A common method to prevent radio frequency and microwave energy from going through windows is to apply a microscopic layer of silver to the surface of the glass, either by attaching a film coated with the silver or by vapor deposition.

How does this film of silver affect the visible and near visible parts of the spectrum (ultraviolet and infrared)? Does it block any of this radiation and if so, by how much?

Is there any standard reference source that contains information like this?

  • 1
    $\begingroup$ Display screens in electronic devices often are coated with a film of Indium Tin Oxide (ITO) as a means to minimize radio frequency emissions. An ITO layer is both electrically conductive and, unlike silver, highly transparent to visible light. $\endgroup$ Commented Jul 31, 2018 at 22:18
  • $\begingroup$ But, look up the optical constants of silver. It is a good metal with high reflectivity up into the near UV once the plasma frequency is exceeded. $\endgroup$
    – Jon Custer
    Commented Jul 31, 2018 at 22:21
  • $\begingroup$ Dear Ambrosse a response or something is appreciated ;) $\endgroup$ Commented Aug 6, 2018 at 18:35

1 Answer 1


So yes there is a very extensive library for this, at https://refractiveindex.info/

However you do need to have a tiny bit of knowledge about optics. Refractive index, usually denoted by $n$, indicates how much light waves get "slowed" down and thus can tell you the diffraction angle when passing through media. (glass is about 1.5 in the visible range, water ~1.3)

Then you have the extinction coefficient, $k$, which indicates how well a material transmits light. You also have a similar value the absorption coefficient, $\alpha$, which is a bit easier to use.

Keep in mind that both values are wavelength dependent, the website also often lets you calculate the transmittance for a specific wavelength and material thickness.

Attenuation is usually $ Attenuation = e^{-\alpha * thickness} $ (e-base). To get from $k$ to $\alpha$ you can use this $\alpha=\frac{4\cdot \pi\cdot k}{\lambda} $. (with $\lambda$ being the wavelength)


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.