## Abstract

We propose and demonstrate multiple shearing interferometry for measuring two-dimensional phase object. Multi-shear interference can effectively eliminate the problem of spectral leakage that results from the single-shear interference. The Fourier coefficients of a two-dimensional wavefront are computed from phase differences obtained from multiple shearing interferograms, which are acquired by a shearing interferometer, and the desired phase is then reconstructed. Numerical and optical tests have confirmed that the multiple shearing interferometry has a higher recovery accuracy than single-shear interferometry and the reconstruction precision increases as the number of shear steps increases.

© 2012 OSA

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### Equations (20)

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(1)
$$\phi (m,n)={\displaystyle \sum _{p=0}^{N-1}{\displaystyle \sum _{q=0}^{N-1}{\alpha}_{pq}{Z}_{pq}}}(m,n),m,n=0,1,2,\mathrm{...},N-1,$$
(2)
$${Z}_{pq}(m,n)=\frac{1}{N}\mathrm{exp}\left[\frac{i2\pi}{N}(pm+qn)\right],p,q=0,1,\mathrm{2...},N-1,$$
(3)
$${D}_{x}(m,n)=\phi (m,n)-\phi (m-s,n),$$
(4)
$${D}_{y}(m,n)=\phi (m,n)-\phi (m,n-s),$$
(5)
$${D}_{x}(m,n)={\displaystyle \sum _{p=0}^{N-1}{\displaystyle \sum _{q=0}^{N-1}{\alpha}_{pq}}}{Z}_{pq}(m,n)\left[1-\mathrm{exp}\left(\frac{i2\pi ps}{N}\right)\right],$$
(6)
$${D}_{y}(m,n)={\displaystyle \sum _{p=0}^{N-1}{\displaystyle \sum _{q=0}^{N-1}{\alpha}_{pq}}}{Z}_{pq}(m,n)\left[1-\mathrm{exp}\left(\frac{i2\pi qs}{N}\right)\right].$$
(7)
$$F={\displaystyle \sum _{m=0}^{N-1}{\displaystyle \sum _{n=0}^{N-1}\{[{D}_{x}{}^{\prime}(m,n)-}}{D}_{x}(m,n){]}^{2}+{[{D}_{y}{}^{\prime}(m,n)-{D}_{y}(m,n)]}^{2}\}.$$
(8)
$$\frac{\partial F}{\partial {\alpha}_{pq}}=0\text{.}$$
(9)
$$\begin{array}{l}{\alpha}_{pq}=\frac{1}{4[{\mathrm{sin}}^{2}(\pi ps/N)+{\mathrm{sin}}^{2}(\pi qs/N)]}\left\{\left[1-\mathrm{exp}(\frac{i2\pi ps}{N})\right]\cdot F{T}_{pq}\{{{D}^{\prime}}_{x}(m,n)\}\right\}\\ \text{+}\left\{\left[1-\mathrm{exp}(\frac{i2\pi qs}{N})\right]\cdot F{T}_{pq}\{{{D}^{\prime}}_{y}(m,n)\}\right\},\end{array}$$
(10)
$$\epsilon ={\displaystyle \sum _{m=0}^{N-1}{\displaystyle \sum _{n=0}^{N-1}\left\{{\displaystyle \sum _{j=1}^{K}{[{D}_{x}^{\text{'}}{}^{j}(m,n)-{D}_{x}{}^{j}(m,n)]}^{2}}+{\displaystyle \sum _{j=1}^{K}[{D}_{y}^{\text{'}}{}^{j}(m,n)-{D}_{y}{}^{j}(m,n)}{]}^{2}\right\}}}\text{,}j=1,\mathrm{2...}k.$$
(11)
$$\alpha (p,q)=\frac{{\displaystyle \sum _{j=1}^{K}(1-{e}^{i2\pi p{s}^{j}/N})F{T}_{pq}}\{{D}_{x}^{\text{'}}{}^{j}(m,n)\}+{\displaystyle \sum _{j=1}^{K}(1-{e}^{i2\pi q{s}^{j}/N})F{T}_{pq}\{{D}_{y}^{\text{'}}{}^{j}(m,n)\}}}{{\displaystyle \sum _{j=1}^{K}\left[4{\mathrm{sin}}^{2}(\frac{\pi p{s}^{j}}{N})+4{\mathrm{sin}}^{2}(\frac{\pi q{s}^{j}}{N})\right]}},$$
(12)
$$\alpha (p,q)={\displaystyle \sum _{j=1}^{K}\left[{b}_{pq}^{xj}F{T}_{pq}\{{D}_{x}^{\text{'}}{}^{j}(m,n)\}+{b}_{pq}^{yj}F{T}_{pq}\{{D}_{y}^{\text{'}}{}^{j}(m,n)\}\right]}$$
(13)
$$({b}_{pq}{}^{xj})=\frac{1-\mathrm{exp}(\frac{i2\pi p{s}_{j}}{N})}{{\displaystyle \sum _{j=1}^{K}\left\{4[{\mathrm{sin}}^{2}(\pi p{s}_{j}/N)+{\mathrm{sin}}^{2}(\pi q{s}_{j}/N)]\right\}}},$$
(14)
$$({b}_{pq}{}^{yj})=\frac{1-\mathrm{exp}(\frac{i2\pi q{s}_{j}}{N})}{{\displaystyle \sum _{j=1}^{K}\left\{4[{\mathrm{sin}}^{2}(\pi p{s}_{j}/N)+{\mathrm{sin}}^{2}(\pi q{s}_{j}/N)]\right\}}}.$$
(15)
$$\u3008{n}_{u}^{i}(k,l){n}_{v}^{j}(m,n)\u3009={\sigma}_{n}^{2}\delta \left(u,v\right)\delta \left(i,j\right)\delta \left(k,m\right)\delta \left(l,n\right),$$
(16)
$${\sigma}_{a}^{2}(p,q)={\displaystyle \sum _{j=1}^{K}\left({\left|{b}_{pq}^{xj}\right|}^{2}+{\left|{b}_{pq}^{yj}\right|}^{2}\right)}{\sigma}_{n}^{2}.$$
(17)
$${\sigma}_{\phi}^{2}=\frac{1}{{N}^{2}}{\displaystyle \sum _{m,n}\u3008{\left|{n}_{\phi}\left(m,n\right)\right|}^{2}\u3009=}\frac{1}{{N}^{2}}{\displaystyle \sum _{m,n}{\sigma}_{a}^{2}\left(p,q\right)}.$$
(18)
$$C=\frac{{\sigma}_{\phi}^{2}}{{\sigma}_{n}^{2}}=\frac{1}{{N}^{2}}{\displaystyle \sum _{p,q}{\displaystyle \sum _{j=1}^{K}\left({\left|{b}_{pq}^{xj}\right|}^{2}+{\left|{b}_{pq}^{yj}\right|}^{2}\right)}},$$
(19)
$$C=\frac{1}{{N}^{2}}{\displaystyle \sum _{p,q}\frac{1}{{\displaystyle \sum _{j=1}^{K}[4{\mathrm{sin}}^{2}(\frac{\pi p{s}^{j}}{N})}+4{\mathrm{sin}}^{2}(\frac{\pi q{s}^{j}}{N})]}}.$$
(20)
$$\phi (x,y)=1.014\times [({x}^{2}-3\times {y}^{2})x+y+{x}^{2}-{y}^{2}].$$