## Abstract

A method to reduce the sensitivity of phase-shifting interferometry to external vibrations is described. The returning interferogram is amplitude split to form two series of interferograms, taken simultaneously and with complementary properties, one with high temporal and low spatial resolution and the other with low temporal and high spatial resolution. The high-temporal-resolution data set is used to calculate the true phase increment between interferograms in the high-spatial-resolution data set, and a generalized phase-extraction algorithm then includes these phase increments when the topographical phases in the high-spatial-resolution data set are calculated. The measured topography thereby benefits from the best qualities of both data sets, providing increased vibration immunity without sacrificing high spatial resolution.

© 1996 Optical Society of America

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### Equations (13)

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(1)
$${I}^{\prime}\left[x,y,\delta \left(t\right)\right]=I\left(x,y\right)+V\left(x,y\right)\text{cos}\left[\varphi \left(x,y\right)-\delta \left(t\right)\right],$$
(2)
$${{I}_{I}}^{\prime}\left(x,y,\delta \right)=I\left(x,y\right)+V\left(x,y\right)\text{cos}\left[\varphi \left(x,y\right)-\delta i\right].$$
(3)
$${{I}_{j}}^{\u2033}\left(x,y,{\Delta}_{j}\right)={{I}_{S\left(j\right)}}^{\prime}\left(x,y,\delta \right),$$
(4)
$$S\left(j\right)=2+5j,$$
(5)
$${\Phi}_{i}\left(x,y\right)=\text{PSI}\left[I\left(x,y,\delta \right),i\right],$$
(6)
$$\text{PSI}\left[I\left(x,y,\delta \right),i\right]={\text{tan}}^{-1}\left\{\frac{2\left[{I}_{i-1}\left(x,y,\delta \right)-{I}_{i+1}\left(x,y,\delta \right)\right]}{2{I}_{i}\left(x,y,\delta \right)-{I}_{i-2}\left(x,y,\delta \right)-{I}_{i+2}\left(x,y,\delta \right)}\right\}$$
(7)
$$\begin{array}{r}{\Phi}_{i}={\Phi}_{i}+2\pi \phantom{\rule{0.2em}{0ex}}\text{round}\left[\frac{{\Phi}_{i}\left(x,y\right)-{\Phi}_{i-1}\left(x,y\right)}{2\pi}\right]\\ \text{for}\phantom{\rule{0.2em}{0ex}}i=1,\dots ,{N}_{F},\end{array}$$
(8)
$${\Delta}_{j}={\Phi}_{S\left(j\right)}-{\Phi}_{S\left(0\right)}\phantom{\rule{0.2em}{0ex}}\text{for}\phantom{\rule{0.2em}{0ex}}j=0,\dots ,{N}_{S}-1.$$
(9)
$$\left|\begin{array}{ccc}{N}_{S}& {\displaystyle \sum \text{cos}\left({\Delta}_{j}\right)}& {\displaystyle \sum \text{sin}\left({\Delta}_{j}\right)}\\ {\displaystyle \sum \text{cos}\left({\Delta}_{j}\right)}& {\displaystyle \sum {\text{cos}}^{2}\left({\Delta}_{j}\right)}& {\displaystyle \sum \text{cos}\left({\Delta}_{j}\right)\text{sin}\left({\Delta}_{j}\right)}\\ {\displaystyle \sum \text{sin}\left({\Delta}_{j}\right)}& {\displaystyle \sum \text{cos}\left({\Delta}_{j}\right)\text{sin}\left({\Delta}_{j}\right)}& {\displaystyle \sum {\text{sin}}^{2}\left({\Delta}_{j}\right)}\end{array}\right|\left|\begin{array}{c}{a}_{0}\left(x,y\right)\\ {a}_{1}\left(x,y\right)\\ {a}_{2}\left(x,y\right)\end{array}\right|=\left|\begin{array}{c}{\displaystyle \sum {{I}_{j}}^{\u2033}\left(x,y\right)}\\ {\displaystyle \sum {{I}_{j}}^{\u2033}\left(x,y\right)\text{cos}\left({\Delta}_{j}\right)}\\ {\displaystyle \sum {{I}_{j}}^{\u2033}\left(x,y\right)\text{sin}\left({\Delta}_{j}\right)}\end{array}\right|,$$
(10)
$$\varphi \left(x,y\right)={\text{tan}}^{-1}\left[\frac{{a}_{2}\left(x,y\right)}{{a}_{1}\left(x,y\right)}\right].$$
(11)
$${N}_{F}=A+{\displaystyle \sum _{j=1}^{{N}_{S}-1}\frac{{\delta}_{j}-{\delta}_{j-1}}{\delta}=A}+{\displaystyle \sum _{j=0}^{{N}_{S}-1}{M}_{j}},$$
(12)
$$\begin{array}{cc}{I}_{mj}=& {\displaystyle {\int}_{{t}_{j}}^{{t}_{j}+P}{I}_{0}\{1+C\phantom{\rule{0.2em}{0ex}}\text{cos}[\frac{4\nu t\pi}{\lambda}+{\theta}_{m}}\\ & +F\phantom{\rule{0.2em}{0ex}}\text{sin}\left(2\pi \omega t+\phi \right)]\}\mathrm{d}t,\end{array}$$
(13)
$${E}_{\text{rms}}={\left[\frac{1}{M}\frac{1}{N}{\displaystyle \sum _{m=1}^{M}{\displaystyle \sum _{n=1}^{N}{\left({{\theta}_{m}}^{2C}-{{\theta}_{m}}^{T}\right)}^{2}}}\right]}^{1/2},$$