# 模拟物体在基板上的光散射

2020年 4月 14日

### 背景和概述

\nabla \times \left( \mu_r \nabla \times \mathbf{E}_{total} \right) – \frac{\omega^2}{c_0^2}\left( \epsilon_r – \frac{i \sigma}{\omega\epsilon_0}\right) \mathbf{E}_{total}= 0

\mathbf{E}_{total} = \mathbf{E}_{relative} + \mathbf{E}_{background}

### 使用菲涅耳方程定义背景场

\frac{\sin \theta_t}
{\sin \theta_i}= \frac{n_a}{n_b}

\begin{array}
rr_{TE} = \frac{n_a \cos \theta_i – n_b \cos \theta_t}{n_a \cos \theta_i + n_b \cos \theta_t} \\
t_{TE} = \frac{2 n_a \cos \theta_i}{n_a \cos \theta_i + n_b \cos \theta_t} \\
r_{TM} = \frac{n_b \cos \theta_i – n_a \cos \theta_t}{n_b \cos \theta_i + n_a \cos \theta_t} \\
t_{TM} = \frac{ 2 n_a \cos \theta_i }{n_b \cos \theta_i + n_a \cos \theta_t}
\end{array}

\mathbf{k}_i = \frac{2 \pi n_a}{\lambda_0} \left

\mathbf{k}_t = \frac{2 \pi n_b}{\lambda_0}\left

E_{i,x} = (E_{i,TM} \cos \theta_i \cos \phi – E_{i,TE} \sin \phi )\exp (-i\mathbf{k}_i \cdot \mathbf{x})
E_{i,y} = (E_{i,TM} \cos \theta \sin \phi + E_{i,TE} \cos \phi )\exp (-i\mathbf{k}_i \cdot \mathbf{x})
E_{i,z} = E_{i,TM} \sin \theta_i \exp (-i\mathbf{k}_i \cdot \mathbf{x})

E_{r,x} = (-r_{TM} E_{i,TM} \cos \theta_i \cos \phi – r_{TE} E_{i,TE} \sin \phi )\exp (-i\mathbf{k}_r \cdot \mathbf{x})
E_{r,y} = (-r_{TM} E_{i,TM} \cos \theta_i \sin \phi + r_{TE} E_{i,TE} \cos\phi )\exp (-i\mathbf{k}_r \cdot \mathbf{x})
E_{r,z} = r_{TM} E_{i,TM} \sin \theta_i \exp (-i\mathbf{k}_r \cdot \mathbf{x})

E_{t,x} = (t_{TM} E_{i,TM} \cos \theta_t \cos \phi – t_{TE} E_{i,TE} \sin \phi )\exp (-i\mathbf{k}_t \cdot \mathbf{x})
E_{t,y} = (t_{TM} E_{i,TM} \cos \theta_t \sin \phi + t_{TE} E_{i,TE} \cos \phi )\exp (-i\mathbf{k}_t \cdot \mathbf{x})
E_{t,z} = t_{TM} E_{i,TM} \sin \theta_t \exp (-i\mathbf{k}_t \cdot \mathbf{x})