## Abstract

An orthogonal form of a general three-term dispersion equation is presented which is useful for determination of the refractive index of transparent optical materials from measured spectral data. The orthogonal basis functions are dependent on the spectral range of the data and whether the spacing of the data is uniform with respect to wavelength or wavenumber. The application of the technique to the reduction of ellipsometric measurements of a single layer thin film of zirconia on a fused silica substrate is presented. The measured data were fit to a Conrady dispersion equation and to the corresponding orthogonal basis set. The convergence to the solution was an order of magnitude faster for the orthogonal form of the dispersion equation.

© 1989 Optical Society of America

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### Equations (34)

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(1)
$$n(\lambda )=A+B{\lambda}^{-2}+C{\lambda}^{-4},$$
(2)
$$\text{MF}={\displaystyle \sum _{i}{\left[{n}_{i}-n\left({\lambda}_{i}\right)\right]}^{2}}$$
(3)
$$\text{MF}={\displaystyle \sum _{i}{\left[{\Delta}_{i}-\Delta \left({\lambda}_{i}\right)\right]}^{2}+{\left[{\Psi}_{i}-\Psi \left({\lambda}_{i}\right)\right]}^{2}},$$
(4)
$$n(\lambda )=A+B{\lambda}^{p}+C{\lambda}^{q}.$$
(7)
$${\chi}_{2}={\lambda}^{p}-b,$$
(8)
$${\chi}_{3}={\lambda}^{q}-c{\lambda}^{p}+d.$$
(9)
$$0={\displaystyle {\int}_{{\lambda}_{1}}^{{\lambda}_{2}}{\chi}_{i}(\lambda ){\chi}_{j}(\lambda )d\lambda}.$$
(10)
$$0<{\lambda}_{1}<{\lambda}_{2}<\infty .$$
(11)
$${Z}_{m}\left({\lambda}_{1},{\lambda}_{2}\right)\equiv \frac{1}{{\lambda}_{2}-{\lambda}_{1}}{\displaystyle {\int}_{{\lambda}_{1}}^{{\lambda}_{2}}{\lambda}^{m}d\lambda}$$
(12)
$$\begin{array}{ll}{Z}_{m}\left({\lambda}_{1},{\lambda}_{2}\right)=\frac{1}{\left(m+1\right)}\left(\frac{{\lambda}_{2}^{m+1}-{\lambda}_{1}^{m+1}}{{\lambda}_{2}-{\lambda}_{1}}\right)\hfill & \text{for}\phantom{\rule{0.2em}{0ex}}m\ne -1\hfill \end{array}$$
(13)
$$\begin{array}{ll}{Z}_{m}\left({\lambda}_{1}{,\lambda}_{2}\right)=\text{ln}\left({\lambda}_{2}/{\lambda}_{1}\right)/\left({\lambda}_{2}-{\lambda}_{1}\right)\hfill & \text{for}\phantom{\rule{0.2em}{0ex}}m=-1\hfill \end{array}.$$
(15)
$$c=\left({Z}_{q}{Z}_{p}-{Z}_{p+q}\right)/D.$$
(16)
$$d=\left({Z}_{2p}{Z}_{q}-{Z}_{p}{Z}_{p+q}\right)/D.$$
(17)
$$D={Z}_{p}^{2}-{Z}_{2p}.$$
(18)
$$n(\lambda )={A}^{\prime}{\chi}_{1}+{B}^{\prime}{\chi}_{2}+{C}^{\prime}{\chi}_{3},$$
(19)
$$A={A}^{\prime}-{B}^{\prime}b+{C}^{\prime}d,$$
(20)
$$B={B}^{\prime}-{C}^{\prime}c,$$
(22)
$$n\left(\sigma \right)=A+B{\sigma}^{p}+C{\sigma}^{q},$$
(24)
$${\chi}_{2}={\sigma}^{p}-b,$$
(25)
$${\chi}_{3}={\sigma}^{q}-c{\sigma}^{p}+d.$$
(26)
$$0={\displaystyle {\int}_{{\sigma}_{1}}^{{\sigma}_{2}}{\chi}_{i}\left(\sigma \right){\chi}_{j}\left(\sigma \right)d\sigma},$$
(27)
$${\sigma}_{1}=1/{\lambda}_{2},$$
(28)
$${\sigma}_{2}=1/{\lambda}_{1}.$$
(29)
$$b=1/{\lambda}_{1}{\lambda}_{2},$$
(30)
$$c=\left(3a-2\right)/\left({5\lambda}_{1}{\lambda}_{2}\right),$$
(31)
$$d=\left(4a-1\right)/\left({15\lambda}_{1}^{2}{\lambda}_{2}^{2}\right),$$
(32)
$$a={\left({\lambda}_{1}+{\lambda}_{2}\right)}^{2}/{\lambda}_{1}{\lambda}_{2}.$$
(33)
$$n\left(\sigma \right)=A+B\sigma +C{\sigma}^{3.5}.$$
(34)
$$b=\left({\sigma}_{1}+{\sigma}_{2}\right)/2.$$