during reflection of light energy transfer is not 100%. when light from a rarer medium like air strikes a denser medium like glass slab, some part of light is reflected back and some of it is refracted.

But during total internal reflection (TIR) when a beam of light from a denser medium like water strikes a rarer medium like air, it is 100% reflected and there is no loss of energy. why?

what is the reason that no energy is lost in TIR but lost in glass slab?

  • $\begingroup$ Don't make edits which entirely change the subject matter of the question, like the one you did here. It invalidated my answer and subjected it to downvotes. $\endgroup$ Oct 30, 2015 at 18:19
  • $\begingroup$ Also please check out the help center for asking and answering questions, and associated reputation changes therein. $\endgroup$ Oct 30, 2015 at 18:20

2 Answers 2


Firstly, your point about (look into the tag wiki for further info about the phenomenon) resulting in 100% transmission of light energy and hence 100% intensity of incident light being preserved during reflection is conventionally considered true, but is not entirely correct.

Let me explain:

Reflection of any type, even if it is total internal reflection, cannot reflect 100% of the incident light energy. Total Internal Reflection nearly 100% (99.9999% maybe, but not 100%) efficiency compared to conventional reflection from a surface.

An interface between 2 optical media has the property of a "critical angle", the angle of incidence beyond which if light is incident on the interface from the optically denser medium into the optically rarer medium, it will be reflected back entirely into the denser medium, with no refraction into the rarer medium. Normally, at an interface between 2 transparent media of different optical densities, the light wave will be partially reflected into the medium from which it was incident and partially refracted into the medium into which it was incident, until the angle of incidence goes beyond this critical angle. This phenomenon of total internal reflection is usually observed for light, but is also applicable for other types of waves, like sound and waves on a string. This phenomenon is explained by Huygen's Wave Theory of Light in a classical sense. It occurs at the interface between 2 media of different densities in which the wave travels.

Hence, the intensity of incident light is preserved nearly entirely in total internal reflection, except for that minor fraction which may be absorbed by the denser medium itself, which is low due to the medium being transparent. There is also some energy loss due to photon tunneling across the interface, but again, it is not conventionally significant (total about $10^{-3} \%$). There are also across the interface, but they do not result in net energy transfer across the interface. Refer Wikipedia here.

In conventional reflection, 2 surfaces are involved:

1) A transparent unsilvered surface

2) An opaque silvered surface

For light transmission through the unsilvered surface, similar energy losses are applicable as in case of total internal reflection.

However, for light reflection at the silvered surface, the surface being opaque, will absorb a significant portion (say about 1%-2%) of the incident light energy, while it will reflect most of the light energy due to it being silvered and reflective.

This 2nd significant energy loss of incident light at the silvered surface is not applicable for total internal reflection, hence we commonly say that total internal reflection forms images at 100% the brightness (intensity) of the incident light.

  • $\begingroup$ when light falls on any dense medium like glass, some of light is reflected and some is refracted but in TIR only reflected none refracted or lost. you said negligible loss due to refraction. why is it negligible and not major? this is what I want to know $\endgroup$ Oct 30, 2015 at 9:52
  • $\begingroup$ BTW, what is the best rate of internal reflection known ? NB: counting absorption on the way in the media is not fair, we should consider only a neighborhood of one wavelenght (or half ?) around the interface. BTW², what are the first obstacle for perfect internal reflection (once already near perfect) ? the quality of the interface itself ? (perfectly flat, perfect nano-joints of materials) The materials themselve ? something else ? $\endgroup$ Oct 30, 2015 at 14:08
  • 1
    $\begingroup$ Energy loss in TIR is purely at the quantum level, i.e. photons "tunneling" thru a disallowed physical region. I don't think it's accurate to say that TIR is not 100% even if you're claiming that momentum conservation transfers energy to the medium (glass) itself. At the very least, you should mention the equations which show why TIR occurs. $\endgroup$ Oct 30, 2015 at 14:13
  • $\begingroup$ @CarlWitthoft It's arguable that it's not purely a quantum phenomenon. There is a classical evanescent wave near the interface, and energy can be absorbed from the evanescent wave. OTOH, I don't understand the answer that we're commenting on. It's not very clear, so maybe your comments are applicable to a better understanding of the answer than I can muster. (answer down-voted) $\endgroup$
    – garyp
    Oct 30, 2015 at 14:39
  • $\begingroup$ If any values mentioned are off please edit the answer to include the correct values. I was rushed for time when I first wrote the answer and have now made it more thorough. $\endgroup$ Oct 30, 2015 at 16:08

I got my answer.

When light enters in glass from air, it likes to reflect back by wave front theory (although 1-2%). In case of from denser to rarer medium first below critical angle some part of it is reflected back and some is refracted, but above critical angle some is reflected and other part (which was earlier refracted) is even reflected back causing 100% reflection back.

  • 3
    $\begingroup$ I addressed this in my answer, including your question before the edit to it, which ended up changing the subject matter. Why have you added an answer which is already in an answer to the question? $\endgroup$ Oct 30, 2015 at 18:04
  • $\begingroup$ my approach is different.I had tried to explained why no part is refracted out (100% conservatiob of energy) while you had just explained TIR. $\endgroup$ Oct 30, 2015 at 18:12
  • $\begingroup$ No. It is absolutely the same, probably a bit less technical than my answer. It had been downvoted for lack of citations, which has been since remedied. also, if you could find the naswer yourself, why ask the question? You should close the question. Also, please bother reading through the answer and associated links therein. $\endgroup$ Oct 30, 2015 at 18:14
  • $\begingroup$ 12 class student. my teacher asked to search on this as he was also confused. $\endgroup$ Oct 30, 2015 at 18:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.