## Abstract

Fresnel diffraction calculation on an arbitrary shape surface is proposed. This method is capable of calculating Fresnel diffraction from a source surface with an arbitrary shape to a planar destination surface. Although such calculation can be readily calculated by the direct integral of a diffraction calculation, the calculation cost is proportional to *O*(*N*^{2}) in one dimensional or *O*(*N*^{4}) in two dimensional cases, where *N* is the number of sampling points. However, the calculation cost of the proposed method is *O*(*N* log *N*) in one dimensional or *O*(*N*^{2} log *N*) in two dimensional cases using non-uniform fast Fourier transform.

© 2012 Optical Society of America

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### Equations (9)

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(1)
$${u}_{2}({\mathbf{x}}_{2})=\frac{{z}_{0}}{i\lambda}\int \int {u}_{I}({\mathbf{x}}_{1}){u}_{1}({\mathbf{x}}_{1})\frac{\text{exp}(ikr)}{{r}^{2}}d{\mathbf{x}}_{1},$$
(2)
$$u({\mathbf{x}}_{\mathbf{2}})=\frac{{z}_{0}}{i\lambda}\int \int {u}_{I}({\mathbf{x}}_{1},{d}_{1}){u}_{1}({\mathbf{x}}_{1},{d}_{1})\frac{\text{exp}(ikr)}{{r}^{2}}d{\mathbf{x}}_{1},$$
(3)
$$r=\sqrt{{\left|{\mathbf{x}}_{\mathbf{2}}-{\mathbf{x}}_{\mathbf{1}}\right|}^{2}+{({z}_{0}-{d}_{1})}^{2}}.$$
(4)
$$r\approx {r}_{0}+\frac{{\mathbf{x}}_{1}^{2}}{2{r}_{0}}-\frac{{\mathbf{x}}_{1}{\mathbf{x}}_{2}}{{r}_{0}}+\frac{{\mathbf{x}}_{2}^{2}}{2{r}_{0}}.$$
(5)
$$\begin{array}{ll}u({\mathbf{x}}_{2})\hfill & =\frac{\text{exp}(ik{z}_{0})}{i\lambda {z}_{0}}\int \int {u}_{I}({\mathbf{x}}_{1},{d}_{1}){u}_{1}({\mathbf{x}}_{1},{d}_{1})\hfill \\ \hfill & \text{exp}(ik(-{d}_{1}+\frac{{\mathbf{x}}_{1}^{2}}{2{r}_{0}}))\text{exp}(-ik\frac{{\mathbf{x}}_{1}{\mathbf{x}}_{2}}{{r}_{0}})\text{exp}(ik\frac{{\mathbf{x}}_{2}^{2}}{2{r}_{0}})d{\mathbf{x}}_{\mathbf{1}}\hfill \end{array}$$
(6)
$$\text{exp}(ik\frac{{\mathbf{x}}_{2}^{2}}{2{r}_{0}})\approx \text{exp}(ik\frac{{\mathbf{x}}_{2}^{2}}{2{z}_{0}}).$$
(7)
$$\begin{array}{ll}u({\mathbf{x}}_{2})\hfill & =\frac{\text{exp}(ik({z}_{0}+\frac{{\mathbf{x}}_{2}^{2}}{2{z}_{0}}))}{i\lambda {z}_{0}}\hfill \\ \hfill & \int \int {u}_{I}({\mathbf{x}}_{1},{d}_{1}){u}_{1}({\mathbf{x}}_{1},{d}_{1})\text{exp}(ik(-{d}_{1}+\frac{{\mathbf{x}}_{1}^{2}}{2({z}_{0}-{d}_{1})}))\hfill \\ \hfill & \text{exp}(ik\frac{{\mathbf{x}}_{1}{\mathbf{x}}_{2}}{{z}_{0}-{d}_{1}})d{\mathbf{x}}_{\mathbf{1}}.\hfill \end{array}$$
(8)
$$\begin{array}{l}u({\mathbf{x}}_{2})=\frac{\text{exp}(ik({z}_{0}+\frac{{\mathbf{x}}_{2}^{2}}{2{z}_{0}}))}{i\lambda {z}_{0}}\\ \mathit{NUF}\left[{u}_{I}({\mathbf{x}}_{1},{d}_{1}){u}_{1}({\mathbf{x}}_{1},{d}_{1})\text{exp}(ik(-{d}_{1}+\frac{{\mathbf{x}}_{1}^{2}}{2({z}_{0}-{d}_{1})}))\right],\end{array}$$
(9)
$$F({\mathbf{x}}_{2})=\mathit{NUF}\left[f({\mathbf{x}}_{1})\right]=\int \int f({\mathbf{x}}_{1})\text{exp}(-i\pi {\mathbf{x}}_{1}{\mathbf{x}}_{2})d{\mathbf{x}}_{1}.$$