# 线性光学笔记（17）：菲涅耳衍射

$U(P_0) = -\frac{1}{2\pi}\iint_\Sigma U(P_1) \left(ik-\frac{1}{r_{01}}\right)\frac{e^{ikr_{01}}}{r_{01}}\cos\theta\,\mathrm{d} s.$

$U(P_0) = \frac{1}{i\lambda}\iint_\Sigma U(P_1) \frac{e^{ikr_{01}}}{r_{01}}\cos\theta\,\mathrm{d} s\\ \qquad\qquad\qquad = \frac{z}{i\lambda}\iint_\Sigma U(P_1) \frac{e^{ikr_{01}}}{r_{01}^2}\,\mathrm{d}\xi\,\mathrm{d}\eta,$

$r_{01} = \sqrt{z^2+(x-\xi)^2+(y-\eta)^2} \\ \qquad= z\sqrt{1+\left(\frac{x-\xi}{z}\right)^2+\left(\frac{y-\eta}{z}\right)^2} \\ \qquad\approx z\left[1+\frac{1}{2}\left(\frac{x-\xi}{z}\right)^2+\frac{1}{2}\left(\frac{y-\eta}{z}\right)^2\right].$

$U(x,y) = \frac{e^{ikz}}{i\lambda z}\iint_{-\infty}^\infty U(\xi,\eta) e^{i\frac{k}{2z}\left[(x-\xi)^2+(y-\eta)^2\right]}\,\mathrm{d}\xi\,\mathrm{d}\eta.$

$h(x,y) = \frac{e^{ikz}}{i\lambda z}e^{i\frac{k}{2z}\left(x^2+y^2\right)}.$

$H(f_X,f_Y) = \mathcal{F}\left\{\frac{e^{ikz}}{i\lambda z}e^{i\frac{k}{2z}\left(x^2+y^2\right)}\right\}\\ \qquad = e^{ikz}e^{-i\pi\lambda z\left(f_X^2+f_Y^2\right)}.$

$U(x,y) = \frac{e^{ikz}}{i\lambda z}e^{i\frac{k}{2z}(x^2+y^2)}\iint_{-\infty}^\infty \left[U(\xi,\eta)e^{i\frac{k}{2z}(\xi^2+\eta^2)}\right] e^{-i\frac{2\pi}{\lambda z}(x\xi+y\eta)}\,\mathrm{d}\xi\,\mathrm{d}\eta.$

http://blog.sciencenet.cn/blog-373392-739375.html

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