circle that way it won't have to hit anything. http://news.sciencemag.org/2012/04/light-bends-itself – Muhammad Umer Nov 01 '13 at 04:38