## Abstract

Tomographic reconstruction techniques used to obtain three-dimensional refractive index fields from interferometric data all assume that the correct fringe number has been assigned. In confined fields or where the imaging beam is smaller than the object field, only relative fringe numbers can be determined from the interferogram because no reference (undisturbed) region is visible. An algorithm that can be applied to iterative reconstruction techniques is described that permits the quantitative reconstruction of three-dimensional fields by using only relative fringe numbers provided that *a priori* data are supplied. This method is applied to an existing reconstruction technique and is demonstrated by using simulated interferometric data.

© 1993 Optical Society of America

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### Equations (14)

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(1)
$${g}_{e}(\mathrm{\rho},\mathrm{\theta})=L\{{f}_{e}(x,y)\},$$
(2)
$$g(\mathrm{\rho},\mathrm{\theta})=\{\overline{N}(\mathrm{\rho},\mathrm{\theta})+{N}_{s}(\mathrm{\theta})\}\mathrm{\lambda}/2,$$
(3)
$$\overline{g}(\mathrm{\rho},\mathrm{\theta})=\overline{N}(\mathrm{\rho},\mathrm{\theta})\mathrm{\lambda}/2.$$
(4)
$${N}_{s,i}=[2{g}_{e}({\mathrm{\rho}}_{i},{\mathrm{\theta}}_{i})/\mathrm{\lambda}]-{\overline{N}}_{i}.$$
(5)
$$\mathrm{\beta}=\text{abs}\left(\sum _{i=1}^{n}{N}_{s,i}\right)/\sum _{i=1}^{n}\text{abs}({N}_{s,i}),$$
(6)
$${N}_{s}=\left\{\frac{1}{n}\sum _{i=1}^{n}{N}_{s,i}\right\}\mathrm{\beta}.$$
(7)
$$g(\mathrm{\rho},\mathrm{\theta})=({N}_{s}+\overline{N})\mathrm{\lambda}/2.$$
(8)
$${g}_{c}(\mathrm{\rho},\mathrm{\theta})=g(\mathrm{\rho},\mathrm{\theta})-{g}_{e}(\mathrm{\rho},\mathrm{\theta}).$$
(9)
$${f}_{c}(x,y)=R\{{g}_{c}(\mathrm{\rho},\mathrm{\theta})\},$$
(10)
$${f}_{r}(x,y)={f}_{e}(x,y)+\mathrm{\alpha}{f}_{c}(x,y).$$
(11)
$${{f}_{r}}^{\text{new}}(x,y)=C\{{f}_{r}(x,y)\},$$
(12)
$$f(x,y)=\text{exp}\{1.0-[{(x-0.2)}^{2}+{(y-0.2)}^{2}]\},$$
(13)
$$\text{Average}\hspace{0.17em}\text{error}=\frac{\left\{{\displaystyle \sum _{i=1}^{I}}{\displaystyle \sum _{j=1}^{J}}\text{abs}[{f}_{r}({x}_{i},{y}_{j})-f({x}_{i},{y}_{j})]\right\}}{\{(I\times J)\text{max}[f({x}_{i},{y}_{j})]\}},$$
(14)
$$f(x,y)=\text{exp}\{1.0=[3.5{(x-0.6)}^{2}+{y}^{2}]\}+0.7\hspace{0.17em}\text{exp}\{1.0-[3.5{(x+0.6)}^{2}+{y}^{2}]\}$$