2
$\begingroup$

Silver is metal with highest electrical conductivity.I read that it have highest optical reflectivity,that it shines most from all metals.

Is there link between conductivity and how shiny the metal appears? Please assume clean surface,no surface oxidation.

$\endgroup$
  • $\begingroup$ There are several kinds of transparent conductors, such as indium tin oxide and graphene. $\endgroup$ – S. McGrew Apr 22 '18 at 1:36
1
$\begingroup$

Materials with high index of refraction have higher reflectivity than materials with lower reflectivity.Because index of refraction varies with the wavelength of photons and so does reflectivity.

Also the metallic reflectance can be related to the conductivity by the Hagens-Ruben equation where ν is the light frequency, ε0 is the vacuum permitivity (8.85 x 10-12 F/m), and σ is the conductivity. In the infrared region (small frequencies), this equation shows that metals with high reflectance also are good conductors.

$R$=$1$-$4 \sqrt{\frac{\nu \pi \epsilon_0} {\sigma}} $

$\endgroup$
  • $\begingroup$ So in the infrared the retlectivity is correlated to conductivity but at higher frequencies in the visible and UV spectrum,it becomes more complex and its no longer correlated? $\endgroup$ – wav scientist Apr 22 '18 at 6:14
  • 1
    $\begingroup$ It's not possible to generalise because there are such a variety of material structures. You will have heard of Refractive Index which is often used to characterise the way (visible) Light is affected by transparent materials. RI in that case is a Real Number. To characterise many substances, you need the Complex Refractive Index (a complex number) and the predominant part of RI for metals is Imaginary; EM waves do not propagate through metals (i.e. mostly reflected) at optical frequencies. $\endgroup$ – Abhinav Apr 22 '18 at 16:58
  • 1
    $\begingroup$ By the time you get to X rays, both the real and imaginary parts of the RI are significant and metals stop being easy to describe. X rays and gamma rays will not interact with the charges that affect light and radio waves because the photon energies are too high $\endgroup$ – Abhinav Apr 22 '18 at 16:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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