# 夫琅禾费衍射

## 夫琅禾费近似

${\displaystyle U(x,y)={\frac {e^{ikz}e^{{\frac {ik}{2z}}(x^{2}+y^{2})}}{i\lambda z}}\iint _{-\infty }^{\infty }\,u(x',y')e^{-i{\frac {2\pi }{\lambda z}}(x'x+y'y)}dx'\,dy'}$ [3]

## 形式

### 解釋

 符合下式就會產生菲涅耳衍射： ${\displaystyle F={\frac {a^{2}}{L\lambda }}\geq 1}$ 符合下式就會產生夫琅禾费衍射： ${\displaystyle F={\frac {a^{2}}{L\lambda }}\ll 1}$ ${\displaystyle a}$  － 圓孔半徑或狹縫寬度， ${\displaystyle \lambda }$  － 波長， ${\displaystyle L}$  － 離圓孔的距離

### 振幅透射率

${\displaystyle \psi _{0}(x,y,z,t)=\ e^{i(kz-\omega t)}}$

${\displaystyle \psi _{\mathrm {rad} }(\theta ,\phi ,r)\,\propto \,{\frac {e^{ikr}}{r}}\int _{-\infty }^{\infty }\int _{-\infty }^{\infty }\psi _{0}(x,y)\,T(x,y)\,e^{ik\sin \theta (x\cos \phi +y\sin \phi )}\,dx\,dy}$

${\displaystyle \ e^{ik\sin \theta (x\cos \phi +y\sin \phi )}}$

${\displaystyle \psi _{\mathrm {rad} }(\theta )\propto \int _{-\infty }^{\infty }\psi _{0}(x)\,T(x)\,e^{i(k\theta )x}\,dx,}$

${\displaystyle \int |\psi _{\mathrm {rad} }(\theta ,\phi ,r)|^{2}r^{2}\sin \theta \,d\theta \,d\phi =\int |\psi _{0}\,T(x,y)|^{2}\,dx\,dy}$

## 例子

### 狹縫衍射

${\displaystyle \psi _{rad}(\theta )\propto \mathrm {sinc} \left({\frac {\pi a\theta }{\lambda }}\right),}$

### 高斯剖面

${\displaystyle \psi (\theta )=\exp \left({\frac {-k^{2}\theta ^{2}}{4a}}\right)}$

${\displaystyle \psi (\theta )=\exp \left(-{\frac {\pi ^{2}W^{2}}{2\lambda ^{2}\ln 2}}\theta ^{2}\right),}$

## 注释

1. ^ 在中国大陆，物理学家Joseph von Fraunhofer译作“约瑟夫·冯·夫琅费”，但在学界指该衍射时译作“夫琅费衍射”

## 參考資料

### 註釋

1. ^ Hecht, E. (1987), p396 -- Definition of Fraunhofer diffraction and explanation of forms.
2. Hecht, E. (1987), p397 -- diagram and explanation of Fraunhofer diffraction with reference to an opaque shield w/ aperture.
3. ^ Goodman, Joseph. Introduction to Fourier Optics. Englewood, Co: Roberts & Company. 2005. ISBN 0-9747077-2-4.
4. Hecht, E. (1987), p396 - description of the Fraunhofer diffraction through an aperture; details the main equations for the identification of Fresnel and Fraunhofer diffraction.
5. ^ 菲涅耳與夫琅禾费間的一般計算例子：${\displaystyle F_{\mathrm {fraunhofer} }={\frac {a^{2}}{L\lambda }}={\frac {3^{2}}{6\times 6}}=0.25}$
${\displaystyle F_{\mathrm {Fresnel} }={\frac {3^{2}}{2\times 2}}=2.25}$