TD 3:模式频率,模式频率间隔,有折射率导致的能量衰减,高斯光束
TD 3:模式频率,模式频率间隔,有折射率导致的能量衰减,高斯光束
- $\nu_m = m(\frac{c}{2nL})$
- $\Delta \nu_c = \frac{c}{2nL} = 8.54\times10^8s^{-1}$
- $m = \frac 1T\frac 1{\Delta\nu_c} = \frac {c}{\lambda}\frac 1{\Delta\nu_c} = 329929.5\approx 329930$
- $\nu_m = 329930\Delta\nu_c =2.818\times10^{14}Hz$
- $Q = mF = m\frac{\pi\sqrt{R_e}}{1-R_e}$
- $R_e = \sqrt{R_1R_2}e^{-\alpha L}$
接下来从复折射率转换到吸收系数$\alpha$:
- 复折射率:$\widetilde w = n-i\kappa$
- 考虑电场:$\vec E = E_0e^{i(\omega t-kz)} = E_0e^{i(\omega t-\frac \omega cn z)}e^{-\frac \omega c \kappa z}\vec e_z$
- 能量强度:$I = |E_0|^2e^{-\frac {2\omega \kappa z}{c}}$
- 得到:$\alpha = \frac{2\omega\kappa}{c} = \frac{4\pi\kappa}{\lambda} = 0.0118$
继续计算:
- $R_e = 0.995\cdot e^{-\frac{4\pi\kappa L}{\lambda}} = 0.9938,\ \ Q = 1.67\times 10^{8}$
- $\tau_c = \frac{Q}{\omega}= \frac{Q\lambda}{c2\pi} = 9.444\times10^{-8}s$
- 高斯光束:$I = I_0e^{-a(\nu-\nu_0)^2}$
- 计算$a$与$\Delta\nu$的关系:在$\nu_0\pm\frac{\Delta\nu}2$的范围内,光强等于$I_0\frac 1{e^2}$:
- 已知$\Delta \lambda$,考虑$\Delta \nu$与$\Delta \lambda$的关系:
- 注意这里$Å$代表埃米,是$10^{-10}m$
- 计算模式间距$\Delta\nu_c$:$\nu = m\times\Delta\nu_c$
- 注意区分模式间距和光束宽度
- 光强表达式转换为:
- 得到可选模数:$329886,329887,329888$
- 反推频率
- 损耗导致的峰变宽:$\Delta\nu_m = \frac {\Delta \nu_c}F$
- 最后考虑到激光的偏振状态,最终模数为$2\times3 = 6$
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