Nov 27, 2014

This note discusses the single-layer antireflection (AR) coating for arbitrary angles of incidence. The incident medium N1 and the AR coating layer N2 are assumed lossless. The substrate medium N3 may be lossy. Synthesis calculators are provided for computing the AR coating layer index N2 and thickness T2 and the formal mathematics is included.

The layer refractive index N2 and thickness T2 are different for TE (s pol) and TM (p pol) cases except for normal incidence θ=0. For normal incidence, the results reduce to the well-known N2 = Sqrt(N1*N3) and a layer thickness of 1/4 optical wavelength in N2. For non-zero angles of incidence, the layer thickness is half an optical thickness in the x direction, that is, the layer thickness is greater by a factor of 1/cos(θ2) since the phase front in N2 is tilted by this angle, and the phase delay normal to the layer for the AR condition must always be π. If N3<N1, θ must be less than the total internal reflection (TIR) angle sin

For comparison purposes, the calculator below can be used to compute reflectance and transmittance for any single-layer and substrate indices and layer thickness h, including losses in N2 and N3. If the incident medium n1 is lossy, meaningful R, T power reflectance values with (R + T )= 1 cannot be defined in terms of power in an incident and reflected wave (Macleod 1986).

It is easy to see that there is a solution for a lossless coating with zero reflectance value even with a lossy substrate by inspecting the general 3 region reflectivity equation:

where r12 and r23 are the Fresnel interface complex reflection coefficients for the respective interfaces. For most values of N3, we can find a real index N2 so that the magnitudes of r12 and r23 are equal (for the lossless case at normal incidence, that index is simply the geometric mean of N1 and N3). However, since r23 is complex (for an absorbing substrate), the phase of r23 will not be zero (or 180°). Therefore the phase delay introduced by the (summed) return reflections in the layer will not cancel for the usual "quarter wavelength normal to layer" condition. However the layer thickness can simply be reduced to compensate for the phase angle of r23 and this will provide an ideal zero reflectivity condition. Furthermore, if the substrate loss is small enough, it is a very good approximation to take the layer index N2, for any θ, as that for the lossless case at that θ, and the layer thickness simply lowered to get the correct phase delay.

The contour plot below shows a reflectance map with a lossy substrate for the P pol. case with three AR coating solutions shown. Two of the solutions are for the same N2=2.21 value between N1 and N3 and are separated in thickness by half an optical thickness normal to the layer. The other solution is N2 = 1.16, lower than both N1 and N3 as discussed below:

where consecutive zeros for a given N2 differ in thickness by:

The P polarized case has two different AR coating solutions N2p when θ >0. The first N2 solution (+) always has a value between N1 and N3. The second solution (-) has a value that is less than both N1 and N3 if the incident angle is less than the Brewster angle for the N1/N3 interface, and has a value between N1 and N3 if θ is greater the Brewster angle.

Assume that the incident medium N1 and the AR coating layer N2 are lossless but medium N3 = (N3_re + jN3_im) may be lossy with N3_im <=0.

With XZ being the plane of incidence, θ

and the propogation constant along the interfaces and normal components in the layers are:

so that:

where N3 and therefore θ3 and kx3 are in general all complex.

For a given solution N2 (or kx2) at a specific angle of incidence, as in the lossless case, there are consecutive AR coating thicknesses

but the actual thickness values T2 are shifted from the lossless values for a lossy N3 due to the phase shift of r23.

A real solution for kx2 and therefore N2 will exist if:

For normal incidence, the solution for both the S and P polarization cases reduces to:

and a real N2 solution exists if:

η1 and η3 values are given. η2 is determined first and then N2 is obtained from the quadratic in N

A real solution for kx2 and therefore N2 will exist if:

If the conditions for a REAL AR solution for N2 are not satisfied (either S or P case), an AR solution may exist, but with a complex (lossy) N2 value. In this more complicated situation with attenuation in the layer, numerical computation can be used.

**Optical Properties of Thin Solid Films**, O. S. Heavens, 1965, Dover**Thin-Film Optical Filters**, H. A. Macleod, 2nd Edn., 1986, Adam Hilger Ltd., Bristol pp 28-29**Principles Of Optics**, M. Born and E. Wolf, 5th Edn. 1975, Pergamon Press, pp. 61-63**Electromagnetic Theory**, J. Stratton, 1941, McGraw Hill

**Field Theory of Guided Waves**, R. E. Collin, 1991, IEEE Press

**Fields and Waves in Communication Electronics**, S. Ramo, J. Whinnery, T. Van Duzer, 1984, J. Wiley & Sons