## Abstract

The absolute optical thickness of a transparent plate 6-mm thick and 10 mm in diameter was measured by the excess fraction method and a wavelength-tuning Fizeau interferometer. The optical thickness, defined by the group refractive index at the central wavelength, was measured by wavelength scanning. The optical thickness deviation, defined by the ordinary refractive index, was measured using the phase-shifting technique. Two kinds of optical thicknesses, measured by discrete Fourier analysis and the phase-shifting technique, were synthesized to obtain the optical thickness with respect to the ordinary refractive index using Sellmeier equation and least-square fitting.

© 2015 Optical Society of America

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### Equations (17)

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(1)
$${n}_{1,2}T=\frac{{\lambda}_{1,2}}{2}\left({N}_{1,2}+{p}_{1,2}\right),$$
(2)
$$\frac{{n}_{1}+{n}_{2}}{2}\left(1-\frac{{\lambda}_{1}+{\lambda}_{2}}{{n}_{1}+{n}_{2}}\cdot \frac{{n}_{2}-{n}_{1}}{{\lambda}_{2}-{\lambda}_{1}}\right)T=\frac{{\lambda}_{1}{\lambda}_{2}}{2\left({\lambda}_{2}-{\lambda}_{1}\right)}\left({N}_{1}-{N}_{2}+{p}_{1}-{p}_{2}\right).$$
(3)
$${n}_{g}\left({\lambda}_{c}\right)=\frac{{n}_{1}+{n}_{2}}{2}\left(1-\frac{{\lambda}_{1}+{\lambda}_{2}}{{n}_{1}+{n}_{2}}\cdot \frac{{n}_{2}-{n}_{1}}{{\lambda}_{2}-{\lambda}_{1}}\right)\approx n\left(1-\frac{\lambda}{n}\cdot \frac{\text{d}n}{\text{d}\lambda}\right).$$
(4)
$${\left[{n}_{g}\left({\lambda}_{c}\right)T\right]}_{meas}=\frac{{\lambda}_{s}}{2}\left({N}_{1}-{N}_{2}+{p}_{1}-{p}_{2}\right).$$
(5)
$${n}^{2}-1=\frac{0.6961663{\lambda}^{2}}{{\lambda}^{2}-{0.0684043}^{2}}+\frac{0.4079426{\lambda}^{2}}{{\lambda}^{2}-{0.1162414}^{2}}+\frac{0.8974794{\lambda}^{2}}{{\lambda}^{2}-{9.896161}^{2}}.$$
(6)
$${\left({n}_{1}T\right)}_{meas}=\frac{{n}_{1}}{{n}_{g}}\cdot {\left({n}_{g}T\right)}_{meas}.$$
(7)
$${p}_{1}=\frac{1}{2\text{\pi}}\mathrm{arc}\mathrm{tan}\frac{{\displaystyle \sum _{r=1}^{77}{w}_{r}{I}_{r}\mathrm{sin}\frac{\text{\pi}r}{10}}}{{\displaystyle \sum _{r=1}^{77}{w}_{r}{I}_{r}\mathrm{cos}\frac{\text{\pi}r}{10}}},$$
(8)
$$\begin{array}{l}{w}_{r}=\left[\frac{1}{6}r\left(r+1\right)\left(r+2\right)\right]\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(1\le r\le 20\right),\\ {w}_{r}=\left[5340+\frac{1}{2}\left|r-39\right|\left({\left|r-39\right|}^{2}-40\left|r-39\right|-1\right)\right]\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(21\le r\le 57\right),\\ {w}_{r}=\left[\frac{1}{6}\left(80-r\right)\left(79-r\right)\left(78-r\right)\right]\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(58\le r\le 77\right).\end{array}$$
(9)
$$F\left(f\right)=\sqrt{{\left\{\left[{\displaystyle \sum _{j=1}^{593}{I}_{j}h\left(j\right)\mathrm{cos}2\text{\pi}f\left(j-1\right)}\right]/593\right\}}^{2}+{\left\{\left[{\displaystyle \sum _{j=1}^{593}{I}_{j}h\left(j\right)\mathrm{sin}2\text{\pi}f\left(j-1\right)}\right]/593\right\}}^{2}},$$
(10)
$$h\left(j\right)=\frac{2}{297}{\mathrm{cos}}^{2}\frac{j-297}{593}.$$
(11)
$$M=\text{round}\left({N}_{1}-{N}_{2}+{p}_{1}-{p}_{2}\right)$$
(12)
$${\left({n}_{1}T\right)}_{dev}=\frac{{\lambda}_{1}}{4\text{\pi}}\cdot \text{unwrap}\left[{p}_{1}\left(x,y\right)\right],$$
(13)
$${n}_{1}{T}_{0}=\frac{{\lambda}_{1}}{2}\cdot {A}_{1}.$$
(14)
$$\frac{1}{P}{{\displaystyle \sum _{i=1}^{P}\left\{{\left[{\left({n}_{1}T\right)}_{meas}\right]}_{i}-{\left[{\left({n}_{1}T\right)}_{dev}\right]}_{i}-{n}_{1}{T}_{0}\right\}}}^{2}=\mathrm{min},$$
(15)
$${n}_{1}T={n}_{1}{T}_{0}+{\left({n}_{1}T\right)}_{dev}.$$
(16)
$${N}_{1}=\text{round}\left(\frac{2{n}_{1}T}{{\lambda}_{1}}-{p}_{1}\right),$$
(17)
$$\frac{\partial \left({n}_{1}T\right)}{{n}_{1}T}=\frac{\partial {\lambda}_{1}}{{\lambda}_{1}}+\frac{\partial \left({N}_{1}+{p}_{1}\right)}{{N}_{1}+{p}_{1}}\approx \frac{\partial {\lambda}_{1}}{{\lambda}_{1}}+\frac{\partial {p}_{1}}{{N}_{1}}.$$