UA OPTI512R 傅立叶光学导论衍射例题

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

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与Fraunhofer field中都是Fresnel近似更准确。

例2 Fresnel衍射基础
Scalar衍射的传递函数为
Hz(ξ,η)=ejkz1−λ2(ξ2+η2)H_z(\xi,\eta)=e^{jkz\sqrt{1-\lambda^2(\xi^2+\eta^2)}}Hz(ξ,η)=ejkz1−λ2(ξ2+η2)

Fresnel衍射的传递函数为
Hz(ξ,η)=ejkze−jπλz(ξ2+η2)H_z(\xi,\eta)=e^{jkz}e^{-j\pi \lambda z(\xi^2+\eta^2)}Hz(ξ,η)=ejkze−jπλz(ξ2+η2)

假设η=0\eta=0η=0，

用scalar diffraction，计算入射波最大可能的频率；
计算Fresnel衍射与scalar衍射的传递函数的相位差；
给定ξ\xiξ，推导zzz的条件使得Fresnel衍射可以近似scalar衍射

答案
第一问，入射波的频率需要满足
1−λ2(ξ2+η2)=1−λ2ξ2≥0⇒ξ≤1λ1-\lambda^2(\xi^2+\eta^2)=1-\lambda^2 \xi^2 \ge 0 \Rightarrow \xi \le \frac{1}{\lambda}1−λ2(ξ2+η2)=1−λ2ξ2≥0⇒ξ≤λ1

第二问，scalar衍射传递函数的相位为kz1−λ2(ξ2+η2)kz\sqrt{1-\lambda^2(\xi^2+\eta^2)}kz1−λ2(ξ2+η2)，Fresnel衍射传递函数的相位为kz−πλz(ξ2+η2)kz-\pi \lambda z(\xi^2+\eta^2)kz−πλz(ξ2+η2)，相位差为
kz1−λ2(ξ2+η2)−[kz−πλz(ξ2+η2)]=2πλz1−λ2(ξ2+η2)−[2πzλ−πλz(ξ2+η2)]=2πzλ[1−λ2(ξ2+η2)+λ2(ξ2+η2)−22]=2πzλ(1−λ2ξ2+λ2ξ2−22)\begin{aligned} & 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{aligned}===kz1−λ2(ξ2+η2)−[kz−πλz(ξ2+η2)]λ2πz1−λ2(ξ2+η2)−[λ2πz−πλz(ξ2+η2)]λ2πz[1−λ2(ξ2+η2)+2λ2(ξ2+η2)−2]λ2πz(1−λ2ξ2+2λ2ξ2−2)

当λ2ξ2<<1\lambda^2\xi^2<<1λ2ξ2<<1时，(1−λ2ξ2+λ2ξ2−22)≈0\left( \sqrt{1-\lambda^2\xi^2}+\frac{\lambda^2\xi^2-2}{2} \right) \approx 0(1−λ2ξ2+2λ2ξ2−2)≈0，此时用Fresnel衍射近似的效果最好；

第三问，Fresnel衍射近似scalar衍射的条件为
∣2πzλ(1−λ2ξ2+λ2ξ2−22)∣≤2π⇒z≤λ∣1−λ2ξ2+λ2ξ2−22∣\begin{aligned} & \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 \\ \Rightarrow & z \le \frac{\lambda}{|\sqrt{1-\lambda^2\xi^2}+\frac{\lambda^2\xi^2-2}{2}|}\end{aligned}⇒∣∣∣∣λ2πz(1−λ2ξ2+2λ2ξ2−2)∣∣∣∣≤2πz≤∣1−λ2ξ2+2λ2ξ2−2∣λ

例3 三缝Fraunhofer衍射，考虑入射光ejkbxe^{jkbx}ejkbx，代表三缝的pupil函数为
P(x,y)=rect(x−dLx,yLy)+rect(xLx,yLy)+rect(x+dLx,yLy)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)P(x,y)=rect(Lxx−d,Lyy)+rect(Lxx,Lyy)+rect(Lxx+d,Lyy)

计算衍射条纹分布；

答案 Fraunhofer衍射公式为
uz(xz,yz)=ejkzjλzejπλz(xz2+yz2)F[uin(xi,yi)]ξ=xzλz,η=yzλzu_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}}uz(xz,yz)=jλzejkzejλzπ(xz2+yz2)F[uin(xi,yi)]ξ=λzxz,η=λzyz

所以光强公式为
Iz(xz,yz)=1λ2z2∣F[uin(xi,yi)]ξ=xzλz,η=yzλz∣2I_z(x_z,y_z)=\frac{1}{\lambda^2z^2}\left| \mathcal{F}[u_{in}(x_i,y_i)]_{\xi=\frac{x_z}{\lambda z},\eta=\frac{y_z}{\lambda z}}\right|^2Iz(xz,yz)=λ2z21∣∣∣F[uin(xi,yi)]ξ=λzxz,η=λzyz∣∣∣2

入射光为
uin(xi,yi)=ejkbxiP(xi,yi)u_{in}(x_i,y_i)=e^{jkbx_i}P(x_i,y_i)uin(xi,yi)=ejkbxiP(xi,yi)

计算它的Fourier变换，
F[uin(xi,yi)]ξ=xzλz,η=yzλz=F[ejkb0xiP(xi,yi)]ξ=xzλz,η=yzλz=F[P(xi,yi)]ξ=xzλz−b0λ,η=yzλz=LxLysinc(Lx(xzλz−b0λ,Lyy))[1+2cos⁡2πdλz(x−b0z)]\begin{aligned} & \mathcal{F}[u_{in}(x_i,y_i)]_{\xi=\frac{x_z}{\lambda z},\eta=\frac{y_z}{\lambda z}} \\ = & \mathcal{F}[e^{jkb_0x_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_xL_ysinc \left( L_x \left( \frac{x_z}{\lambda z}-\frac{b_0}{ \lambda},L_yy \right) \right) \left[ 1+ 2\cos \frac{2 \pi d}{\lambda z} \left( x- b_0z\right) \right] \end{aligned}===F[uin(xi,yi)]ξ=λzxz,η=λzyzF[ejkb0xiP(xi,yi)]ξ=λzxz,η=λzyzF[P(xi,yi)]ξ=λzxz−λb0,η=λzyzLxLysinc(Lx(λzxz−λb0,Lyy))[1+2cosλz2πd(x−b0z)]

所以光强分布为
Iz(xz,yz)=Lx2Ly2λ2z2sinc2(Lx(xzλz−b0λ,Lyy))[1+2cos⁡2πdλz(x−b0z)]2I_z(x_z,y_z)=\frac{L_x^2L_y^2}{\lambda^2z^2}sinc^2 \left( L_x \left( \frac{x_z}{\lambda z}-\frac{b_0}{ \lambda},L_yy \right) \right) \left[ 1+ 2\cos \frac{2 \pi d}{\lambda z} \left( x- b_0z\right) \right]^2 Iz(xz,yz)=λ2z2Lx2Ly2sinc2(Lx(λzxz−λb0,Lyy))[1+2cosλz2πd(x−b0z)]2

例4 有黑点的方孔Fraunhofer衍射，考虑均匀入射光，方孔的中心有一个圆形黑斑，所以pupil函数为
P(x,y)=rect(xL1,yL1)−circ(ρL2),L2<<L1P(x,y)=rect \left( \frac{x}{L_1},\frac{y}{L_1} \right)-circ\left( \frac{\rho}{L_2} \right),L_2<<L_1P(x,y)=rect(L1x,L1y)−circ(L2ρ),L2<<L1

计算far field；

答案
经过方孔后，入射光为
uin(xi,yi)=rect(xL1,yL1)−circ(ρL2)u_{in}(x_i,y_i)=rect \left( \frac{x}{L_1},\frac{y}{L_1} \right)-circ\left( \frac{\rho}{L_2} \right)uin(xi,yi)=rect(L1x,L1y)−circ(L2ρ)

计算Fourier变换，
F[uin(xi,yi)]ξ=xzλz,η=yzλz=L12sinc(L1ξ,L1η)−L22J1(πL2ξ2+η2)2L2ξ2+η2∣ξ=xzλz,η=yzλz=L12sinc(L1xzλz,L1yzλz)−L22J1(πL2(xzλz)2+(yzλz)2)2L2(xzλz)2+(yzλz)2\begin{aligned} & \mathcal{F}[u_{in}(x_i,y_i)]_{\xi=\frac{x_z}{\lambda z},\eta =\frac{y_z}{\lambda z}} \\ = & \left.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}} \right|_{\xi=\frac{x_z}{\lambda z},\eta =\frac{y_z}{\lambda z}} \\ = & L_1^2 sinc\left(\frac{L_1x_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{aligned}==F[uin(xi,yi)]ξ=λzxz,η=λzyzL12sinc(L1ξ,L1η)−L222L2ξ2+η2J1(πL2ξ2+η2)∣∣∣∣∣ξ=λzxz,η=λzyzL12sinc(λzL1xz,λzL1yz)−L222L2(λzxz)2+(λzyz)2J1(πL2(λzxz)2+(λzyz)2)