## Abstract

Phase errors that arise in phase-stepping interferometry are
discussed. Investigations were performed by use of a Twyman–Green
interferometer equipped with a compensation plate with a variable and
servo-controlled tilt angle. With this instrument, phase-stepping
errors can be reduced to a negligible level. There are, however,
phase errors that are caused by camera nonlinearities. Two methods
for minimizing these errors are presented. The first method is
based on the simple idea that the interference intensity at the output
of a two-beam interferometer has an exact cosine shape. The camera
signals were monitored as a function of the tilt angle of the
compensation plate, and the deviation from the cosine form was used to
produce a correction. The second method is based on the idea that,
under specific conditions, errors of an average of two phase
measurements may compensate for each other. Numerical calculations
were performed and give evidence of this hypothesis. Each method,
the signal-correction and the averaging method, drastically reduces
errors in evaluation of phases. The combination of both methods is
a powerful tool that allows precise phase data to be obtained with an
uncertainty, in the range λ/2000 ≈ 0.3 nm, that is caused
mainly by signal noise.

© 2002 Optical Society of America

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### Equations (8)

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(1)
$$\mathrm{\alpha}=\pm 2\mathrm{arctan}\left\{{\left[\frac{3\left({I}_{2}-{I}_{3}\right)-\left({I}_{1}-{I}_{4}\right)}{\left({I}_{2}-{I}_{3}\right)+\left({I}_{1}-{I}_{4}\right)}\right]}^{1/2}\right\},$$
(2)
$$\mathrm{\varphi}=\mathrm{arctan}\left\{tan\left(\frac{\mathrm{\alpha}}{2}\right){\left[\frac{\left({I}_{1}-{I}_{4}\right)+\left({I}_{2}-{I}_{3}\right)}{\left({I}_{2}+{I}_{3}\right)-\left({I}_{1}+{I}_{4}\right)}\right]}^{1/2}\right\}.$$
(3)
$${I}^{\mathrm{low}}=c\left({I}^{\mathrm{high}}-d\right)+d,$$
(4)
$${I}_{i}\left(\mathrm{\varphi},\mathrm{\alpha}\right)={I}_{0}\left\{1+\mathrm{\gamma}cos\left[\mathrm{\varphi}+\left(i-\frac{5}{2}\right)\mathrm{\alpha}\right]\right\},i=1,2,3,4,$$
(5)
$${S}_{i}\left(\mathrm{\varphi},\mathrm{\alpha}\right)=\left\{\begin{array}{l}{I}_{i}\\ {I}_{i}-\left({a}_{0}+{a}_{1}{I}_{i}+{a}_{2}{I}_{i}^{2}+{a}_{3}{I}_{i}^{3}\right)\end{array}\begin{array}{l}{I}_{i}410\\ 410\le {I}_{i}4096\end{array}\right.,$$
(6)
$${a}_{0}=-254.3,{a}_{1}=0.754,{a}_{2}=-3.624\times {10}^{-4},{a}_{3}=4.813\times {10}^{-8},$$
(7)
$${\mathrm{\varphi}}_{{A}_{12}}^{*}=\left[{\mathrm{\varphi}}_{A}^{*}+{\mathrm{\varphi}}_{B}^{*}\left({\mathit{\nu}}_{0}\right)\right]/2,$$
(8)
$${\mathrm{\varphi}}_{{B}_{12}}^{*}=\left[{\mathrm{\varphi}}_{{B}_{1}}^{*}\left(\mathit{\nu}\right)+{\mathrm{\varphi}}_{{B}_{2}}^{*}\left(\mathit{\nu}+{\mathit{\nu}}_{0}\right)\right]/2,$$