## Abstract

A novel phase-step calibration technique is presented on the basis of a two-run-times-two-frame phase-shift method. First the symmetry factor *M* is defined to describe the distribution property of the distorted phase due to phase-shifter miscalibration; then the phase-step calibration technique, in which two sets of two interferograms with a straight fringe pattern are recorded and the phase step is obtained by calculating *M* of the wrapped phase map, is developed. With this technique, a good mirror is required, but no uniform illumination is needed and no complex mathematical operation is involved. This technique can be carried out *in situ* and is applicable to any phase shifter, whether linear or nonlinear.

© 2006 Optical Society of America

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### Equations (15)

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(1)
$${I}_{i}\left(x,y\right)=A\left(x,y\right)+B\left(x,y\right)\mathrm{cos}\left[\varphi \left(x,y\right)+{\alpha}_{i}\right]\text{,}$$
(2)
$${I}_{1}=A+B\text{\hspace{0.17em}}\mathrm{cos}\left(\varphi -\alpha /2\right)\text{,}$$
(3)
$${I}_{2}=A+B\text{\hspace{0.17em}}\mathrm{cos}\left(\varphi +\beta -\alpha /2\right)\text{,}$$
(4)
$${I}_{3}=A+B\text{\hspace{0.17em}}\mathrm{cos}\left(\varphi +\alpha /2\right)\text{,}$$
(5)
$${I}_{4}=A+B\text{\hspace{0.17em}}\mathrm{cos}\left(\varphi +\beta +\alpha /2\right)\text{,}$$
(6)
$$\mathrm{tan}\text{\hspace{0.17em}}\varphi =\frac{\left({I}_{3}-{I}_{1}\right)\mathrm{sin}\text{\hspace{0.17em}}\beta}{\left({I}_{4}-{I}_{2}\right)-\left({I}_{3}-{I}_{1}\right)\mathrm{cos}\text{\hspace{0.17em}}\beta}\text{.}$$
(7)
$$\mathrm{tan}\text{\hspace{0.17em}}\varphi =\frac{{I}_{3}-{I}_{1}}{{I}_{4}-{I}_{2}}\text{.}$$
(8)
$$\varphi \prime =f\left(\varphi \right)=\mathrm{arctan}\left(\frac{\mathrm{sin}\text{\hspace{0.17em}}\varphi}{\mathrm{sin}\left(\varphi +\beta \right)}\right)\text{.}$$
(9)
$${S}_{-}={\displaystyle {\int}_{-\beta}^{0}\varphi \prime \mathrm{d}\varphi}\text{, \hspace{1em}}{S}_{+}={\displaystyle {\int}_{0}^{\pi -\beta}\varphi \prime \mathrm{d}\varphi}\text{.}$$
(10)
$${S}_{-}=\frac{\pi}{P}\left(\frac{\pi}{4}+{\displaystyle \sum {\varphi}_{-}\prime}\right)\text{, \hspace{1em}}{S}_{+}=\frac{\pi}{P}\left(-\frac{\pi}{4}+{\displaystyle \sum {\varphi}_{+}\prime}\right)\text{,}$$
(11)
$$M=\left(\frac{\pi}{4}+\frac{{\displaystyle \sum {\varphi}_{-}\prime}}{L}\right)/\left(-\frac{\pi}{4}+\frac{{\displaystyle \sum {\varphi}_{+}\prime}}{L}\right)\text{, \hspace{1em}}\left(\left|{S}_{-}\right|\le {S}_{+}\right)\text{.}$$
(12)
$$M=\left(\frac{\pi}{4}+\frac{{\displaystyle \sum {\varphi}_{-}\prime}}{h}\right)/\left(-\frac{\pi}{4}+\frac{{\displaystyle \sum {\varphi}_{+}\prime}}{h}\right)\text{, \hspace{1em}}\left(\left|{S}_{-}\right|\le {S}_{+}\right)\text{.}$$
(13)
$$\overline{X*X}=t\cong {r}^{2}/2R\text{,}$$
(14)
$$\overline{X*X}=t\cong r\theta +{r}^{2}/2R\text{.}$$
(15)
$$\overline{C*D}\cong {\overline{A*B*}}^{2}/8R\cong {\overline{AB}}^{2}/8R\mathrm{.}$$