UA OPTI512R 傅立葉光學導論 衍射例題

例1 Fresnel衍射與Fraunhofer衍射的基本概念：在Fresnel field中可以使用Fraunhofer近似嗎？在Fraunhofer field中可以使用Fresnel近似嗎？

答案 Fresnel近似是二次波近似，Fraunhofer近似是平面波近似，所以在Fresnel field與Fraunhofer field中都是Fresnel近似更準確。

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例2 Fresnel衍射基礎
Scalar衍射的傳遞函數為
$$H_z(\xi ,\eta )=e^{jkz\sqrt{1?{\lambda }^2({\xi }^2+{\eta }^2)}}$$

Fresnel衍射的傳遞函數為
$$H_z(\xi ,\eta )=e^{jkz}{e}^{?j\pi \lambda z({\xi }^2+{\eta }^2)}$$

假設$\eta =0$，

用scalar diffraction，計算入射波最大可能的頻率；

計算Fresnel衍射與scalar衍射的傳遞函數的相位差；

給定$\xi$，推導$z$的條件使得Fresnel衍射可以近似scalar衍射

答案
第一問，入射波的頻率需要滿足
$$1?{\lambda }^2({\xi }^2+{\eta }^2)=1?{\lambda }^2{\xi }^2\ge 0?\xi \le \frac{1}{\lambda }$$

第二問，scalar衍射傳遞函數的相位為$kz\sqrt{1?{\lambda }^2({\xi }^2+{\eta }^2)}$，Fresnel衍射傳遞函數的相位為$kz?\pi \lambda z({\xi }^2+{\eta }^2)$，相位差為
$$\begin{array}{rl}& kz\sqrt{1?{\lambda }^2({\xi }^2+{\eta }^2)}?[kz?\pi \lambda z({\xi }^2+{\eta }^2)]\\ =& \frac{2\pi }{\lambda }z\sqrt{1?{\lambda }^2({\xi }^2+{\eta }^2)}?\left[\frac{2\pi z}{\lambda }?\pi \lambda z({\xi }^2+{\eta }^2)\right]\\ =& \frac{2\pi z}{\lambda }\left[\sqrt{1?{\lambda }^2({\xi }^2+{\eta }^2)}+\frac{{\lambda }^2({\xi }^2+{\eta }^2)?2}{2}\right]\\ =& \frac{2\pi z}{\lambda }\left(\sqrt{1?{\lambda }^2{\xi }^2}+\frac{{\lambda }^2{\xi }^2?2}{2}\right)\end{array}$$

當${\lambda }^2{\xi }^2<<1$時，$\left(\sqrt{1?{\lambda }^2{\xi }^2}+\frac{{\lambda }^2{\xi }^2?2}{2}\right)\approx 0$，此時用Fresnel衍射近似的效果最好；

第三問，Fresnel衍射近似scalar衍射的條件為
$$\begin{array}{rl}& \left|\frac{2\pi z}{\lambda }\left(\sqrt{1?{\lambda }^2{\xi }^2}+\frac{{\lambda }^2{\xi }^2?2}{2}\right)\right|\le 2\pi \\ ?& z\le \frac{\lambda }{|\sqrt{1?{\lambda }^2{\xi }^2}+\frac{{\lambda }^2{\xi }^2?2}{2}|}\end{array}$$

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例3 三縫Fraunhofer衍射，考慮入射光$e^{jkbx}$，代表三縫的pupil函數為
$$P(x,y)=rect\left(\frac{x?d}{L_x},\frac{y}{L_y}\right)+rect\left(\frac{x}{L_x},\frac{y}{L_y}\right)+rect\left(\frac{x+d}{L_x},\frac{y}{L_y}\right)$$

計算衍射條紋分布；

答案 Fraunhofer衍射公式為
$$u_z(x_z,y_z)=\frac{e^{jkz}}{j\lambda z}{e}^{j\frac{\pi }{\lambda z}(x_z^2+y_z^2)}\mathcal{F}[u_{in}(x_i,y_i){]}_{\xi =\frac{x_z}{\lambda z},\eta =\frac{y_z}{\lambda z}}$$

所以光強公式為
$$I_z(x_z,y_z)=\frac{1}{{\lambda }^2{z}^2}{\left|\mathcal{F}[u_{in}(x_i,y_i){]}_{\xi =\frac{x_z}{\lambda z},\eta =\frac{y_z}{\lambda z}}\right|}^2$$

入射光為
$$u_{in}(x_i,y_i)=e^{jkb{x}_i}P(x_i,y_i)$$

計算它的Fourier變換，
$$\begin{array}{rl}& \mathcal{F}[u_{in}(x_i,y_i){]}_{\xi =\frac{x_z}{\lambda z},\eta =\frac{y_z}{\lambda z}}\\ =& \mathcal{F}[e^{jk{b}_0{x}_i}P(x_i,y_i){]}_{\xi =\frac{x_z}{\lambda z},\eta =\frac{y_z}{\lambda z}}\\ =& \mathcal{F}[P(x_i,y_i){]}_{\xi =\frac{x_z}{\lambda z}?\frac{b_0}{\lambda },\eta =\frac{y_z}{\lambda z}}\\ =& L_x{L}_{y}sinc\left(L_x\left(\frac{x_z}{\lambda z}?\frac{b_0}{\lambda },L_{y}y\right)\right)\left[1+2\cos \frac{2\pi d}{\lambda z}\left(x?b_{0}z\right)\right]\end{array}$$

所以光強分布為
$$I_z(x_z,y_z)=\frac{L_x^2{L}_y^2}{{\lambda }^2{z}^2}sin{c}^2\left(L_x\left(\frac{x_z}{\lambda z}?\frac{b_0}{\lambda },L_{y}y\right)\right){\left[1+2\cos \frac{2\pi d}{\lambda z}\left(x?b_{0}z\right)\right]}^2$$

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例4 有黑點的方孔Fraunhofer衍射，考慮均勻入射光，方孔的中心有一個圓形黑斑，所以pupil函數為
$$P(x,y)=rect\left(\frac{x}{L_1},\frac{y}{L_1}\right)?circ\left(\frac{\rho }{L_2}\right),L_2<<L_1$$

計算far field；

答案
經過方孔后，入射光為
$$u_{in}(x_i,y_i)=rect\left(\frac{x}{L_1},\frac{y}{L_1}\right)?circ\left(\frac{\rho }{L_2}\right)$$

計算Fourier變換，
$$\begin{array}{rl}& \mathcal{F}[u_{in}(x_i,y_i){]}_{\xi =\frac{x_z}{\lambda z},\eta =\frac{y_z}{\lambda z}}\\ =& {L_1^{2}sinc(L_1\xi ,L_1\eta )?L_2^2\frac{J_1(\pi L_2\sqrt{{\xi }^2+{\eta }^2})}{2L_2\sqrt{{\xi }^2+{\eta }^2}}|}_{\xi =\frac{x_z}{\lambda z},\eta =\frac{y_z}{\lambda z}}\\ =& L_1^{2}sinc\left(\frac{L_1{x}_z}{\lambda z},\frac{L_1{y}_z}{\lambda z}\right)?L_2^2\frac{J_1(\pi L_2\sqrt{(\frac{x_z}{\lambda z}{)}^2+(\frac{y_z}{\lambda z}{)}^2})}{2L_2\sqrt{(\frac{x_z}{\lambda z}{)}^2+(\frac{y_z}{\lambda z}{)}^2}}\end{array}$$

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