Theory of interference

原子的发光机制

自发辐射,激发态寿命决定波列的发光时间.

规定

• 波列的发光时间 \(\Delta t\), 或原子在激发态的寿命
• 波列长度/相干长度 \(\Delta x\)
• 光波列的频谱宽度 \(\Delta\lambda\)/\(\Delta\nu\)

理论可证明: \[\Delta x\cdot\Delta k\approx 2\pi,\;k= \frac{2\pi}{\lambda}.\] \[\Rightarrow \Delta\cdot\Delta\nu = 1\] 显然

• \(\Delta x\rightarrow\infty\), \(\Delta\lambda\rightarrow 0\)
• \(\Delta t\rightarrow\infty\), \(\Delta\nu\rightarrow\infty\)

两类实现光波干涉的装置

• 分波前干涉;
• 分振幅干涉.

实验装置如下图: (tikz code)

由前述, \(d^{\prime}\gg d\), 故波程差有: \[\Delta r = r_2 - r_1 \approx d\sin\theta = d \frac{x}{d^{\prime}}\]

• \(\Delta r =\pm m\lambda\), 相干加强, \(x = \pm m \frac{d^{\prime}}{d}\lambda\), 明纹;
• \(\Delta r = \pm (2m+1) \frac{\lambda}{2}\), 相干减弱, \(x = \pm \frac{d^{\prime}}{d}(2m+1) \frac{\lambda}{2}\) 暗纹.

此处 \(m\) 均取 \(0, 1, 2, \cdots\) 另外我们得到明/暗纹间距为: \[\Delta x = \frac{d^{\prime}}{d}\lambda\] 当 \(d, d^{\prime}\) 一定时, 若 \(\lambda\) 增大 (\(\omega\) 减小), 则 \(\Delta x\) 增大.

如果采用不同波长的光作杨氏干涉实验,形成的干涉条纹间距不同,但零级亮纹位置相同.

光源偏移对干涉条纹影响

光源宽度对干涉条纹的影响

衬比度

\[\gamma = \frac{\max I -\min I}{\max I + \min I}\]

若光程差大于光的相干长度, \(P\) 点的干涉图案变模糊.

• 双镜,
• 劳埃德镜(半波损失),
• 菲涅耳双棱镜.

实验装置如下: (tikz code)

反射光 2、3 的光程差近似为 \(AD\): \[\Delta_r = 2d \sqrt{n_2^2 - n_1^2\sin^2 i} + \frac{\lambda}{2}\]

• \(\Delta_r = m\lambda\), (\(m = 1, 2, \cdots\)), 相干加强.
• \(\Delta_r = (2m+1)\frac{\lambda}{2}\), (\(m = 0, 1, 2, \cdots\)), 相干减弱.

用 (点光源透过凸透镜产生的) 平行光垂直入射一劈尖, 其劈尖角度为 \(\theta\)、另一侧棱边高度为 \(D\). 显微镜观察到反射光有平行于棱边的干涉条纹. 明/暗纹间隔为 \(b\). 如下图, 劈尖角度有夸大: (tikz code)

玻片是光密介质, 显然 \(n < n_1\), 光程差有: \[\Delta = 2nd + \frac{\lambda}{2}\]

• \(\Delta = m\lambda\), \(m =1, 2, \cdots\), 明纹.
• \(\Delta = (2m+1)\frac{\lambda}{2}\), \(m = 0, 1,\cdots\), 暗纹.

有推论:

• 劈尖处 \(d = 0\), \(\Delta = \frac{\lambda}{2}\), 中心暗纹.
• 相邻明暗纹间厚度差 \[\Delta d = \frac{\lambda}{2n}\]
• 由于劈尖角 \(\theta\) 很小, 故 \(\theta\approx\sin\theta\approx\tan\theta\), 即 \[\frac{D}{L}=\frac{\lambda/2n}{b}\] 故明/暗条纹间距 \(b\) 有: \[b = \frac{\lambda}{2n\theta}.\]

将一块平凸透镜凸面朝下放在一块平面透镜上，将单色光直射向凸镜的平面，可以观察到一个个明暗相间的圆环条纹。若使用白光，则可以观察到彩虹状的圆环彩色条纹。实验装置如下图: (tikz code)

光程差有: \[\Delta = 2d + \frac{\lambda}{2}\]

• \(\Delta = m \lambda\), 明纹
• \(\Delta = \left (m+\frac{1}{2} \right )\lambda\), 暗纹.

设生成环半径为 r, 凸透镜曲率半径为 \(R\), 易得: (tikz code)

\[r^2 + (R-d)^2 = R^2\Rightarrow r^2 = 2dR -d^2\] 由于 \[R \ll d, d^2\approx 0\], 可得: \[r = \sqrt{2dR}\]

• \(r = \sqrt{\left( m - \frac{1}{2} \right) R\lambda}\), 明纹半径
• \(r = \sqrt{mR\lambda}\), 暗纹半径.

• 测量相干长度,
• 引力波.

• 一次反射 \(\tilde{E}_{1r} = E_0 r \mathrm{e}^{\mathrm{i}\omega t}\)
• 二次反射 \(\tilde{E}_{2r} = E_0 tr^{\prime}t^{\prime} \mathrm{e}^{\mathrm{i}(\omega t-\delta)}\)
• 三次反射 \(\tilde{E}_{3r} = E_0 t{r^{\prime}}^3t^{\prime} \mathrm{e}^{\mathrm{i}(\omega t-2\delta)}\)

如是, \(N\) 次反射有: \[\tilde{E}_{Nr} = E_0 t{r^{\prime}}^{2N-3}t^{\prime} \mathrm{e}^{\mathrm{i}[\omega t-(N-1)\delta]}\] 其中 \(\delta\) 为 \[\delta = \frac{2\pi}{\lambda}\Delta = \frac{4\pi n h\cos\theta_t}{\lambda}\]

注意半波损失计入反射系数中.

反射电场叠加有 \[\tilde{E}_r = \sum_{n=1}^{\infty}\tilde{E}_{nr} = E_0 \mathrm{e}^{\mathrm{i}\omega t} \left[ \frac{r(1 - \mathrm{e}^{\mathrm{i}\delta})}{1 - r^2 \mathrm{e}^{-\mathrm{i}\delta}} \right]\] 已知入射光强 \(I_i = \frac{E_0^2}{2}\), 反射光强有: \[I_r = \left\langle \vec{E}_r^2 \right\rangle_T = \frac{2r^2(1-\cos\delta)}{(1+r^4-2r^2\cos\delta)}\]

• \(\cos\delta = 1\), 反射光强最小, 为零;
• \(\cos\delta =-1\), 反射光强最大, \(\max I_r = \frac{4}{(1+r^2)^2}\)

• \(\tilde{E}_{1t} = E_0 tt^{\prime}\mathrm{e}^{\mathrm{i}\omega t}\)
• \(\tilde{E}_{2t} = E_0 tt^{\prime}{r^{\prime}}^2\mathrm{e}^{\mathrm{i}(\omega t-\delta)}\)
• \(\tilde{E}_{3t} = E_0 tt^{\prime}{r^{\prime}}^4\mathrm{e}^{\mathrm{i}(\omega t-2\delta)}\)

N 次透射有: \[\tilde{E} = E_0tt^{\prime}{r^{\prime}}^{2(N-1)} \mathrm{e}^{\mathrm{i}[\omega - (N-1)\delta]}\] 透射电场叠加有: \[\tilde{E}_t = E_0\mathrm{e}^{\mathrm{i}\omega t}\left[ \frac{tt^{\prime}}{1-r^2\mathrm{e}^{-\mathrm{i}\delta}} \right]\] 透射光强有: \[I_t = \frac{I_i (tt^{\prime})^2}{(1+r^4)-2r^2\cos\delta}\]

显然 \(I_i = I_r + I_t\). 能量守恒.

引入锐度系数 \(F\) \[F\equiv \left( \frac{2r}{1-r^2} \right)^2\Rightarrow \begin{cases}\dfrac{I_r}{I_i} = \dfrac{F\sin^2(\delta/2)}{1+F\sin^2(\delta/2)}\\ \dfrac{I_t}{I_i} = \dfrac{1}{1+F\sin^2(\delta/2)} \end{cases}.\] 显然 \(\frac{I_r}{I_i}+\frac{I_t}{I_i} = 1\)

已知 \[\delta = \frac{2\pi}{\lambda}\Delta = \frac{4\pi nd\cos\theta}{\lambda}, \;\frac{I_t}{I_i} = \frac{1}{1+ F\sin^2(\delta/2)}\in \left[ \frac{1}{1+F}, 1 \right]\]

当透射光强等于反射光强时有: \[\frac{I_t}{I_i} = \frac{1}{2}\Rightarrow\delta_{1/2} = 2\arcsin \left( \frac{1}{\sqrt{F}} \right)\] 故半峰宽 (类似于品质系数 \(Q\)) \(\gamma\) 为: \[\gamma = 2\delta_{1/2} = 4\arcsin \left( \frac{1}{\sqrt{F}} \right)\] 当 \(F\ll 1\) 时, \(\arcsin F^{-1/2}\Theta F^{-1/2}\), 此时 \(\gamma = \frac{4}{\sqrt{F}}\).

• 锐度 finesse \(\mathcal{F}\) \[\mathcal{F} = \frac{2\pi}{\gamma} = \frac{\pi \sqrt{F}}{2}\]

反射率越高, \(F\) 越大,透射峰越尖锐,相位分辨能力越强.

\[\delta = \frac{4\pi nd\cos\theta_t}{\lambda}\]

• 固定 \(n,d,\lambda\), 则 \(\delta = \delta(\theta_t)\). \[\Rightarrow\Delta\delta = -\frac{4\pi nd\sin\theta_t}{\lambda}\Delta\theta_t\] 由于 \(\gamma = \Delta\delta\), 可得角分辨率 \(|\theta_t|\) 为 \[|\Delta\theta_t| = \frac{\lambda}{\pi nd\sin\theta_t}\frac{1}{\sqrt{F}}\]

• 固定 \(n,d,\theta_t\), \(\delta = \delta(\lambda)\).

  考虑第 \(m\) 个亮线, 入射光为两个单色波 \(\lambda, \lambda+\delta\lambda\) \[\Rightarrow\begin{cases}2nd\cos\theta_t=m\lambda\\ 2nd\cos(\theta_t+\delta\theta_t) = m(\lambda+\delta\lambda) \end{cases}\Rightarrow|\delta\theta_t| = \frac{m}{2nd\sin\theta_t}\delta\lambda\] 欲使两条 \(m\) 级亮线能够分清, 则亮线错开角度 \(\delta\theta_t\) 需大于对应波长的角分辨率. 即: \[|\delta\theta_t|>|\Delta\theta_t|\Rightarrow\delta\lambda>\frac{2\lambda}{\pi m \sqrt{F}} \]

高透射时有:(频率等间距的透射峰) \[\delta = \frac{4\pi nd\cos\theta_t}{\lambda} = 2\pi m \Rightarrow v_m=\frac{c}{\lambda_m} = \frac{mc}{2nd\cos\theta_t}\]

稳态时有: \(E_{\text{circ}} = E_{\text{laun}} + E_{RT} = E_{\text{laun}} + r_1r_2\mathrm{e}^{\mathrm{i}2\phi}E_{\text{circ}}\) 故有: \[\frac{E_{\text{circ}}}{E_{\text{laun}}} = \frac{1}{1-r_1r_2\mathrm{e}^{\mathrm{i}2\phi}}\] \[\Rightarrow \frac{I_{\text{circ}}}{I_{\text{laun}} / (1-R_1)} = (1-R_1)\left( \frac{E_{\text{circ}}}{E_{\text{laun}}} \right)^2\] \[\Rightarrow \frac{I_{\text{circ}}}{I_{\text{inc}}} = \frac{1-R_1}{(1-\sqrt{R_1R_2})^2 + 4 \sqrt{R_1R_2}\sin^2\phi}\]