## Abstract

A self-calibrating algorithm for phase-shift interferometry is described that is able to cancel the effect of accidental relative tilts that may occur during phase stepping. The algorithm is able to retrieve both the phase steps and the tilts that accompany them. Only three phase-shifted interferograms are needed, and no other information about the intentional phase shifts or possible tilts has to be supplied. This purpose is achieved by division of the interferogram space into blocks on which a previously reported self-calibrating algorithm is applied and the actual values of the local phase shifts are calculated. The information thus obtained is used for extracting the global shift and tilt values. Further improvement in the results is achieved by means of a fitting routine.

© 2002 Optical Society of America

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### Equations (7)

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(1)
$${I}_{k}\left(x,y\right)={I}_{m}\left(x,y\right)\left\{1+V\left(x,y\right)cos\left[\mathrm{\varphi}\left(x,y\right)-{\mathrm{\beta}}_{k}\right]\right\},k=0,1,2,$$
(2)
$${\mathrm{\beta}}_{k}\left(x,y\right)={\mathrm{\beta}}_{k0}+{t}_{\mathit{xk}}x+{t}_{\mathit{yk}}y,k=1,2,$$
(3)
$$V=\frac{{\left({c}^{2}+{s}^{2}\right)}^{1/2}}{{I}_{0}-c},$$
(4)
$$c=\frac{\left({I}_{1}-{I}_{0}\right)sin{\mathrm{\beta}}_{2}-\left({I}_{2}-{I}_{0}\right)sin{\mathrm{\beta}}_{1}}{\left(cos{\mathrm{\beta}}_{1}-1\right)sin{\mathrm{\beta}}_{2}-\left(cos{\mathrm{\beta}}_{2}-1\right)sin{\mathrm{\beta}}_{1}},$$
(5)
$$s=\frac{\left(cos{\mathrm{\beta}}_{1}-1\right)\left({I}_{2}-{I}_{0}\right)-\left(cos{\mathrm{\beta}}_{2}-1\right)\left({I}_{1}-{I}_{0}\right)}{\left(cos{\mathrm{\beta}}_{1}-1\right)sin{\mathrm{\beta}}_{2}-\left(cos{\mathrm{\beta}}_{2}-1\right)sin{\mathrm{\beta}}_{1}}.$$
(6)
$$W=\frac{{\displaystyle \sum _{{N}_{\mathrm{pixels}}}}{\left[V\left(x,y\right)-{V}_{0}\right]}^{2}}{{N}_{\mathrm{pixels}}}.$$
(7)
$$\begin{array}{c}{t}_{x1}=-2.43\times {10}^{-4}\times 2\mathrm{\pi}/\mathrm{pixel}=-0.091\times 2\mathrm{\pi}/\left(x\mathrm{side}\right),\\ {t}_{x2}=-5.94\times {10}^{-4}\times 2\mathrm{\pi}/\mathrm{pixel}=-0.219\times 2\mathrm{\pi}/\left(x\mathrm{side}\right),\\ {t}_{y1}=+2.98\times {10}^{-5}\times 2\mathrm{\pi}/\mathrm{pixel}=+0.009\times 2\mathrm{\pi}/\left(y\mathrm{side}\right),\\ {t}_{y2}=+7.15\times {10}^{-5}\times 2\mathrm{\pi}/\mathrm{pixel}=+0.021\times 2\mathrm{\pi}/\left(y\mathrm{side}\right).\end{array}$$