## Abstract

The use of the term differential intensity measurements implies that two or more intensity measurements are made, with a slight change in a parameter of the measurement-taking system between measurements. For example, the position of the focal plane or the transmission of the aperture might be changed. In either of these cases it is possible to find a closed-form solution to the phase-retrieval problem, as opposed to the usual iterative solution. The first case corresponds to a quadratic phase shift in the unknown phase and has been studied previously by several authors. This examination of the problem leads to the discovery of some details that may be useful. In the second case a new result leads to a simpler solution for the phase. The results in this paper apply to the one-dimensional phase-retrieval problem. An extension to the more important two-dimensional problem can be made by means described in this paper.

© 1987 Optical Society of America

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### Equations (26)

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(1)
$$S(u)=K{u}^{-(11/3)},$$
(2)
$$w(f)=\sum a(k)g(f,k),$$
(3)
$$g(f,k)=2\pi \hspace{0.17em}\text{cos}(k\pi f/D),$$
(4)
$$H(f)=A(f){e}^{iw(f)},$$
(5)
$$h(x)=\int H(f)\text{exp}(i2\pi fx)\text{d}f,$$
(6)
$$p(x)=\mid h(x){\mid}^{2}.$$
(7)
$$A(f,B)=A(f)\text{exp}(2\pi Bf),$$
(8)
$$h(x,B)={\int}_{-\infty}^{\infty}A(f)\text{exp}[2\pi Bf+iw(f)+i2\pi fx]\text{d}f,$$
(9)
$$p(x,B)=\mid h(x,B){\mid}^{2}=h(x,B){h}^{*}(x,B).$$
(10)
$${\frac{\text{d}p(x,B)}{\text{d}B}|}_{B=0}=\frac{p(x,B)-p(x)}{B}\mathrm{\hspace{0.17em}\u200a\u200a}\mathrm{\hspace{0.17em}\u200a\u200a}\mathrm{\hspace{0.17em}\u200a\u200a}\text{for}\hspace{0.17em}B\hspace{0.17em}\text{small}.$$
(11)
$${\frac{\text{d}p(x,B)}{\text{d}B}|}_{B=0}=2\hspace{0.17em}\text{Re}\left[{{h}^{*}(x)\frac{\text{d}h(x,B)}{\text{d}B}|}_{B=0}\right].$$
(12)
$$\begin{array}{l}{\frac{\text{d}h(x,B)}{\text{d}B}|}_{B=0}={\int}_{-\infty}^{\infty}A(f)2\pi f\hspace{0.17em}\text{exp}(iw(f)+i2\pi fx)\text{d}f\\ =-i\frac{\text{d}h(x)}{\text{d}x},\end{array}$$
(13)
$${\frac{\text{d}p(x,B)}{\text{d}B}|}_{B=0}=2\hspace{0.17em}\text{Re}\left[-i{h}^{*}(x)\frac{\text{d}h(x)}{\text{d}x}\right].$$
(14)
$$h(x)=r(x){e}^{it(x)}.$$
(15)
$$\frac{\text{d}h(x)}{\text{d}x}={h}^{\prime}(x)={r}^{\prime}(x){e}^{it(x)}+i{t}^{\prime}(x)r(x){e}^{it(x)},$$
(16)
$$\begin{array}{l}{\frac{\text{d}p(x,B)}{\text{d}B}|}_{B=0}=2\hspace{0.17em}\text{Re}\{-ir(x){e}^{-it(x)}[{r}^{\prime}(x)+ir(x){t}^{\prime}(x)]{e}^{it(x)}\}\\ =2{r}^{2}(x){t}^{\prime}(x).\end{array}$$
(17)
$${r}^{2}(x)=p(x),$$
(18)
$${\frac{\text{d}p(x,B)}{\text{d}B}|}_{B=0}=2p(x){t}^{\prime}(x).$$
(19)
$${t}^{\prime}(x)=\frac{1}{2p(x)}\frac{p(x,B)-p(x)}{B},\mathrm{\hspace{0.17em}\u200a\u200a}\mathrm{\hspace{0.17em}\u200a\u200a}\mathrm{\hspace{0.17em}\u200a\u200a}B\hspace{0.17em}\text{small}.$$
(20)
$$t(x)={\int}_{-\infty}^{x}\frac{1}{2p(y)}\frac{p(y,B)-p(y)}{B}\text{d}y.$$
(21)
$$h(x)=r(x){e}^{it(x)},$$
(22)
$$h(x,B)={\int}_{-\infty}^{\infty}A(f)\text{exp}[-i2\pi B{f}^{2}+iw(f)+i2\pi fx]\text{d}f.$$
(23)
$$t(x)={\int}_{-\infty}^{x}\frac{1}{\pi p(y)}{\int}_{-\infty}^{y}\frac{p(z)-p(z,B)}{B}\text{d}z\text{d}y.$$
(24)
$$\frac{\partial}{\partial x}p\frac{\partial t}{\partial x}+\frac{\partial}{\partial y}p\frac{\partial t}{\partial y}=\frac{1}{2\mathrm{\Pi}}\frac{p(x,y)-p(x,y,B)}{B},$$
(25)
$${d}_{1}(x,y)=\frac{\partial t(x,y)}{\partial x},\mathrm{\hspace{0.17em}\u200a\u200a}\mathrm{\hspace{0.17em}\u200a\u200a}\mathrm{\hspace{0.17em}\u200a\u200a}{d}_{2}(x,y)=\frac{\partial t(x,y)}{\partial y}.$$
(26)
$$T(f,l)=\frac{(i2\mathrm{\Pi}f){D}_{1}(f,l)+(i2\mathrm{\Pi}l){D}_{2}(f,l)}{{(2\mathrm{\Pi}f)}^{2}+{(2\mathrm{\Pi}l)}^{2}}.$$