In a discussion with a friend, he seemed to be saying that reflection of light happens because of the difference in refractive index between two media, implying that that is the only reason reflection ever happens.

Now, I understand that when light crosses an air/glass boundary, some light is reflected because of the change in refractive index.

But what about when light reflects from polished silver? Does the light reflect because silver has a very different refractive index to air, or is there some other mechanism happening?


1 Answer 1


Yes, silver reflects because of the difference of its refractive index from that of air. However, silver is different from air or vacuum in that it is a lossy medium: it is conductive, so electromagnetic waves propagating in it induce currents, thereby losing energy and decaying in amplitude. This is conveniently modeled by a complex refractive index: $$\tilde n= n-i\kappa$$ where $\kappa$ is called the extinction coefficient. It describes how "lossy" the medium is.

With refractive index defined as such, the reflectance of an interface is given by Fresnel's equations. For normal incidence on the interface of media $1$ and $2$, the reflectance is

$$R = \left|\frac{\tilde n_2 - \tilde n_1}{\tilde n_2 + \tilde n_1}\right|^2.$$ For incidence from air on silver, $$R = \frac{\kappa^2+(1-n)^2}{\kappa^2+(1+n)^2}$$ where $n$ and $\kappa$ refer to the refractive index of silver. Silver is highly reflective at visible wavelengths in part because of its large extinction coefficient, and in part because of its small real refractive index.

Intuitively, since silver is highly conductive, the incident electromagnetic radiation can freely move charge within it, so it induces a current at the surface, which "shields" the rest of the material and keeps the wave from penetrating deep into silver. This current also emits radiation that appears to reflect off the surface in the direction you would expect.

  • $\begingroup$ Is the extinction really due to losses? Isn't the exponential decay of amplitude simply due to evanescent mode of propagation, rather than due to absorption? $\endgroup$
    – Ruslan
    Commented Jun 23, 2020 at 21:41
  • 1
    $\begingroup$ +1 on this answer and question for introducing me to the complex refractive index. $\endgroup$ Commented Jun 23, 2020 at 21:49
  • $\begingroup$ Question: Is “lossy” a typo of glossy or is it an adjective of loss/lose? $\endgroup$ Commented Jun 23, 2020 at 21:52
  • $\begingroup$ @Ruslan A medium with a positive extinction coefficient is indeed lossy. Waves do lose power as they propagate in a lossy medium, usually in the form of heat. Evanescent waves that occur e.g. in the case of total internal reflection at the interface of two dielectrics is different in that while the evanescent wave has an exponentially decaying amplitude, no power is lost to the medium. $\endgroup$
    – Puk
    Commented Jun 23, 2020 at 21:56
  • $\begingroup$ @gen-zreadytoperish I did mean lossy: the medium absorbs power from the wave. $\endgroup$
    – Puk
    Commented Jun 23, 2020 at 21:57

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