平面波在双层介质的传播学性质

Optics

目的

目前我们课题组的实验基本都是片上的一些实验，实际的实验中片上会有一些光学微纳元件，如量子点、金纳米颗粒等。我最近需要计算片上的散射、透射性质，为了增加结果的精度，背景场最好是输入解析的表达式。将解析表达式输入进去COMSOL也得格外注意。

关系推导

边界条件以及平面波的基本性质

对于任意的平面波，当其在介质表面发生透射、反射时，由于其偏振方向和入射面的方向的不同，可以分两种情况来讨论

• S偏振， 电场与入射面垂直，如下面的左图所示
• P偏振，电场与入射面平行，如下面的右图所示

注意下面的图的正方向规定。因为不同的规定最后的表达式会有正负号的区别，我接下来的所有推导都是按照下面的图来的。

即任意的平面波可以写为S偏振和P偏振的合成

$$\vec{E}=\vec{E}^{s}+\vec{E}^{p}$$

我们假设一平面波从介质1入射到介质2，我们规定的方向为由介质2指向介质1，则电场和磁场需要满足对应的边界条件

$$\vec{n}\times (\vec{E}_{1}-\vec{E}_{2})=0$$ $$\vec{n}\cdot (\vec{D}_{1}-\vec{D}_{2})=\sigma$$ $$\vec{n}\times (\vec{H}_{1}-\vec{H}_{2})=\vec{K}$$ $$\vec{n}\cdot (\vec{B}_{1}-\vec{B}_{2})=0$$

根据上图所示的定义，我们可以将场和波矢分别分解。无论是P偏振还是S偏振，都可以将波矢写为

$$\vec{k}_{1}=(k_x,k_y,k_{z1})$$ $$\vec{k}_{2}=(k_x,k_y,k_{z2})$$

横向的波矢是连续的，可以定义一个横向的波数

$$k_{t}=\sqrt{k_{x}^2+k_{y}^2}=k_{1}\sin{\theta_{1}}$$

为了得出地磁场的反射、透射性质，我们需要根据边界条件来分析。我们定义在介质1的入射电场为

$$\vec{E}_{1}=\vec{E}_{1}^{\prime}e^{i\vec{k}\vec{r}-i\omega t}$$

另外，需要强调的是折射率的定义、速度的定义，平面波在介质中的传播速度为

$$\nu=\sqrt{\frac{1}{\varepsilon\mu}}=c/n$$

即折射率可以表示为

$$n=\sqrt{\varepsilon\mu}\cdot c$$

对于平面波来说，其电场和磁场的关系为

$$E=B\nu=B\sqrt{\frac{1}{\varepsilon\mu}}$$

而平面波的功率密度可以表示为

$$S=E\times H=E\cdot B/\mu=E^2\sqrt{\frac{\varepsilon}{\mu}}$$

平均功率密度可以表示为

$$\langle S \rangle=\frac{1}{2}\text{Re}\{E\times H^{*}\}=\frac{1}{2}\sqrt{\frac{\varepsilon}{\mu}}|E|^2$$

S偏振

对于S偏振，首先是电场切向分量连续

$$E_{1}+E_{1r}=E_{2}=E_{1t}$$

然后是磁场切向分量连续

$$-\frac{B_{1}}{\mu_{1}}\cos(\theta_{1})+\frac{B_{1r}}{\mu_{1}}\cos(\theta_{1})=-\frac{B_{2}}{\mu_{2}}\cos(\theta_{2})=-\frac{B_{1t}}{\mu_{2}}\cos(\theta_{2})$$

根据电场、磁场的关系，（需要注意的是，）我们可以将磁场的关系转化为电场的关系

$$-\frac{E_{1}}{\mu_{1}\nu_{1}}\cos(\theta_{1})+\frac{E_{1r}}{\mu_{1}\nu_{1}}\cos(\theta_{1})=-\frac{E_{2}}{\mu_{2}\nu_{2}}\cos(\theta_{2})=-\frac{E_{1t}}{\mu_{1}\nu_{2}}\cos(\theta_{2})$$ $$-\frac{E_{1}n_{1}}{\mu_{1}}\cos(\theta_{1})+\frac{E_{1r}n_{1}}{\mu_{1}}\cos(\theta_{1})=-\frac{E_{2}n_{2}}{\mu_{2}}\cos(\theta_{2})=-\frac{E_{1t}n_{2}}{\mu_{1}}\cos(\theta_{2})$$

综合可得，最后$E{1},E{1r},E_{1t}$的关系

$$\frac{E_{r}^{s}}{E_{1}^{s}}=\frac{\frac{n_{1}}{\mu_{1}}\cos(\theta_{1})-\frac{n_{2}}{\mu_{2}}\cos(\theta_{2})}{\frac{n_{1}}{\mu_{1}}\cos(\theta_{1})+\frac{n_{2}}{\mu_{2}}\cos(\theta_{2})}$$ $$\frac{E_{t}^{s}}{E_{1}^{s}}=\frac{2\frac{n_{1}}{\mu_{1}}\cos(\theta_{1})}{\frac{n_{1}}{\mu_{1}}\cos(\theta_{1})+\frac{n_{2}}{\mu_{2}}\cos(\theta_{2})}$$

上面的表达式还可以进一步化为波矢表示的形式，，因为$k{z1}=k{1}\sin(\theta{1}),k{z2}=k{2}\sin(\theta{2})$,因此

$$\frac{E_{r}^{s}}{E_{1}^{s}}=\frac{\mu_{2}k_{z1}-\mu_{1}k_{z2}}{\mu_{2}k_{z1}+\mu_{1}k_{z2}}$$ $$\frac{E_{t}^{s}}{E_{1}^{s}}=\frac{2\mu_{2}k_{z1}}{\mu_{2}k_{z1}+\mu_{1}k_{z2}}$$

P偏振

对于P偏振，首先是切向的电场连续，即

$$-E_{1}\cos(\theta_{1})+E_{1r}\cos(\theta_{1})=-E_{2}\cos(\theta_{2})=-E_{1t}\cos(\theta_{2})$$

然后是磁场切向连续

$$-H_{1}-H_{1r}=-H_{2}=-H_{1t}$$

进一步转化

$$\frac{E_{1}}{\nu_{1}\mu_{1}}+\frac{E_{1r}}{\nu_{1}\mu_{1}}=\frac{E_{2}}{\nu_{2}\mu_{2}}=\frac{E_{1t}}{\nu_{2}\mu_{2}}$$ $$n_{1}E_{1}/\mu_{1}+n_{1}E_{1r}/\mu_{1}=n_{2}E_{2}/\mu_{2}=n_{2}E_{1t}/\mu_{2}$$

综合可得，最后$E{1},E{1r},E_{1t}$的关系

$$\frac{E_{r}^{p}}{E_{1}^{p}}=\frac{\frac{n_{2}}{\mu_{2}}\cos(\theta_{1})-\frac{n_{1}}{\mu_{1}}\cos(\theta_{2})}{\frac{n_{2}}{\mu_{2}}\cos(\theta_{1})+\frac{n_{1}}{\mu_{1}}\cos(\theta_{2})}=$$ $$\frac{E_{t}^{p}}{E_{1}^{p}}=\frac{2\frac{n_{1}}{\mu_{1}}\cos(\theta_{1})}{\frac{n_{2}}{\mu_{2}}\cos(\theta_{1})+\frac{n_{1}}{\mu_{1}}\cos(\theta_{2})}$$

上面的表达式还可以进一步化为波矢表示的形式，因为$k{z1}=k{1}\sin(\theta{1}),k{z2}=k{2}\sin(\theta{2})k{1}=k{0}n{1},k{2}=k{0}n{2}$,因此

$$\begin{aligned} \frac{E_{r}^{p}}{E_{1}^{p}}&=\frac{\frac{n_{2}}{\mu_{2}}\cos(\theta_{1})-\frac{n_{1}}{\mu_{1}}\cos(\theta_{2})}{\frac{n_{2}}{\mu_{2}}\cos(\theta_{1})+\frac{n_{1}}{\mu_{1}}\cos(\theta_{2})}\\ &=\frac{\frac{n_{2}}{\mu_{2}n_{1}}k_{z1}-\frac{n_{1}}{\mu_{1}n_{2}}k_{z2}}{\frac{n_{2}}{\mu_{2}n_{1}}k_{z1}+\frac{n_{1}}{\mu_{1}n_{2}}k_{z2}}\\ &=\frac{n_{2}^2\mu_{1}k_{z1}-n_{1}^2\mu_{2}k_{z2}}{n_{2}^2\mu_{1}k_{z1}+n_{1}^2\mu_{2}k_{z2}}\\ &=\frac{\varepsilon_{2} k_{z1}-\varepsilon_{1}k_{z2}}{\varepsilon_{2} k_{z1}+\varepsilon_{1}k_{z2}}\\ \end{aligned}$$

按照同样的变换方式，我们可以得到

$$\begin{aligned} \frac{E_{t}^{p}}{E_{1}^{p}}&=\frac{2\frac{n_{1}}{\mu_{1}}\cos(\theta_{1})}{\frac{n_{2}}{\mu_{2}}\cos(\theta_{1})+\frac{n_{1}}{\mu_{1}}\cos(\theta_{2})}\\ &=\frac{2\varepsilon_{2} k_{z1}}{\varepsilon_{2} k_{z1}+\varepsilon_{1}k_{z2}}\frac{n_{1}\mu_{2}}{n_{2}\mu_{1}}\\ &=\frac{2\varepsilon_{2} k_{z1}}{\varepsilon_{2} k_{z1}+\varepsilon_{1}k_{z2}}\sqrt{\frac{\mu_{2}\varepsilon_{1}}{\mu_{1}\varepsilon_{2}}} \end{aligned}$$

此结果与书Principle of Nano Optics结果是一致的。结果此次推导，我算是更加熟悉了所谓的菲列尔公式了。需要注意的是，我们的推导都是基于特定的图写出来的，不同的图会有不同的符号。

::: alert-warning

Nonetheless, be aware that the literature is not standardized, and all possible sign variations have been labeled the Fresnel Equations. To avoid confusion they must be related to the specific field directions from which they were derived.

:::

特殊情况-非磁性材料

现实中大部分材料的磁性可以忽略，我们认为$\mu{1}=\mu{2}$，因此

$$\frac{E_{r}^{s}}{E_{1}^{s}}=\frac{\frac{n_{1}}{\mu_{1}}\cos(\theta_{1})-\frac{n_{2}}{\mu_{2}}\cos(\theta_{2})}{\frac{n_{1}}{\mu_{1}}\cos(\theta_{1})+\frac{n_{2}}{\mu_{2}}\cos(\theta_{2})}=\frac{n_{1}\cos(\theta_{1})-n_{2}\cos(\theta_{2})}{n_{1}\cos(\theta_{1})+n_{2}\cos(\theta_{2})}=-\frac{\sin(\theta_{1}-\theta_{2})}{\sin (\theta_{1}+\theta_{2})}$$ $$\frac{E_{t}^{s}}{E_{1}^{s}}=\frac{2\frac{n_{1}}{\mu_{1}}\cos(\theta_{1})}{\frac{n_{1}}{\mu_{1}}\cos(\theta_{1})+\frac{n_{2}}{\mu_{2}}\cos(\theta_{2})}=\frac{2n_{1}\cos(\theta_{1})}{n_{1}\cos(\theta_{1})+n_{2}\cos(\theta_{2})}=+\frac{2\sin(\theta_{2})\cos(\theta_{1})}{\sin (\theta_{1}+\theta_{2})}$$ $$\frac{E_{r}^{p}}{E_{1}^{p}}=\frac{\frac{n_{2}}{\mu_{2}}\cos(\theta_{1})-\frac{n_{1}}{\mu_{1}}\cos(\theta_{2})}{\frac{n_{2}}{\mu_{2}}\cos(\theta_{1})+\frac{n_{1}}{\mu_{1}}\cos(\theta_{2})}=\frac{n_{2}\cos(\theta_{1})-n_{1}\cos(\theta_{2})}{n_{2}\cos(\theta_{1})+n_{1}\cos(\theta_{2})}=+\frac{\tan(\theta_{1}-\theta_{2})}{\tan (\theta_{1}+\theta_{2})}$$ $$\frac{E_{t}^{p}}{E_{1}^{p}}=\frac{2\frac{n_{1}}{\mu_{1}}\cos(\theta_{1})}{\frac{n_{2}}{\mu_{2}}\cos(\theta_{1})+\frac{n_{1}}{\mu_{1}}\cos(\theta_{2})}=\frac{2n_{1}\cos(\theta_{1})}{n_{2}\cos(\theta_{1})+n_{1}\cos(\theta_{2})}=+\frac{2\sin(\theta_{2})\cos(\theta_{1})}{\sin (\theta_{1}+\theta_{2})\sin (\theta_{1}-\theta_{2})}$$

特殊情况-垂直入射

S,P偏振在垂直入射时的情况将会一样，为了简便我们认为$\mu{1}=\mu{2}\theta{1}=\theta{2}=0$,此时

$$\frac{E_{r}^{s}}{E_{1}^{s}}=\frac{\frac{n_{1}}{\mu_{1}}\cos(\theta_{1})-\frac{n_{2}}{\mu_{2}}\cos(\theta_{2})}{\frac{n_{1}}{\mu_{1}}\cos(\theta_{1})+\frac{n_{2}}{\mu_{2}}\cos(\theta_{2})}=\frac{n_{1}-n_{2}}{n_{1}+n_{2}}$$ $$\frac{E_{t}^{s}}{E_{1}^{s}}=\frac{2\frac{n_{1}}{\mu_{1}}\cos(\theta_{1})}{\frac{n_{1}}{\mu_{1}}\cos(\theta_{1})+\frac{n_{2}}{\mu_{2}}\cos(\theta_{2})}=\frac{2n_{1}}{n_{1}+n_{2}}$$ $$\begin{aligned} \frac{E_{r}^{p}}{E_{1}^{p}}&=\frac{\frac{n_{2}}{\mu_{2}}\cos(\theta_{1})-\frac{n_{1}}{\mu_{1}}\cos(\theta_{2})}{\frac{n_{2}}{\mu_{2}}\cos(\theta_{1})+\frac{n_{1}}{\mu_{1}}\cos(\theta_{2})}\\ &=\frac{n_{2}-n_{1}}{n_{2}+n_{1}} \end{aligned}$$ $$\begin{aligned} \frac{E_{t}^{p}}{E_{1}^{p}}&=\frac{2\frac{n_{1}}{\mu_{1}}\cos(\theta_{1})}{\frac{n_{2}}{\mu_{2}}\cos(\theta_{1})+\frac{n_{1}}{\mu_{1}}\cos(\theta_{2})}\\ &=\frac{2n_{1}}{n_{2}+n_{1}} \end{aligned}$$

我们发现s偏振和p偏振的反射刚好有一个负号，这是由与s和p偏振对于电场的正方向定义有区别。

• 对于s偏振应该是
• 对于p偏振应该是

特殊情况：全反射

根据菲涅尔公式，

$$\sin(\theta_{2})n_{2}=\sin(\theta_{1})n_{1}\implies \sin(\theta_{2})=\frac{\sin(\theta_{1})n_{1}}{n_{2}}$$

考虑$n{1}>n{2}\theta{1}<\theta{2}\theta{c}=\arcsin(\frac{n{1}}{n_{2}})$。再继续增加角度，透过界面后的波会沿着z方向指数衰减。

although there must be a transmitted wave, it cannot, on the average, carry energy across the boundary.

下面我们来分析光再全反射下的具体物理过程。将前面的透射、反射系数进一步表示为

$$r^{s}=\frac{\cos\theta_{1}-\sqrt{(\frac{n_{2}^2}{n_{1}^2}-\sin^2\theta_{1})}}{\cos\theta_{1}+\sqrt{(\frac{n_{2}^2}{n_{1}^2}-\sin^2\theta_{1})}}$$ $$r^{s}=\frac{\frac{n_{2}^2}{n_{1}^2}\cos\theta_{1}-\sqrt{(\frac{n_{2}}{n_{1}}-\sin^2\theta_{1})}}{\frac{n_{2}^2}{n_{1}^2}\cos\theta_{1}+\sqrt{(\frac{n_{2}}{n_{1}}-\sin^2\theta_{1})}}$$

在$\theta{1}\ge\theta{c}$时，根号里就是一个虚数。这样的话反射系数的分子分母只是一个复共轭的差别，但是模的平方等于1。即此时所有的能量其实是全部反射了的，当然，这不代表没有能量透过界面，只是透过之后在短距离衰减了。我们可以将透射的波写为

$$\vec{\vec{E}_{2}}=\vec{\vec{E}_{02}}e^{i\vec{k}_{2}\cdot \vec{r}-i\omega t}$$

其中，$\vec{k}{2}\cdot \vec{r}=k{2x}x+k{2z}z=k{2}\sin\theta{2}+k{2}\cos\theta_{2}$，根据菲涅尔定律

$$k_{2}\cos\theta_{2}=\pm k_{2}\sqrt{1-\frac{\sin^2\theta_{1}}{n_{2}^2/n_{1}^2}}=\pm ik_{2}\sqrt{\frac{\sin^2\theta_{1}}{n_{2}^2/n_{1}^2}-1}=\pm i\beta_{z}$$

而

$$k_{2x}=\frac{k_{2}}{n_{2}/n_{1}}\sin\theta_{1}$$

即最终的场可以表示为

$$\vec{\vec{E}_{2}}=\vec{\vec{E}_{02}}e^{\pm \beta_{z}z}e^{ik_{x}x\sin\theta_{1}n_{1}/n_{2}-i\omega t}$$

其中为了的符号必须让最终的场衰减，即取。