Bases of calculation


Light falls from a medium (n1) onto a thin film (n2) which is laying on a substrate (n3).

At each interface one part oft the light will be reflected and a se­cond transmitted. Every part can be calculated by the Fresnel equations. For the interface n1/n2 you get:

For the s-polarization:  

For the p-polarization:

N1 = n1 + ik1 ; N2 = n2 + ik2 are the complex indices of refraction, the extinction coefficients k1 and k2 relate to absorption.

α is the angle of incidence, β the angle of refraction and can be calculated using Snell’s law: N1sinα = N2sinβ.

With normal incidence (α = 0°) and zero absorption the terms become very simple:

The same formulas can be used for the interface n2/n3.

All reflected parts and the transmitted parts will interfere. The conditions for interfe­rence depend on the optical path difference OPD. For the reflected parts: OPD = 2n2dcosβ;  where d describes the phy­sical thickness of the layer.

The interference for a wavelength λ will be  constructive for OPD = mλ and destructive for OPD = (m+12) λ ; (m = 0,1,2,3,…)

So, the total reflected and transmitted light can be calculated by the angle of incidence, refractive indices and film thickness.

The reflection with normal incidence becomes zero for a wavelength λ by using a monolayer with an optical thickness of λ/4 if the medium is air (n1 = 1) and n2 = √n3. This is the easiest way to produce an anti-reflection coating. For substrates where it is not possible to find a coating material with matching refractive index you need more layers (see item "Number of layers").

If more than one layer is deposited the descri­bed process occurs at every interface between the layers and the total mathematical calculation becomes more difficult but it is still solvable.

The most effective way to get a high reflective coating for the wavelength l is to stack alternately two different coating materials with an optical layer thickness of l/4 each. With an increasing number of layer pairs the reflection will grow up steadily to 100%, only limited by the los­ses through absorption and scattering, while the transmission falls down to 0% (see item "Number of layers").

One frequently asked question is: How thick is a coating? To give you an estimation: A normal human hair has a thickness of about 50-100 µm, a dielectric mirror coating HR 1064 nm / 0° has about 5 µm and an EAR 248 nm / 0° about 0.05 µm.