## Abstract

A new method for analytically obtaining the form of an interferogram is proposed. The method permits simple implementation for a computerized treatment. We can use it to determine easily the thickness of a thin film.

© 1993 Optical Society of America

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### Equations (6)

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(1)
$$I\phantom{\rule{0.1em}{0ex}}(y)\phantom{\rule{0.2em}{0ex}}={a}^{2}\phantom{\rule{0.2em}{0ex}}{\left|\phantom{\rule{0.2em}{0ex}}{r}_{1}\frac{exp\phantom{\rule{0em}{0ex}}(ik{R}_{0})}{{R}_{0}}\phantom{\rule{0.2em}{0ex}}+\phantom{\rule{0.1em}{0ex}}\sum _{N\phantom{\rule{0.1em}{0ex}}=1}^{\infty}{{r}_{1}}^{N\phantom{\rule{0.1em}{0ex}}-1}\phantom{\rule{0.1em}{0ex}}{{r}_{2}}^{N}\phantom{\rule{0.1em}{0ex}}{t}^{2}\phantom{\rule{0.2em}{0ex}}\frac{exp\phantom{\rule{0em}{0ex}}(ik{R}_{N})}{{R}_{N}}\phantom{\rule{0.2em}{0ex}}\right|}^{2},$$
(2)
$${R}_{N}\phantom{\rule{0.1em}{0ex}}={[\phantom{\rule{0.1em}{0ex}}{l}^{2}\phantom{\rule{0.2em}{0ex}}+\phantom{\rule{0.1em}{0ex}}{y}^{2}\phantom{\rule{0.2em}{0ex}}+\phantom{\rule{0.1em}{0ex}}{N}^{2}{E}^{2}\phantom{\rule{0.2em}{0ex}}+\phantom{\rule{0.1em}{0ex}}\surd \phantom{\rule{0em}{0ex}}2NE\phantom{\rule{0.1em}{0ex}}(l\phantom{\rule{0.2em}{0ex}}+\phantom{\rule{0.1em}{0ex}}y)]}^{1/2},$$
(3)
$${E}_{\text{pl}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{2d}{{(2{n}^{2}\phantom{\rule{0.1em}{0ex}}-1)}^{1/2}}\phantom{\rule{0.2em}{0ex}},$$
(4)
$$\alpha ={sin}^{-1}\phantom{\rule{0.1em}{0ex}}\left(qsin\frac{1}{\surd \phantom{\rule{0em}{0ex}}2n}\right)\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.1em}{0ex}}\frac{1}{\surd \phantom{\rule{0em}{0ex}}2n}\phantom{\rule{0.1em}{0ex}},$$
(5)
$$\begin{array}{ll}{E}_{\text{cyl}}\phantom{\rule{0.2em}{0ex}}=& \phantom{\rule{0em}{0ex}}\frac{\rho}{sin\phantom{\rule{0.2em}{0ex}}2\phantom{\rule{0.1em}{0ex}}\alpha \phantom{\rule{0.2em}{0ex}}+\phantom{\rule{0.2em}{0ex}}cos\phantom{\rule{0.2em}{0ex}}2\phantom{\rule{0.1em}{0ex}}\alpha}\\ & \phantom{\rule{0em}{0ex}}\times \phantom{\rule{0.1em}{0ex}}\{\phantom{\rule{0em}{0ex}}sin\phantom{\rule{0em}{0ex}}2\phantom{\rule{0em}{0ex}}\alpha \phantom{\rule{0em}{0ex}}[2\phantom{\rule{0.1em}{0ex}}({n}^{2}\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.1em}{0ex}}1)\phantom{\rule{0em}{0ex}}sin\phantom{\rule{0em}{0ex}}\alpha \phantom{\rule{0em}{0ex}}-q\phantom{\rule{0.1em}{0ex}}{(2{n}^{2}\phantom{\rule{0.1em}{0ex}}-\phantom{\rule{0.1em}{0ex}}1)}^{1/2}]\\ & \phantom{\rule{0em}{0ex}}+\phantom{\rule{0em}{0ex}}cos\phantom{\rule{0em}{0ex}}2\phantom{\rule{0em}{0ex}}\alpha \phantom{\rule{0em}{0ex}}[2sin\phantom{\rule{0em}{0ex}}\alpha \phantom{\rule{0em}{0ex}}{(2{n}^{2}\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.1em}{0ex}}1)}^{1/2}]\phantom{\rule{0.1em}{0ex}}\}\phantom{\rule{0.2em}{0ex}}.\end{array}$$
(6)
$${d}_{e}\phantom{\rule{0.2em}{0ex}}(\phantom{\rule{0em}{0ex}}\rho \phantom{\rule{0em}{0ex}})\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.1em}{0ex}}\frac{{E}_{\text{cyl}}\phantom{\rule{0.1em}{0ex}}(\phantom{\rule{0em}{0ex}}\rho \phantom{\rule{0em}{0ex}})}{{E}_{\text{pl}}}\phantom{\rule{0.2em}{0ex}}d\phantom{\rule{0.1em}{0ex}},$$