大物Ⅱ复习

静电力

点电荷相互作用力

$$\vec{F}_{21} = \frac{1} {4 \pi \varepsilon_0} \frac{q_1q_2} {r_{12} ^2} \vec{e}_{12}$$

电力叠加原理

$$\vec{F} = \sum_{i=1} ^{n} \vec{F}_i$$

静电场

电场强度

$$\vec{E} = \frac{ \vec{F} } {q_0}$$

场强叠加原理

$$\vec{E} = \sum_{i=1} ^{n} \vec{E}_i = \frac{1} {4 \pi \varepsilon_0} \sum_{i=1} ^{n} \frac{q_i} {r_i^2} \vec{e}_i$$

带电体场强

$$\vec{E} = \int \mathrm{d} \vec{E} = \frac{1} {4 \pi \varepsilon_0} \int_V \frac{\mathrm{d} q} {r^2} \vec{e}_r$$

常见场强模型

$$\vec{E}_{点电荷} = \frac{q} {4 \pi \varepsilon_0} \frac{1} {r^2} \vec{e}_r$$

$$E_{直线} = \frac{\lambda} {2 \pi \varepsilon_0} \frac{1} {a}$$

$$E_{平面} = \frac{\sigma} {2 \varepsilon_0}$$

$$E_{环} = \frac{q} {4 \pi \varepsilon_0} \frac{1}{\sqrt{R^2 + x^2}^{3} x^{-1}}$$

$$E_{盘} = \frac{q} {4 \pi \varepsilon_0} \frac{2}{ R^2}(1 - \frac{x}{\sqrt{R^2 + r^2}})$$

$$E_{球面} = \begin{cases} 0 & r < R \\ \frac{q} {4 \pi \varepsilon_0} \frac{1}{r^2} & r > R\end{cases}$$

$$E_{球体} = \begin{cases}\frac{q} {4 \pi \varepsilon_0} \frac{1}{R^3 r^{-1}} & r < R \\ \frac{q} {4 \pi \varepsilon_0} \frac{1}{r^2} & r > R \end{cases}$$

电荷密度

$$\lambda = \frac{\mathrm{d} q}{\mathrm{d}l}$$

$$\sigma = \frac{\mathrm{d} q}{\mathrm{d}S}$$

$$\rho = \frac{\mathrm{d} q}{\mathrm{d}V}$$

电通量

$$\varPhi_e = \int_S \mathrm{d} \varPhi_e = \int_S E \cos\theta \mathrm{d} S = \int_S \vec{E} \cdot \mathrm{d} \vec{S}$$

高斯定理

$$\varPhi_e = \oint_S \vec{E} \cdot \mathrm{d} \vec{S} = \frac{1}{\varepsilon_0} \sum_i q_i$$

静电场环路定理

$$\oint_L \vec{E} \cdot \mathrm{d} \vec{l} = 0$$

电势

电势能

$$W_p = A_{P \infty} = \int^{\infty}_P q_0 \vec{E} \cdot \mathrm{d} \vec{l}$$

电势

$$U_P = \frac{W_P}{q_0} = \int^{\infty}_P \vec{E} \cdot \mathrm{d} \vec{l}$$

电势差

$$U_{a b} = U_a - U_b = \int^b_a \vec{E} \cdot \mathrm{d} \vec{l}$$

电场力所做的功

$$A_{a b} = q_0\int^b_a \vec{E} \cdot \mathrm{d} \vec{l} = q_0 (U_a - U_b)$$

电势叠加原理

$$U_P = \int \mathrm{d} U = \int \frac{\mathrm{d} q}{4\pi\varepsilon_0 r}$$

常见电势模型

$$U_{点电荷} = \frac{q} {4 \pi \varepsilon_0} \frac{1} {r} \vec{e}_r$$

$$U_{直线} = \frac{\lambda} {2 \pi \varepsilon_0} \ln \frac{d}{a}$$

$$U_{平面} = \frac{\sigma} {2 \varepsilon_0} (S - x)$$

$$U_{环} = \frac{q} {4 \pi \varepsilon_0} \frac{1}{\sqrt{R^2 + x^2}}$$

$$U_{盘} = \frac{q} {4 \pi \varepsilon_0} \frac{2}{ R^2}(\sqrt{R^2 + x^2} - x)$$

$$U_{球面} = \begin{cases} \frac{q} {4 \pi \varepsilon_0} \frac{1}{R} & r < R \\ \frac{q}{4 \pi \varepsilon_0} \frac{1}{r} & r > R\end{cases}$$

$$U_{球体} = \begin{cases} \frac{q} {4 \pi \varepsilon_0} \frac{1}{2R} ( 3 - \frac{r^2}{R^2}) & r < R \\ \frac{q} {4 \pi \varepsilon_0} \frac{1}{r} & r > R \end{cases}$$

场强与电势关系

$$\vec{E} = - \frac{\mathrm{d}U}{\mathrm{d}n} \vec{e}_n = - \nabla U$$

$$\vec{E} = - (\partial_x U \vec{i} + \partial_y U \vec{j} + \partial_z U \vec{k})$$

电容

静电平衡与尖端放电

$$E_内 = 0 \quad E_{表面} = \frac{\sigma}{\varepsilon_0}$$

$$外电场 \nLeftrightarrow 内电场(接地)$$

电容

$$C = \frac{Q}{U}$$

$$C_球 = 4 \pi \varepsilon_0 R$$

常见电容器模型

$$C = \frac{Q}{U_A - U_B}$$

$$C_板 = \varepsilon_r \frac{\varepsilon_0 S}{d}$$

$$C_柱 = \varepsilon_r \frac{2 \pi \varepsilon_0 l}{ln \frac{R_B}{R_A}}$$

$$C_球 = \varepsilon_r \frac{4 \pi \varepsilon_0 }{\frac{1}{R_A} - \frac{1}{R_B}}$$

求电容步骤

$$假设带电 +q -q \rightarrow 求场强 \rightarrow 求电势差 \rightarrow C = \frac{q}{U_A - U_B} \quad C_0 = \varepsilon_r C_0$$

电介质

介质效应

$$U = \frac{U_0}{\varepsilon_r}$$

$$C = \varepsilon_r C_0$$

$$E = \frac{E_0}{\varepsilon_r} \quad \vec{E} = \vec{E}_0 + \vec{E'} \quad \varepsilon_r = 1 + \chi_e$$

电极化强度矢量

$$\vec{P} = \frac{\sum \vec{P}_i}{\Delta V} = \varepsilon_0 \chi_e \vec{E} \quad 单位:C/m^2$$

$$\oiint_S \vec{P} \cdot \mathrm{d} \vec{S} = - \sum_{S内} q'_{极化}$$

极化电荷面密度

$$\sigma'_{极化} = P \cos \theta = P_n$$

电位移矢量

$$\vec{D} = \varepsilon_0 \vec{E} + \vec{P}$$

$$\oiint_S \vec{D} \cdot \mathrm{d} \vec{S} = \sum q_{自由}$$

电物理量关系

$$\vec{D} \xrightarrow{\vec{E} = \frac{\vec{D}}{\varepsilon}} \vec{E} \xrightarrow{\vec{P} = \varepsilon_0 \chi_e \vec{E} } \vec{P} \xrightarrow{\sigma' = P \cos \theta}\sigma'$$

$$其中，\varepsilon = \varepsilon_0 \varepsilon_r \quad \chi_e = \varepsilon_r -1 \quad \sigma'为极化电荷密度$$

静电场的能量

$$W_{点电荷} = \frac{1}{2} \sum q_i U_i$$

$$W_{带电体} = \frac{1}{2} \int U \mathrm{d} q$$

$$W_{电容器} = \frac{1}{2} C U^2$$

$$能量密度：\mathcal{w}_e = \frac{1}{2} D E$$

$$W_{电场} = \int \mathcal{w}_e \mathrm{d} V = \int \frac{\varepsilon_0 \varepsilon_r E^2}{2} \mathrm{d} V$$

$$其中，\mathrm{d} V = \begin{cases} 4 \pi r^2 \mathrm{d} r & 球 \\2 \pi r l \mathrm{d} r & 柱 \\ S \mathrm{d} r & 板 \\ \end{cases}$$

磁场作用力

洛伦兹力

$$\vec{F}_磁 = q \vec{v} \times \vec{B}$$

电磁场作用力

$$\vec{F}_总 = q \vec{v} \times \vec{B} + q \vec{E}$$

安培力

$$\mathrm{d} \vec{F} = I \mathrm{d} \vec{l} \times \vec{B}$$

$$\vec{F} = \int \mathrm{d} \vec{F} = \int \mathrm{d} \vec{F}_x + \int \mathrm{d} \vec{F}_y$$

磁场

磁感应强度

$$B = \frac{F_{max}}{qv} \quad 方向：\vec{F}_{max} \times \vec{v} \quad 单位：T$$

毕奥-萨伐尔定律

$$\mathrm{d} \vec{B} = \frac{\mu_0 I \mathrm{d} \vec{l} \times \vec{r}}{4 \pi r^3}$$

$$\mathrm{d} B = \frac{\mu_0 I \mathrm{d} l \sin \theta}{4 \pi r^2}$$

$$I = \frac{\mathrm{d} q}{\mathrm{d} t}$$

运动电荷的磁场

$$\vec{B} = \frac{\mathrm{d} B}{\mathrm{d} N} = \frac{\mu_0 q \vec{v} \times \vec{r}}{4 \pi r^3}$$

磁通量

$$\varPhi_m = \int_S \mathrm{d} \varPhi_m = \int_S B \cos\theta \mathrm{d} S = \int_S \vec{B} \cdot \mathrm{d} \vec{S}$$

常见磁感应强度模型

$$B_{线} = \frac{\mu_0 I} {4 \pi a} (\cos \theta_1 - \cos \theta_2)$$

$$B_{直线} = \frac{\mu_0 I} {2 \pi a}$$

$$B_{平面} = \frac{\mu_0 j} {2}$$

$$B_{螺线管} = \frac{\mu_0}{2} n I (\cos \beta_2 - \cos \beta_1) \quad B_外 = 0$$

$$B_{螺绕环} = \begin{cases} 0 & r < R_1 \\ \mu_0 n I & R_1 < r < R_2 \\ 0 & r > R_2\end{cases}$$

$$B_{环} = \frac{\mu_0 I}{2} \frac{1}{\sqrt{R^2 + x^2}^3 R^{-2}} \quad 圆心(x = 0)处，B = \frac{\mu_0 I}{2 R}$$

$$B_{弧} = \frac{\mu_0 I}{2 R} \frac{\theta}{2 \pi}$$

$$B_{柱面} = \begin{cases} 0 & r < R \\ \frac{\mu_0 I}{2 \pi r} & r > R\end{cases}$$

$$B_{柱体} = \begin{cases} \frac{\mu_0 I}{2 \pi \frac{R^2}{r}} & r < R \\ \frac{\mu_0 I}{2 \pi r} & r > R \end{cases}$$

电流面密度

$$\sigma = \frac{I_总}{S_\perp}$$

高斯定理

$$\varPhi_m = \oint_S \vec{B} \cdot \mathrm{d} \vec{S} = 0$$

安培环路定理

$$\oint_L \vec{B} \cdot \mathrm{d} \vec{l} = \mu_0 \sum I_i$$

$$注意 I 的正负与闭合回路环绕方向的关系$$

磁力矩

磁矩

$$\vec{P}_m = I \vec{S} = I S \vec{n}$$

磁力矩

$$\vec{M} = \vec{P}_m \times \vec{B} = N I S \vec{n} \times \vec{B}$$

$$M = P_m B \sin \theta = N I S B \sin \theta$$

$$非均匀磁场中，\vec{M} = \int \vec{r} \times \mathrm{d} \vec{F}$$

磁力的功

$$A = I \Delta \varPhi$$

带电粒子在磁场中的运动

带电粒子在匀强磁场中的匀速圆周运动

$$q v_{\perp} B =\frac{m v_{\perp}^2}{R} \quad R = \frac{m v_{\perp}}{q B} \quad T = \frac{2 \pi m}{q B}$$

霍尔效应

$$U_H = R_H \frac{I B}{h} \quad R_H = \frac{1}{n q}$$

磁介质

介质效应

$$\begin{cases} \mu_r >> 1 & 铁磁质 \\ \mu_r > 1 & 顺磁质 \\ \mu_r < 1 & 抗磁质 \end{cases}$$

$$B = \mu_r B_0 \quad \vec{B} = \vec{B}_0 + \vec{B'} \quad \mu_r = 1 + \chi_m$$

磁化强度矢量

$$\vec{M} = \frac{\sum \vec{ P_{mi} } }{\Delta V} = \chi_m \vec{H} \quad 单位:A/m$$

$$\oiint_S \vec{P} \cdot \mathrm{d} \vec{S} = - \sum_{S内} q'_{极化}$$

磁化电流

$$\vert \vec{M} \vert = j_m \quad j_m = \frac{I_m}{l} \quad \oint_L \vec{B} \cdot \mathrm{d} \vec{l} = \mu_0(\sum I + \sum I_m)$$

磁场强度

$$\vec{H} = \frac{ \vec{B}_0 }{ \mu_0 } - \vec{P}$$

$$\oint_L \vec{H} \cdot \mathrm{d} \vec{l} = \sum_{in} I_{0}$$

磁物理量关系

$$I \xrightarrow{\oint_L \vec{H} \cdot \mathrm{d} \vec{l} = \sum_L I} \vec{H} \xrightarrow{\vec{B} = \mu \vec{H}} \vec{B} \xrightarrow{\vec{M} = \chi_m \vec{H}} \vec{M} \xrightarrow{j_m = M} j_m \xrightarrow{I_m = j_m l} I_m$$

$$其中，\mu = \mu_0 \mu_r \quad \chi_m = \mu_r -1 \quad j_m为磁化电流 \quad I_m为束缚电流$$

铁磁质

$$铁磁质特点、磁畴、磁滞回线、剩磁、矫顽力、居里点$$

电磁感应

感应电动势

$$\mathcal{E}_i = - \frac{\mathrm{d} \varPhi_B}{\mathrm{d} t} = -\frac{\mathrm{d}}{\mathrm{d}t}\iint_S \vec{B} \cdot \mathrm{d} \vec{S}$$

动生电动势

$$\vec{E}_K = \vec{v} \times \vec{B} \quad \mathrm{d} \mathcal{E}_i = (\vec{v} \times \vec{B}) \cdot \mathrm{d} \vec{l}$$

$$\mathcal{E}_i = \int^b_a \vec{E}_K \cdot \mathrm{d} \vec{l} = \int_a^b (\vec{v} \times \vec{B}) \cdot \mathrm{d} \vec{l}$$

$$\mathcal{E}_总 = \mathcal{E}_i时，\mathcal{E}_i = - \frac{\mathrm{d} \varPhi_B}{\mathrm{d} t} = - B \frac{\mathrm{d} S}{\mathrm{d} t}$$

涡旋电场

$$\oint_L \vec{E}_i \cdot \mathrm{d} \vec{l} = - \iint_S \frac{\partial \vec{B}}{\partial t} \cdot \mathrm{d} \vec{S}$$

$$E_{i内} = - \frac{r}{2} \frac{\mathrm{d} B}{\mathrm{d} t}$$

$$E_{i外} = - \frac{R^2}{2 r} \frac{\mathrm{d} B}{\mathrm{d} t}$$

感生电动势

$$\mathcal{E}_{i} = \int_a^b \vec{E}_i \cdot \mathrm{d} \vec{l}$$

$$\mathcal{E}_总 = \mathcal{E}_i时，\mathcal{E}_i = - \frac{\mathrm{d} \varPhi_B}{\mathrm{d} t} = - S \frac{\mathrm{d} B}{\mathrm{d} t}$$

自感

$$\varPhi = L I$$

$$\mathcal{E}_L = - \frac{\mathrm{d} \varPhi}{\mathrm{d} t} = - L \frac{\mathrm{d} I}{\mathrm{d} t}$$

$$L = \frac{\varPhi}{I}$$

$$L_{螺线管} = μ n^2 V$$

$$L = \frac{2 W_m}{I^2}$$

互感

$$\varPhi_2 = M I_1 \quad \varPhi_2 = M I_2 \quad M = M_{21} = M_{12}$$

$$\mathcal{E}_2 = - M \frac{\mathrm{d} I_{1}}{\mathrm{d} t} \quad \mathcal{E}_1 = - M \frac{\mathrm{d} I_{2}}{\mathrm{d} t}$$

$$\mathcal{E}_2 = - \frac{\mathrm{d} \varPhi_{B21}}{\mathrm{d} t} \vert_{I_1} \quad \mathcal{E}_1 = - \frac{\mathrm{d} \varPhi_{B12}}{\mathrm{d} t} \vert_{I_2}$$

$$M = \frac{\varPhi_m}{I}$$

求自感/互感系数步骤

$$在易求磁场分布的线圈中假设通电流I \rightarrow 求出相应的磁场分布 \rightarrow 计算相应的磁通量(面元法向只能取一个方向) \rightarrow 求出L/M(I一定消去)$$

磁场的能量

均匀磁场

$$磁能密度：w_m = \frac{1}{2} \vec{B} \cdot \vec{H} = \frac{1}{2} \mu H^2$$

$$W_m = \int_V w_m \mathrm{d} V = \int_V \frac{1}{2} \vec{B} \cdot \vec{H} \mathrm{d} V$$

自感/互感磁能

$$W_m = \frac{1}{2} L I_D^2$$

$$W_{21} = M_{21} I_1 I_2$$

电磁场与电磁波

位移电流

$$\vec{I}_d = \frac{\mathrm{d} \varPhi_D}{\mathrm{d} t}$$

$$\vec{j}_d = \frac{\mathrm{d} \vec{D}}{\mathrm{d} t}$$

$$j_{d平行板} = \frac{\mathrm{d} D}{\mathrm{d} t} = \varepsilon_0 \varepsilon_r \frac{\mathrm{d} E}{\mathrm{d} t} = \frac{\varepsilon_0 \varepsilon_r}{d} \frac{\mathrm{d} V}{\mathrm{d} t} \quad I_d = j_d \cdot S$$

全电流安培环路定理

$$\oint_L \vec{H} \cdot \mathrm{d} \vec{l} = I + I_d' \quad I_d' = \pi r^2 j_d$$

麦克斯韦方程组

$$\begin{aligned} & 高斯定律 : 电荷总伴随有电场 \\ &\oiint_S \vec{D} \cdot \mathrm{d} \vec{S} = \sum_{in} q = \iiint_V \rho \mathrm{d} V \\ & 法拉第电磁感应 : 变化的磁场一定伴随有电场 \\ &\oint_L \vec{E} \cdot \mathrm{d} \vec{l} = - \frac{\mathrm{d} \varPhi_m}{\mathrm{d} t} = - \iint_S \frac{\partial \vec{B}}{\partial t} \cdot \mathrm{d} \vec{S} \\ & 高斯磁定律 : 磁感应线是无头无尾的 \\ &\oiint_S \vec{B} \cdot \mathrm{d} \vec{S} = 0 \\ & 麦克斯韦-安培位移电流定律 : 变化的电场一定伴随着磁场 \\ &\oint_L \vec{H} \cdot \mathrm{d} \vec{l} = \sum_{in} (I + I_d) = \iint_S \vec{j} \mathrm{d} \vec{S} + \iint_S \frac{\partial \vec{D}}{\partial t} \cdot \mathrm{d} \vec{S} \\ \end{aligned}$$

电磁波的性质

$$\frac{E}{H} = \frac{E_0}{H_0} = \sqrt{\frac{\mu}{\varepsilon}}$$

$$v = \frac{1}{\sqrt{\varepsilon \mu}}$$

$$c = \frac{1}{\sqrt{\varepsilon_0 \mu_0}}$$

电磁波的能量密度

$$w = w_e + w_m = \frac{1}{2} \varepsilon E^2 + \frac{1}{2} \mu H^2$$

电磁波的能流密度

$$坡印廷矢量：\vec{S} = \vec{E} \times \vec{H}$$

$$S = w v$$

$$\bar{S} = \frac{1}{2} E_0 H_0$$

几何光学

符号规则

$$物距：与入射光同侧，s > 0$$

$$像距：与出射光同侧，s' > 0$$

$$曲率半径：与出射光同侧，R > 0；f与R同符号$$

$$垂直于轴的成像，轴上为正$$

反射成像

$$平面镜：s = s'$$

$$球面镜：f = \frac{R}{2} \quad \frac{1}{s} + \frac{1}{s'} = \frac{1}{f}$$

单球面折射成像

$$\frac{n_1}{s} + \frac{n_2}{s'} = \frac{n_2 - n_1}{R}$$

薄透镜成像公式

$$\frac{1}{s} + \frac{1}{s'} = \frac{1}{f} \quad m = \frac{y'}{y} = - \frac{s'}{s}$$

磨镜者公式

$$\frac{1}{f} = (n - 1)(\frac{1}{R_1} - \frac{1}{R_2})$$

光学器件

$$m_{\theta 放大镜} = \frac{\theta'}{\theta}$$

$$M_{显微镜} = m_{\theta}m = - \frac{s'}{s} \frac{25 cm}{f_e} \quad s' = f_0 + \Delta$$

$$m_{\theta 望远镜} = - \frac{f_0}{f_e}$$

光的干涉

一般步骤

$$找到干涉的两条光线 \rightarrow 计算它们到叠加点 的光程差 \rightarrow 考虑有无半波损失$$

$$\delta = n_2 r_2 - n_1 r_1 + \delta' = \begin{cases} \pm k \lambda & , k = 0, 1, 2, \dots 加强-明纹 \\ \pm (k + \frac{1}{2}) \lambda & , k = 0, 1, 2, \dots 减弱-暗纹\end{cases}$$

$$\Delta \varphi = \frac{2 \pi}{\lambda} \delta$$

杨氏双缝干涉

$$\delta = d \sin \theta = \begin{cases} \pm k \lambda & , k = 0, 1, 2, \dots 明纹 \\ \pm (k - \frac{1}{2}) \lambda & , k = 1, 2, 3, \dots 暗纹(零级条纹在光程差为0处)\end{cases}$$

$$\Delta x = x_{k + 1} - x_k = \frac{D}{d} \lambda$$

$$近轴条件下，d \sin \theta \approx \frac{d}{D} x$$

$$条纹移动意味着光程差改变：\delta = \delta_1 - \delta_2 = N \lambda$$

薄膜干涉一般步骤

$$确定薄膜, 找到哪两条光线干涉 \rightarrow 确定是反射光还是透射光干涉 \rightarrow 确定有无半波损失 \rightarrow 由明暗条纹的干涉条件分析干涉条纹特征$$

等倾干涉

$$干涉条纹：内疏外密的同心圆$$

$$\delta = 2e \sqrt{n_2^2 - n_1^2 \sin^2 i} + \delta' = \begin{cases} k \lambda & , k = 1, 2, 3, \dots 明纹 \\ (k + \frac{1}{2}) \lambda & , k = 0, 1, 2, \dots 暗纹\end{cases}$$

等厚干涉-劈尖

$$\delta = 2 n e + \delta' = \begin{cases} k \lambda & , k = 1, 2, 3,\dots 明纹 \\ (k + \frac{1}{2}) \lambda & , k = 0, 1, 2, \dots 暗纹\end{cases}$$

$$\Delta e = \frac{\lambda}{2n}$$

$$\Delta l = \frac{\Delta e}{\sin \theta} \approx \frac{\lambda}{2 n \theta}$$

等厚干涉-牛顿环

$$干涉条纹：内疏外密的同心圆$$

$$\delta = 2 n e + \delta' = \begin{cases} k \lambda & , k = 1, 2, 3, \dots 明纹 \\ (k + \frac{1}{2}) \lambda & , k = 0, 1, 2, \dots 暗纹\end{cases}$$

$$e = \frac{r^2}{2R}$$

$$r_{明环} = \sqrt{(k - \frac{1}{2})\frac{R \lambda}{n}}, k = 1, 2, 3, \dots$$

$$r_{暗环} = \sqrt{k\frac{R \lambda}{n}}, k = 0, 1, 2, \dots$$

应用

$$迈克尔孙干涉仪：2 d = \Delta N \lambda$$

$$\delta = 2 n_2 e = (k + \frac{1}{2})\lambda , k = 0, 1, 2, \dots 增透膜$$

$$\delta = 2 n_2 e + \frac{\lambda}{2}= k \lambda , k = 1, 2, 3, \dots 高反膜$$

单缝衍射

$$\delta = a \sin \theta = \begin{cases} 0 & , 中央明纹 \\ \pm k \lambda & , k = 1, 2, 3, \dots 暗纹 \\ \pm (k + \frac{1}{2}) \lambda & , k = 1, 2, 3, \dots 明纹\end{cases}$$

$$x_k = f \tan \theta_k \quad \theta \leqslant 5°时，\sin \theta \approx \tan \theta \approx \theta$$

$$\Delta x = \frac{f \lambda}{a} \quad \Delta x_{中央明纹} = \frac{2 f \lambda}{a}$$

光栅衍射

$$\begin{cases}d \sin \theta = \pm k \lambda & , k = 0, 1, 2, \dots 主极大-明纹 \\ N d \sin \theta = \pm k' \lambda & , k = 1, 2, 3, \dots, k' \not= k N 极小-暗纹\end{cases}$$

$$缺极条件：\begin{matrix}d \sin \theta = k_2 \lambda \\ a \sin \theta = k_1 \lambda\end{matrix} \Rightarrow \frac{d}{a} = \frac{k_2}{k_1} \in \mathbb{Z}$$

$$斜入射时的光栅方程：d(\sin \varphi + \sin \theta) = \pm k \lambda \quad k = 0, 1, 2, \dots 主极大$$

$$\varphi 和\theta 在法线的同侧同号，异侧异号；两侧 k_{max} 不同$$

$$分辨本领(分辨波长)：R = \frac{\lambda}{\Delta \lambda} \leqslant k N$$

圆孔衍射

$$\theta = 1.22 \frac{\lambda}{D} \quad D 为圆孔孔径$$

$$分辨本领(分辨波长)：R = \frac{1}{\theta_{min}} = \frac{D}{1.22 \lambda}$$

晶体衍射

$$布拉格方程：2 d \sin \theta = k \lambda \quad k = 1, 2, 3, \dots$$

$$其中\theta 为掠射角$$

光的偏振

马吕斯定律

$$I \propto A^2 \quad A_o \rightarrow A_o \cos \alpha$$

$$I_{出线} = \frac{1}{2} I_{入自}$$

$$I_{出线} = I_{入线} \cos^2 \alpha$$

布儒斯特定律

$$\tan i_0 = \frac{n_2}{n_1} \Rightarrow i_0 + \gamma = \frac{\pi}{2}$$

$$此时有线偏振光$$

双折射现象

$$o 光振动方向垂直于光轴，e 光不符合折射定律$$

$$\delta = \vert n_0 - n_e \vert d$$

$$\Delta \varphi = \frac{2 \pi}{\lambda} \delta$$

$$\delta_{\frac{1}{2}波片} = \vert n_0 - n_e \vert d = \frac{\lambda}{2}$$

$$\delta_{\frac{1}{4}波片} = \vert n_0 - n_e \vert d = \frac{\lambda}{4}$$

偏振光的干涉

$$\Delta \varphi_{正交} = \frac{2 \pi}{\lambda} \vert n_0 - n_e \vert d + \pi$$

$$A_{e2} = A_{o2} = A \cos \alpha \cos \beta$$

$$\Delta \varphi_{平行} = \frac{2 \pi}{\lambda} \vert n_0 - n_e \vert d$$

$$A_{e2} = A \cos^2 \alpha \quad A_{o2} = A \cos^2 \beta$$

$$相强干涉与相消干涉$$

椭圆偏振光

$$\alpha = 0° 或 90° ？\begin{cases}是：出射与入射相同，为线偏振光 \\ 否，使用 \frac{1}{2} 波片 \begin{cases}是：出射与入射振动方向关于光轴对称 \\ 否，使用 \frac{1}{2} 波片 \begin{cases} \alpha = 45°：出射为圆偏振光 \\ \alpha \ne 45°：出射为椭圆偏振光 \end{cases}\end{cases}\end{cases}$$

电磁辐射的量子性

黑体辐射

$$M_B(T) = \sigma T^4$$

$$T \lambda_m = b$$

光电效应

$$e U_a = E_{km} = \frac{1}{2} m v_{max}^2$$

$$红限波长：A = h \nu_0 = h \frac{c}{\lambda_0}$$

$$光的强度：I = N h \nu$$

$$h \nu = E_{km} + A = \frac{1}{2} m v_{max}^2 + A$$

康普顿散射

$$\Delta \lambda = \lambda - \lambda_0 = \lambda_c(1 - \cos \varphi) = 2 \lambda_c \sin^2 \frac{\varphi}{2}$$

$$\lambda_c = \frac{h}{m_0 c} = 0.00243 nm$$

光的波粒二象性

$$E = h \nu = m c^2 = E_k$$

$$m = \frac{h \nu}{c^2} = \frac{h}{\lambda c}$$

$$p = m c = \frac{h \nu}{c} = \frac{h}{\lambda}$$

量子力学简介

德布罗意波

$$E = h \nu = m c^2 \quad m = \frac{m_0}{\sqrt{1-(\frac{v}{c})^2}}$$

$$E_k = m c^2 - m_0 c^2$$

$$m = \frac{h \nu}{c^2} = \frac{h}{\lambda c}$$

$$p = m v = \frac{h}{\lambda} \quad v \ne \lambda \nu$$

$$v << c 时，E_k = m c^2 \approx \frac{1}{2} m_0 v^2 = \frac{p^2}{2 m_0} \quad \lambda = \frac{h}{p} = \frac{h}{m_0 v}$$

不确定关系

$$\Delta x \Delta p_x \geqslant \frac{\hbar}{2} \quad \hbar = \frac{h}{2 \pi}$$

$$由于 p = \frac{h}{\lambda} \quad \Delta p = \frac{h}{\lambda^2} \Delta \lambda$$

$$有 \Delta E \Delta t \geqslant \frac{\hbar}{2} \quad \Delta t 为粒子在某能级的寿命$$

波函数及其统计意义

$$概率密度：f_p(x) = \vert \varphi (x) \vert^2$$

$$在 \mathrm{d} x 范围内出现粒子的概率：\vert \varphi (x) \vert^2 \mathrm{d} x$$

$$波函数单值、有限、连续时，\int_{- \infty}^{+ \infty} \vert \varphi (x) \vert^2 \mathrm{d} x = 1 \Rightarrow A$$

定态薛定谔方程与一维无限深势阱

$$零点能：E_1 = \frac{\pi^2 \hbar^2}{2 m a^2} \quad n能级能量：E_n = n^2 E_1$$

$$\psi_n(x) = \sqrt{\frac{2}{a}} \sin (\frac{n \pi}{a} x) \quad 0 \leqslant x \leqslant a \quad n = 1, 2, 3, \dots$$

$$概率密度：f_p(x) = \vert \psi_n(x) \vert^2 = \frac{2}{a} \sin^2 (\frac{n \pi}{a} x) \quad 0 \leqslant x \leqslant a \quad n = 1, 2, 3, \dots$$

$$x ～ x + \mathrm{d} x 概率：\mathrm{d} P = \vert \psi_n(x) \vert^2 \mathrm{d} x = \frac{2}{a} \sin^2 (\frac{n \pi}{a} x) \mathrm{d} x \quad 0 \leqslant x \leqslant a \quad n = 1, 2, 3, \dots$$

$$x_1 ～ x_2 概率：P = \int_{x_1}^{x_2} \vert \psi_n(x) \vert^2 \mathrm{d} x = \int_{x_1}^{x_2} \frac{2}{a} \sin^2 (\frac{n \pi}{a} x) \mathrm{d} x \quad 0 \leqslant x \leqslant a \quad n = 1, 2, 3, \dots$$

氢原子及原子结构初步

氢光谱的规律

$$\frac{1}{\lambda} = R_H (\frac{1}{k^2} - \frac{1}{n^2}) \quad k = 1, 2, 3, \dots ; n = k+1, k+2, k+3, \dots$$

$$\begin{cases} k = 1，紫外线系 \\ k = 2，巴尔末线系 \\ k = 3，红外线系 \end{cases}$$

玻尔氢原子理论

$$E_n = - \frac{13.6}{n^2} eV \quad n = 1, 2, 3, \dots$$

$$r_n = n^2 r_1 \quad n = 1, 2, 3, \dots$$

$$h \nu = \frac{h c}{\lambda} = E_n - E_k$$

$$横线上密下疏，数量为 n ；同线系箭头向下，起点向右递增，终点相同；能画出 n - 1 个线系$$

氢原子的量子力学描述

$$主量子数 n ：E_n = - \frac{13.6}{n^2} eV \quad n = 1, 2, 3, \dots$$

$$角量子数 l ：L = \sqrt{l(l + 1)} \hbar \quad l = 0, 1, 2, \dots , n - 1$$

$$磁量子数 m_l ：L_Z = m_l \hbar \quad m_l = 0, \pm 1, \pm 2, \dots , \pm l$$

$$自旋磁量子数 m_s ：S = \sqrt{s(s + 1)} \hbar = \frac{\sqrt{3}}{2} \hbar \quad S_Z = m_S \hbar = \pm \frac{1}{2} \hbar$$

$$某一能级可容纳的最多电子数为 2 n^2 个, 某一层可容纳的最多电子数为 2(2 l + 1) 个$$

$$波尔理论：L = m v r = n \hbar \quad 量子力学：L = \sqrt{l(l + 1)} \hbar$$

$$核外电子径向概率密度：P(r) = r^2 \vert R_{nl}(r) \vert^2$$

激光与固体的能带结构

激光产生条件与特性

$$粒子数反转(受激辐射)；光放大(光学谐振腔)$$

$$方向性好；亮度高；单色性好；相干性好$$

固体能带结构

$$p 型半导体，空穴导电，掺入3价金属； n 型半导体，电子导电，掺入5价金属$$

$$禁带宽度与外加光子能量关系： h \nu \geqslant E_g$$