Deterministic Phase Retrieval Using the LCT

Abstract

Phase retrieval is one of the most interesting and widely researched inverse problems in optical sciences. Since all the forward problems based on real-world physical phenomenon involve some loss of information, the corresponding inverse problem is ill-posed unlike its forward counterpart. This makes the solution of inverse problems quite challenging. A majority of the approaches to solve the phase retrieval problem broadly fall into two categories. In the first category, one uses a priori information of the signal to find one of the many possible approximate solutions to the problem within an acceptable error criteria. One of the first and the most widely used method in this category is the Gerchberg–Saxon (GS) algorithm that uses a priori information of the signal at the input and output plane of an optical system that performs a Fourier Transformation. The second category consists of a direct approach based on a deterministic algorithm. The present chapter is concerned with these approaches. Typically, these algorithms are based on linear transformations on a set of intensity measurements of the signal. These intensity measurements are performed at different planes as the signal propagates through an optical system. In this chapter, we discuss the complex signal determination from multiple intensity measurements. We derive a generic algorithm that can retrieve a complex valued signal from two intensity measurements, one at the input plane and the second at the output plane, of an arbitrary optical system that consists of thin lenses (refractive or GRIN elements) separated by sections of free space.

Appendix

The AF can be written as:

$$ A\left(\overline{x},\overline{\nu}\right)={\displaystyle \int }f\left(x+\frac{\overline{x}}{2}\right){f}^{*}\left(x-\frac{\overline{x}}{2}\right) \exp \left(-j2\pi \overline{\nu}x\right)dx. $$

(11.A1)

$$ {\left.\frac{dA\left(\overline{x},\overline{\nu}\right)}{d\overline{x}}\right|}_{\overline{x}=0}={\displaystyle \int }{\left.\frac{d\left[f\left(x+\frac{\overline{x}}{2}\right){f}^{*}\left(x-\frac{\overline{x}}{2}\right)\right]}{d\overline{x}}\right|}_{\overline{x}=0} \exp \left(-j2\pi \overline{\nu}x\right)dx. $$

(11.A2)

Let $f(x)=A(x) \exp \left[i\varphi (x)\right]$, then

$$\begin{array}{lll} f\left(x+\frac{\overline{x}}{2}\right){f}^{*}\left(x-\frac{\overline{x}}{2}\right)=A\left(x+\frac{\overline{x}}{2}\right)A\left(x-\frac{\overline{x}}{2}\right) \nonumber\\ \exp \left\{i\left[\varphi \left(x+\frac{\overline{x}}{2}\right)-\varphi \left(x-\frac{\overline{x}}{2}\right)\right]\right\}. \end{array}$$

(11.A3)

Differentiating both sides with respect to $\overline{x}$and substituting $\overline{x}=0$, we get:

$$ {\left.\frac{d\left[f\left(x+\frac{\overline{x}}{2}\right){f}^{*}\left(x-\frac{\overline{x}}{2}\right)\right]}{d\overline{x}}\right|}_{\overline{x}=0}=i{I}_0(x)\frac{d\varphi (x)}{dx}, $$

(11.A4)

where ${I}_0(x)={\left|A(x)\right|}^2$

$$ {\left.\frac{dA\left(\overline{x},\overline{\nu}\right)}{d\overline{x}}\right|}_{\overline{x}=0}=i{\displaystyle \int }{I}_0(x)\frac{d\varphi (x)}{dx} \exp \left(-i2\pi \overline{\nu}x\right)dx. $$

(11.A5)