# Refractive index of extremely thin film

I need to calculate the reflectivity of a thin film, designed for one wavelength $$\lambda_1$$, at a different wavelength $$\lambda_2$$. The challenge is that $$\lambda_2 \gg \lambda_1$$, for example a factor of 100, so that $$d \ll \lambda_2$$, ($$d$$ is the film thickness). For example, would a $$d=\lambda/1000$$ film still reflect light? And if it would, then what is the refractive index of such thin film? Can I assume that it's the same as for a thicker film? Down to which film thickness would this approximation hold? Is there perhaps something like effective refractive index (as with optical fibres)?

• Generally, the formulae hold well under the condition that the various films are planar and continuous. Your bigger problem may be that the optical constants are likely different, the only question being how different. – Jon Custer Dec 19 '19 at 13:59
• When the medium is the same on both sides, the phase changes for reflection at both sides are opposite, so for a film much thinner than the wavelength both reflected waves will almost cancel (the "black" part of soap films). – Pieter Dec 19 '19 at 14:00
• whats the formula you are using to calulate reflectivity at a given $\lambda$? – lineage Dec 19 '19 at 14:34
• And if it would, then what is the refractive index of such thin film? but you just said that the film has been designed for another $\lambda$ Doesn't that mean you have already fixed $\mu$? – lineage Dec 19 '19 at 14:36

• Say we have blue light $\lambda=400nm$, if the film thickness is $\lambda/1000 = 400pm$ that is less than 1 atomic layer of quartz. Are you sure that Fresnel's formulas are still valid? What is the permittivity of such material? – hyportnex Dec 19 '19 at 16:10
• Obviously the refractive index will vary with wavelength (dispersion). Is that what you mean? Or do you mean that a ~2.5 um film ($\lambda/4$ for 10 um) would have a different refractive index (at the same wavelength) than a ~5 nm film ($\lambda/4$ for 20 nm), due to some topological effects? – texnic Dec 19 '19 at 19:09
• @texnic when you’re using the Fresnel formulas, you’ll plug in wavelength ($\lambda$), the film thickness ($d$), and refractive index ($n(\lambda)$, which is a function of wavelength). Assume the refractive index is constant with film thickness (which is more-or-less accurate). – Gilbert Dec 19 '19 at 19:23