May there be materials yet to be discovered which may have a higher refractive index than today's known materials (for wavelengths within the visible range)?

Is there a theoretical limit for the refractive index of a material?


2 Answers 2


Theoretically, there is no limit to the refractive index. The reason is that, if you go by the definition, $n= c/v$, the more you can slow down light (short of stopping it completely), the higher you refractive index will be. And, mathermatically, we are looking at the following,

$$ n = \lim_{v \to 0^{+}} \frac{c}{v} = \infty $$

and is undefined at 0, which is why the limit is coming from the left.

For example, using a cloud of cold atoms (laser cooled), light can be slowed down light to less that 10 mph. See link.


Practically, there is a limit to refraction imposed by the nature of the refractive medium itself and the nature of the condensed state. In terms of materials, there are advances using metal arrays to increase the refractive index even more. See link.


  • $\begingroup$ My argument is the exact same argument. Yours is better put though. +1 $\endgroup$ Jan 8, 2014 at 22:22
  • $\begingroup$ Thanks! 38.6, while far from infinity, is still amazing (for non-gas). $\endgroup$
    – j-a
    Feb 18, 2014 at 8:49

Since the refractive index is given by $\displaystyle{n_{12}=\frac{\sin \theta_1}{\sin \theta_2}}$, theoretically there is no limit at all on the value of refractive index. You could say that it must be positive, but then check this out: http://en.wikipedia.org/wiki/Negative_refraction

  • 5
    $\begingroup$ Is this Snell's Law? If so, the logic is backward. Just because you can imagine incident and refracted angles to be anything doesn't mean a material must exist that bends light in that way. $\endgroup$
    – user10851
    Jan 8, 2014 at 18:00
  • 1
    $\begingroup$ agreed! This is not the definition but a consequence of the proper definition. This argument is therefore flawed. $\endgroup$
    – mcodesmart
    Jan 8, 2014 at 18:35
  • $\begingroup$ @ChrisWhite, hence the theoretically. Or did you not bother to read that? $\endgroup$ Jan 8, 2014 at 22:20

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