# Bases of calculation

Light falls from a medium (n_{1}) onto a thin film (n_{2}) which is laying on a substrate (n_{3}).

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 n_{1}/n_{2} you get:*For the s-polarization:*

*For the p-polarization:*

N

_{1}= n

_{1}+ ik1 ; N

_{2}= n

_{2}+ ik

_{2}are the complex indices of refraction, the extinction coefficients k

_{1}and k

_{2}relate to absorption. α is the angle of incidence, β the angle of refraction and can be calculated using Snell’s law: N

_{1}sinα = N

_{2}sinβ. With normal incidence (α = 0°) and zero absorption the terms become very simple:

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 = 2n

_{2}dcosβ; 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+

^{1}⁄

_{2}) λ ; (m = 0,1,2,3,…)

The same formulas can be used for the interface n

_{2}/n

_{3}. So, the total reflected and transmitted light can be calculated by the angle of incidence, refractive indices and film thickness.

The reflection becomes zero for a wavelength λ by using a

^{λ}/

_{4}-thick monolayer and normal incidence if the medium is air (n

_{1}= 1) and n

_{2}= √

^{-}n

_{3}. 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 λ is to stack alternately two different coating materials with a layer thickness of

^{λ}/

_{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.