# 线性光学笔记（16）：角谱衍射理论（三）

$\frac{e^{ikr}}{r} = \frac{ik}{2\pi}\iint^\infty_{-\infty}\frac{e^{i\vec{k}\cdot\vec{r}}}{\sqrt{1-\alpha^2-\beta^2}}\,\mathrm{d}\alpha\,\mathrm{d}\beta,\quad(\alpha^2+\beta^2<1, z>0).$

$h(P_0;P_1) = -\frac{1}{2\pi}\frac{\partial}{\partial z}\left(\frac{ik}{2\pi}\iint^\infty_{-\infty}\frac{e^{i\vec{k}\cdot\vec{r}}}{\sqrt{1-\alpha^2-\beta^2}}\,\mathrm{d}\alpha\,\mathrm{d}\beta\right)\\ = -\frac{i\lambda}{2\pi}\iint^\infty_{-\infty}\frac{\partial}{\partial z}\left(\frac{e^{i2\pi\left(f_Xx+f_Yy+\frac{z}{\lambda}\sqrt{1-(\lambda f_X)^2-(\lambda f_Y)^2}\right)}}{\sqrt{1-(\lambda f_X)^2-(\lambda f_Y)^2}}\right)\,\mathrm{d} f_X\,\mathrm{d} f_Y\\ = \iint^\infty_{-\infty}\left(e^{i2\pi\left(\frac{z}{\lambda}\sqrt{1-(\lambda f_X)^2-(\lambda f_Y)^2}\right)}\right)e^{i2\pi(f_Xx+f_Yy)}\,\mathrm{d} f_X\,\mathrm{d} f_Y.$

$H(f_X, f_Y) = e^{i\frac{2\pi}{\lambda}z\sqrt{1-(\lambda f_X)^2-(\lambda f_Y)^2}}\quad \left(\sqrt{f_X^2+f_Y^2}<\frac{1}{\lambda}\right).$

1. 将入射波乘以代表衍射屏的透射函数（例如狭缝可以用 $\operatorname{rect}(x)$ 函数来表示）;

2. 将上一步的结果用傅里叶变换展开为角谱;

3. 将每个角谱分量乘以相应的传递函数;

4. 用傅里叶逆变换获得衍射场。

http://blog.sciencenet.cn/blog-373392-728408.html

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