I am not a physicist, but reading up on interference in thin films. I understood how the relation of two interfering light waves intensities $I_1$ and $I_2$ of a thin film (the first reflected at the air-film surface, the second at the film-substrate surface)

$I(\lambda) = I_1 + I_2 + 2\sqrt{I_1I_2} \cos (\epsilon_1 - \epsilon_2) $

with $\epsilon_1 - \epsilon_2$ being the phase shift between both waves due to the thin film can be derived. I also know that after further derivation, $nd = \frac{1}{2} \frac{\lambda_m \lambda_{m+1}}{\Delta \lambda}$ with $nd$ being the optical thickness and $\Delta \lambda$ being the difference between two maxima $\lambda_m$ and $\lambda_{m+1}$ in the reflection spectrum. This equation is used for determination of thin film thicknesses in thin film reflectometry using white light.

My question is whether this equation only holds assuming the intensities of the light source are equal for all wavelengths (white light) or if other light sources can be used as well. In addition, I am asking myself what happens if the film is absorbing some of the light? Also, some of the light has to be reflected at the air-film surface in order to get any interference. Does this mean that the intensity of interference depends on the reflectance of the surface, i.e. for no or total reflection, this method does not work?


1 Answer 1


The above equation holds only for one specific wavelength of light. However, you are correct that that absorption of the light is not expressly included. It would arise in 'n,' which is the index of refraction and the absorption would manifest itself as an imaginary term. So, in general: n = a + i b. Where 'a' is the usual index of refraction and 'b' is the absorption.

Note, my first statement concerning that the equation is for a single wavelength implies that n = n(wavelength 1). If there are multiple wavelengths present, as in white light, then n will be different (it's called dispersion) for each wavelength and a more general expression will have to be formulated.

  • $\begingroup$ Thanks, expressing the refractive index as a complex number is a good idea. $\endgroup$
    – Iridium
    Jul 28 at 18:34

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