Why no transparent materials with large refractive index?

Click For Summary
The refractive index of materials is influenced by their dielectric constant and magnetic permeability, with few transparent materials exceeding an index of 4. While some materials, like silicon, can achieve high refractive indices at specific wavelengths, they often exhibit significant absorption, limiting their transparency. The Kramers-Kronig relations connect the refractive index to the absorption spectrum, revealing that high indices near resonances come with increased absorption. Sum rules restrict the average refractive index over frequency ranges, complicating the existence of materials that are both transparent and have high refractive indices. Achieving a high index while maintaining transparency typically requires low temperatures to minimize spectral broadening.
jfizzix
Science Advisor
Homework Helper
Insights Author
Messages
757
Reaction score
356
I know that the refractive index is determined by a material's dielectric constant and magnetic permeability.

It's also true that we can treat the refractive index as a complex function with the imaginary part giving you an absorption spectrum.

You can then get the index of refraction from the absorption spectrum with the Kramers-Kronig relations.

My question is, what makes it that there are so few if any transparent materials with an index of refraction greater than 4?
 
Science news on Phys.org
How many materials irrespective of transparancy have an index that high?
 
Silicon at 500 nm has a refractive index of 4.3 or so. Coming up with more specific examples would be a challenge, but there's a whole field of slow light optics for narrowband light sources tuned to particular spectral lines, say of Rubidium vapor
 
Long story short, there are lots of materials that have a high index at one wavelength or another, but not that also are transparent. I don't know why you can't have both, though
 
The index of refraction averaged over some frequency range is seriously restricted by sum rules. This does not preclude n to become very large near a resonance. However, near a resonance, you will also get absorption. The absorption follows a Lorentzian line shape, hence the imaginary part decays like ##1/(\Delta f)^2## with distance from the center of the line. On the other hand, the real part will decay like ##1/(\Delta f)## So for suffiently narrow absorption lines, you can get very close to the center of the line without getting appreciable absorption but a high index of refraction. In solid state this requires low temperatures to reduce spectral broadening.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
Replies
2
Views
1K
Replies
7
Views
20K
  • · Replies 16 ·
Replies
16
Views
6K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 1 ·
Replies
1
Views
7K
  • · Replies 7 ·
Replies
7
Views
5K
Replies
10
Views
6K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
56K