第一章

$\text{§1.1}$

电磁波速度

$$v=\frac{c}{\sqrt{{ϵ}_0{\mu }_r}}$$

折射率

$$n=\frac{c}{v}n=\sqrt{{ϵ}_r{\mu }_r}$$

光强正比于振幅的平方。

$$I=A^2$$

两个频率相同，振幅与相位不同的震动:

$$\begin{array}{rl}{E}_1& =A_1\cos (\omega t+{\varphi }_1)\\ E_2& =A_2\cos (\omega t+{\varphi }_2)\end{array}$$

其叠加结果为

$$\begin{gathered} A^2=A_1^2+A_2^2+2A_1{A}_2\cos ({\varphi }_2-{\varphi }_{1}0) \\ \tan \varphi =\frac{A_1\sin {\varphi }_1+A_2\sin {\varphi }_2}{A_1\cos {\varphi }_1+A_2\cos {\varphi }_2} \end{gathered}$$

可得

$$\begin{gathered} I_{max}=(A_1+A_2{)}^2 \\ {I}_{min}=(A_1-A_2{)}^2 \end{gathered}$$

$\text{§1.2}$

双缝干涉

$$\begin{array}{}\text{暗条纹}& r_2-r_1=(2j+1)\frac{\lambda }{2}\end{array}$$$$\begin{array}{}\text{明条纹}& r_2-r_1=2j\frac{\lambda }{2}\end{array}$$