## Abstract

A method for both image encryption and watermarking by three-step phase-shifting interferometry is proposed. The image to be hidden is stored in three interferograms and then can be reconstructed by use of one random phase mask, several specific geometric parameters, and a certain algorithm. To further increase the security of the hidden image and confuse unauthorized receivers, images with the same or different content can be added to the interferograms, and these images will have no or only a small effect on the retrieval of the hidden image, owing to the specific property of this algorithm. All these features and the utility of this method for image retrieval from parts of interferograms are verified by computer simulations. This technique uses intensity maps as decrypted images for delivery, and both encryption and decryption can be conveniently achieved digitally. It is particularly suitable for the remote transmission of secret information via the Internet.

© 2004 Optical Society of America

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### Equations (8)

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(1)
$$u\prime \left({x}_{2},{y}_{2}\right)=\frac{exp\left({\mathit{ikd}}_{1}\right)}{i\mathrm{\lambda}{d}_{1}}\iint {t}_{0}\left({x}_{1},{y}_{1}\right)\times exp\left[i2\mathrm{\pi}p\left({x}_{1},{y}_{1}\right)\right]exp\left\{\frac{i\mathrm{\pi}}{\mathrm{\lambda}{d}_{1}}\times \left[{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}\right]\right\}\mathrm{d}{x}_{1}\mathrm{d}{y}_{1},$$
(2)
$$u\left(x,y\right)=\frac{exp\left({\mathit{ikd}}_{2}\right)}{i\mathrm{\lambda}{d}_{2}}\iint u\prime \left({x}_{2},{y}_{2}\right)\times exp\left[i2\mathrm{\pi}q\left({x}_{2},{y}_{2}\right)\right]exp\left\{\frac{i\mathrm{\pi}}{\mathrm{\lambda}{d}_{2}}\times \left[{\left(x-{x}_{2}\right)}^{2}+{\left(y-{y}_{2}\right)}^{2}\right]\right\}\mathrm{d}{x}_{2}\mathrm{d}{y}_{2},$$
(3)
$${I}_{n}\left(x,y\right)={A}^{2}\left(x,y\right)+A_{r}{}^{2}+2{A}_{r}A\left(x,y\right)cos\left(\mathrm{\phi}-{\mathrm{\delta}}_{n}\right).$$
(4)
$$u\left(x,y\right)=\frac{1}{4{A}_{r}}\left[{I}_{1}-{I}_{3}+i\left(2{I}_{2}-{I}_{1}-{I}_{3}\right)\right].$$
(5)
$${I}_{1}\prime \left(x,y\right)={I}_{1}\left(x,y\right)+\mathrm{\alpha}a\left(x,y\right),{I}_{2}\prime \left(x,y\right)={I}_{2}\left(x,y\right)+\mathrm{\alpha}a\left(x,y\right),{I}_{3}\prime \left(x,y\right)={I}_{3}\left(x,y\right)+\mathrm{\alpha}a\left(x,y\right),$$
(6)
$${I}_{1}\prime \left(x,y\right)={I}_{1}\left(x,y\right)+{\mathrm{\alpha}}_{1}{a}_{1}\left(x,y\right),{I}_{2}\prime \left(x,y\right)={I}_{2}\left(x,y\right)+{\mathrm{\alpha}}_{2}{a}_{2}\left(x,y\right),{I}_{3}\prime \left(x,y\right)={I}_{3}\left(x,y\right)+{\mathrm{\alpha}}_{3}{a}_{3}\left(x,y\right),$$
(7)
$$u\prime \left(x,y\right)=u\left(x,y\right)+\mathrm{\Delta}u\left(x,y\right),$$
(8)
$$\mathrm{\Delta}u\left(x,y\right)={\mathrm{\alpha}}_{1}{a}_{1}-{\mathrm{\alpha}}_{3}{a}_{3}+i\left(2{\mathrm{\alpha}}_{2}{a}_{2}-{\mathrm{\alpha}}_{1}{a}_{1}-{\mathrm{\alpha}}_{3}{a}_{3}\right)$$