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KTA属正双轴晶体,BGSe属于双轴单斜晶体。它们的具体指标如表1所示。
表 1 BGSe与KTA性能对比
Table 1. Comparison between BGSe and KTA
Name BGSe KTA Characteristic Biaxial, monoclinic, point group m[11] Uniaxial, point group mm2[13] Transmittance range 0.47-18µm[14] 0.35-5.3 µm [10] Damage threshold 557 MW/cm2[14](5 ns,1.064 µm,1 Hz) >600 MW/cm2 Nonzero tensor d11=24.3 pm/V, d13=20.4 pm/V [15] d33=16.2 pm/V, d31=2.8 pm/V, d32=4.2 pm/V, d31=2.8 pm/V
d24=3.2 pm/V, d15=2.3 pm/V[13],deff=4.47 pm/V[10] -
选取参考文献[16]所给出的BGSe晶体Sellmeier方程进行计算:
$$ \left\{\begin{aligned} & {{{n}}}_{{{x}}}^{2}=6.724\;31+\frac{0.263\;75}{{{{\lambda}}}^{2}-0.042\;48}+\frac{602.97}{{{{\lambda}}}^{2}-749.87}\\ & {{{n}}}_{{{y}}}^{2}=6.866\;03+\frac{0.268\;16}{{{{\lambda}}}^{2}-0.042\;59}+\frac{682.97}{{{{\lambda}}}^{2}-781.78}\\ & {{{n}}}_{{{z}}}^{2}=7.167\;09+\frac{0.326\;81}{{{{\lambda}}}^{2}-0.069\;73}+\frac{731.86}{{{{\lambda}}}^{2}-790.16}\\ & 0.901\leqslant \lambda \leqslant 10.591\end{aligned}\right. $$ (1) KTA的Sellmeier方程如公式(2)所示[17]:
$$ {{{n}}}_{i}^{2}={{{A}}}_{i}+\frac{{{{B}}}_{i}{{{\lambda}}}^{{p}_{i}}}{{{{\lambda}}}^{{p}_{i}}-{{{C}}}_{i}}+\frac{{{{D}}}_{i}{{{\lambda}}}^{{q}_{i}}}{{{{\lambda}}}^{{q}_{i}}-{{{E}}}_{i}} $$ (2) 公式(2)中系数的具体数值如表2所示。
表 2 KTA晶体的Sellmeier系数
Table 2. Sellmeier coefficients of KTA
Ax 2.149 5 Ay 2.1308 Az 2.193 1 Bx 1.020 3 By 1.056 4 Bz 1.238 2 Cx 0.042 378 Cy 0.042 523 Cz 0.059 171 Dx 0.553 1 Dy 0.692 7 Dz 0.508 8 Ex 72.304 5 Ey 54.850 5 Ez 53.289 8 px 1.995 1 py 2.001 7 pz 1.892 0 qx 1.956 7 qy 1.726 1 qz 2.000 0 将公式(1)、(2)代入公式(3)[18],即可求出不同入射方向
$({{\theta}},\phi)$ 的${{{n}}}_{{{\rm{e}}}_{1}}({{\theta}},\phi)$ 和${{{n}}}_{{{\rm{e}}}_{2}}\left({{\theta}},\phi\right)$ :$$ \left\{\begin{aligned} &{{{n}}}_{{{{e}}}_{1}}(\theta ,\phi )=\sqrt{2}{[{{d}}+{{A}}-{({{{b}}}^{2}-2{{b}}{{B}}+{{{A}}}^{2})}^{1/2}]}^{-\frac{1}{2}}\\ &{{{n}}}_{{{{e}}}_{2}}\left({{\theta}},\phi\right)=\sqrt{2}{[{{d}}+{{A}}+{({{{b}}}^{2}-2{{b}}{{B}}+{{{A}}}^{2})}^{1/2}]}^{-\frac{1}{2}}\\ & A={{{k}}}_{{{z}}}^{2} \cdot c-{{{k}}}_{{{x}}}^{2} \cdot a,B={{{k}}}_{{{z}}}^{2} \cdot c+{{{k}}}_{{{x}}}^{2} \cdot a\\ &a=\frac{1}{{{{n}}}_{{{x}}}^{2}}-\frac{1}{{{{n}}}_{{{y}}}^{2}},b=\frac{1}{{{{n}}}_{{{x}}}^{2}}-\frac{1}{{{{n}}}_{{{z}}}^{2}},c=\frac{1}{{{{n}}}_{{{y}}}^{2}}-\frac{1}{{{{n}}}_{{{z}}}^{2}},d=\frac{1}{{{{n}}}_{{{x}}}^{2}}+\frac{1}{{{{n}}}_{{{z}}}^{2}}\\ &{{{k}}}_{{{x}}}={{\sin\theta \cos}}\theta ,{{{k}}}_{{{y}}}={{\sin\theta \sin}}\theta ,{{{k}}}_{{{z}}}={{\cos\theta}} \end{aligned}\right. $$ (3) 再代入公式(4),即可得出各相位匹配条件下的相位匹配曲线。
$$ \left\{\!\!\!\begin{array}{c}{\rm{type-I}}\\ \dfrac{{{{n}}}_{{{{e}}}_{2}}^{{{{\omega }}}_{3}}}{{{{\lambda}}}_{3}}=\dfrac{{{{n}}}_{{{{e}}}_{1}}^{{{{\omega }}}_{1}}}{{{{\lambda}}}_{1}}+\dfrac{{{{n}}}_{{{{e}}}_{1}}^{{{{\omega }}}_{2}}}{{{{\lambda}}}_{2}}\\ \dfrac{1}{{{{\lambda}}}_{3}}=\dfrac{1}{{{{\lambda}}}_{1}}+\dfrac{1}{{{{\lambda}}}_{2}}\end{array}\right.\left\{\!\!\!\begin{array}{c}{{{\rm type-II}-A}}\\ \dfrac{1}{{{{\lambda}}}_{3}}=\dfrac{1}{{{{\lambda}}}_{1}}+\dfrac{1}{{{{\lambda}}}_{2}}\\ \dfrac{{{{n}}}_{{{{e}}}_{2}}^{{{{\omega }}}_{3}}}{{{{\lambda}}}_{3}}=\dfrac{{{{n}}}_{{{{e}}}_{2}}^{{{{\omega }}}_{1}}}{{{{\lambda}}}_{1}}+\dfrac{{{{n}}}_{{{{e}}}_{1}}^{{{{\omega }}}_{2}}}{{{{\lambda}}}_{2}}\end{array}\right.\left\{\!\!\!\begin{array}{c}{{{\rm type-II}-B}}\\ \dfrac{1}{{{{\lambda}}}_{3}}=\dfrac{1}{{{{\lambda}}}_{1}}+\dfrac{1}{{{{\lambda}}}_{2}}\\ \dfrac{{{{n}}}_{{{{e}}}_{2}}^{{{{\omega }}}_{3}}}{{{{\lambda}}}_{3}}=\dfrac{{{{n}}}_{{{{e}}}_{1}}^{{{{\omega }}}_{1}}}{{{{\lambda}}}_{1}}+\dfrac{{{{n}}}_{{{{e}}}_{2}}^{{{{\omega }}}_{2}}}{{{{\lambda}}}_{2}}\end{array}\right. $$ (4) 由于双轴晶体的相位匹配曲线与
$({{\theta}},\phi)$ 均有关,为简便起见,仅讨论$ \theta=0^{ \circ } $ 的情况,BGSe和KTA的相位匹配曲线$ (\theta=0^{ \circ }) $ 如图1所示。如图1所示,BGSe(
${\rm{\theta}}=56.3^{ \circ },\;\phi=0^{ \circ }$ )可在I类相位匹配条件下产生~3.5 μm的闲频光,KTA($ {\rm{\theta}}=90^{ \circ } $ ,$\phi=0^{ \circ }$ )可在II-A类相位匹配条件下产生~3.5μm的闲频光。图 1 BGSe与KTA的相位匹配曲线(
$\theta=0^{ \circ } $ )Figure 1. Phase matching curve of BGSe and KTA(
$\theta=0^{ \circ } $ )为了便于实验对比,选取切割角度为(56.3°,0°)的BGSe和切割角度为(90°,0°)的KTA晶体在同一光路中进行对比实验。为了计算两种晶体的OPO振荡阈值,要先计算出BGSe(56.3°,0°)和KTA(90°,0°)的有效非线性系数。
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BGSe晶体
$({\rm{\theta}}=56.3^{ \circ },\;\phi =0^{ \circ })$ 在I类相位匹配条件下的有效非线性系数为:$$\begin{split} {{{d}}}_{\rm{eff}}=\; &\left[{{{b}}}_{1}\left({{{\omega }}}_{3}\right),{{{b}}}_{2}\left({{{\omega }}}_{3}\right),{{{b}}}_{3}\left({{{\omega }}}_{3}\right)\right]\times \\ & \left(\begin{array}{c}{{{d}}}_{11}\\ {{{d}}}_{21}\\ {{{d}}}_{31}\end{array}\begin{array}{c}{{{d}}}_{12}\\ {{{d}}}_{22}\\ {{{d}}}_{32}\end{array}\begin{array}{c}{{{d}}}_{13}\\ {{{d}}}_{23}\\ {{{d}}}_{33}\end{array}\begin{array}{c}{{{d}}}_{14}\\ {{{d}}}_{24}\\ {{{d}}}_{34}\end{array}\begin{array}{c}{{{d}}}_{15}\\ {{{d}}}_{25}\\ {{{d}}}_{35}\end{array}\begin{array}{c}{{{d}}}_{16}\\ {{{d}}}_{26}\\ {{{d}}}_{36}\end{array}\right)\times \\ & \left(\begin{array}{c}{{{a}}}_{1}\left({{{\omega }}}_{1}\right){{{a}}}_{1}\left({{{\omega }}}_{2}\right)\\ {{{a}}}_{2}{\left({{{\omega }}}_{1}\right){{a}}}_{2}\left({{{\omega }}}_{2}\right)\\ {{{a}}}_{3}{\left({{{\omega }}}_{1}\right){{a}}}_{3}\left({{{\omega }}}_{2}\right)\\ {{{a}}}_{2}\left({{{\omega }}}_{1}\right){{{a}}}_{3}\left({{{\omega }}}_{2}\right)+{{{a}}}_{3}\left({{{\omega }}}_{1}\right){{{a}}}_{2}\left({{{\omega }}}_{2}\right)\\ {{{a}}}_{1}{\left({{{\omega }}}_{1}\right){{a}}}_{3}\left({{{\omega }}}_{2}\right)+{{{a}}}_{3}\left({{{\omega }}}_{1}\right){{{a}}}_{1}\left({{{\omega }}}_{2}\right)\\ {{{a}}}_{1}\left({{{\omega }}}_{1}\right){{{a}}}_{2}\left({{{\omega }}}_{2}\right)+{{{a}}}_{2}\left({{{\omega }}}_{1}\right){{{a}}}_{1}\left({{{\omega }}}_{2}\right)\end{array}\right) \end{split} $$ (5) 其中,
$ {{{a}}}_{{{i}}}\left({{{\omega }}}_{{{j}}}\right) $ 和$ {{{b}}}_{{{i}}}\left({{{\omega }}}_{{{j}}}\right) $ 由公式(1)、(6)、(7)得出。$$\begin{split} & \left(\begin{array}{c}{b}_{1}\\ {b}_{2}\\ {b}_{3}\end{array}\right)=\left(\begin{array}{c}{b\_d}_{1}/{{{n}}}_{{{x}}}^{2}\\ {b\_d}_{2}/{{{n}}}_{{{y}}}^{2}\\ {b\_d}_{3}/{{{n}}}_{{{z}}}^{2}\end{array}\right)\dfrac{1}{\sqrt{{\left(\dfrac{{bd}_{1}}{{{{n}}}_{{{x}}}^{2}}\right)}^{2}+{\left(\dfrac{{bd}_{2}}{{{{n}}}_{{{y}}}^{2}}\right)}^{2}+{\left(\dfrac{{bd}_{3}}{{{{n}}}_{{{z}}}^{2}}\right)}^{2}}} , \\ & \left( {\begin{array}{*{20}{c}} {b\_{d_1}}\\ {b\_{d_2}}\\ {b\_{d_3}} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} { - {{\cos\theta \cos}}\theta {{\sin\delta }} - {{\sin}}\theta {{\cos\delta }}}\\ { - {{\cos\theta \sin}}\theta {{\sin\delta }} + {{\cos}}\theta {{\cos\delta }}}\\ {{{\sin\theta \sin\delta }}} \end{array}} \right) \end{split} $$ (6) $$\begin{split} & \left(\begin{array}{c}{{{a}}}_{1}\\ {{{a}}}_{2}\\ {{{a}}}_{3}\end{array}\right)=\left(\begin{array}{c}{{{a}}\_{{d}}}_{1}/{{{n}}}_{{{x}}}^{2}\\ {{{a}}\_{{d}}}_{2}/{{{n}}}_{{{y}}}^{2}\\ {{{a}}\_{{d}}}_{3}/{{{n}}}_{{{z}}}^{2}\end{array}\right)\dfrac{1}{\sqrt{{\left(\dfrac{{{{a}}{{d}}}_{1}}{{{{n}}}_{{{x}}}^{2}}\right)}^{2}+{\left(\dfrac{{{{a}}{{d}}}_{2}}{{{{n}}}_{{{y}}}^{2}}\right)}^{2}+{\left(\dfrac{{{{a}}{{d}}}_{3}}{{{{n}}}_{{{z}}}^{2}}\right)}^{2}}} , \\ & \left( {\begin{array}{*{20}{c}} {{{a}}\_{{{d}}_1}}\\ {{{a}}\_{{{d}}_2}}\\ {{{a}}\_{{{d}}_3}} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {{\rm{cos\theta cos}}\theta {\rm{cos\delta }} - {\rm{sin}}\theta {\rm{sin\delta }}}\\ {{\rm{cos\theta sin}}\theta {\rm{cos\delta }} + {\rm{cos}}\theta {\rm{sin\delta }}}\\ {{\rm{ - sin\theta cos\delta }}} \end{array}} \right) \end{split} $$ (7) 目前,BGSe晶体的
$ {{{d}}}_{{\rm{i}}{\rm{j}}} $ 矩阵尚没有定论,但参考文献[11, 15, 19]通过不同的方法给出了BGSe晶体的$ {d}_{\rm ij} $ 值。参考文献[20]对比分析了现有文献中BGSe晶体的$ {d}_{\rm ij} $ 值,并给出了公式(8),笔者以此来估算BGSe晶体的有效非线性系数。$$ {{{d}}}_{{\rm{i}}{\rm{j}}}=\left(\begin{array}{c}0\\ 5.2\\ 1.2\end{array}\begin{array}{c}0\\ 24.3\\ -3.7\end{array}\begin{array}{c}0\\ -20.4\\ -2.2\end{array}\begin{array}{c}0\\ -3.7\\ -20.4\end{array}\begin{array}{c}1.2\\ 0\\ 0\end{array}\begin{array}{c}5.2\\ 0\\ 0\end{array}\right) $$ (8) 将公式(1)、(6)、(7)、(8)
$({\rm{\theta}}=56.3^{ \circ },\;\phi=0^{ \circ })$ 代入公式(5),求得$ {d}_{\rm eff}=-11.908\;1\;{\rm{p}}{\rm{m}}/{\rm{V}} $ (输出3.64 μm)。KTA晶体
$({{\theta}}=90^{ \circ },\;\phi=0^{ \circ })$ 在II-A类相位匹配条件下,$$\begin{split} {{{d}}}_{\rm eff}=\; &\left[{{{b}}}_{1}\left({{{\omega }}}_{3}\right),{{{b}}}_{2}\left({{{\omega }}}_{3}\right),{{{b}}}_{3}\left({{{\omega }}}_{3}\right)\right]\times \\ & \left(\begin{array}{c}{{{d}}}_{11}\\ {{{d}}}_{21}\\ {{{d}}}_{31}\end{array}\begin{array}{c}{{{d}}}_{12}\\ {{{d}}}_{22}\\ {{{d}}}_{32}\end{array}\begin{array}{c}{{{d}}}_{13}\\ {{{d}}}_{23}\\ {{{d}}}_{33}\end{array}\begin{array}{c}{{{d}}}_{14}\\ {{{d}}}_{24}\\ {{{d}}}_{34}\end{array}\begin{array}{c}{{{d}}}_{15}\\ {{{d}}}_{25}\\ {{{d}}}_{35}\end{array}\begin{array}{c}{{{d}}}_{16}\\ {{{d}}}_{26}\\ {{{d}}}_{36}\end{array}\right)\times \\ & \left(\begin{array}{c}{{{b}}}_{1}\left({{{\omega }}}_{1}\right){{{a}}}_{1}\left({{{\omega }}}_{2}\right)\\ {{{b}}}_{2}\left({{{\omega }}}_{1}\right){{{a}}}_{2}\left({{{\omega }}}_{2}\right)\\ {{{b}}}_{3}\left({{{\omega }}}_{1}\right){{{a}}}_{3}\left({{{\omega }}}_{2}\right)\\ {{{b}}}_{2}\left({{{\omega }}}_{1}\right){{{a}}}_{3}\left({{{\omega }}}_{2}\right)+{{{b}}}_{3}\left({{{\omega }}}_{1}\right){{{a}}}_{2}\left({{{\omega }}}_{2}\right)\\ {{{b}}}_{3}\left({{{\omega }}}_{1}\right){{{a}}}_{1}\left({{{\omega }}}_{2}\right)+{{{b}}}_{1}\left({{{\omega }}}_{1}\right){{{a}}}_{3}\left({{{\omega }}}_{2}\right)\\ {{{b}}}_{1}\left({{{\omega }}}_{1}\right){{{a}}}_{2}\left({{{\omega }}}_{2}\right)+{{{b}}}_{2}\left({{{\omega }}}_{1}\right){{{a}}}_{1}\left({{{\omega }}}_{2}\right)\end{array}\right) \end{split} $$ (9) 其中
$$ {{{d}}_{{{ij}}}} = \left( {\begin{array}{*{20}{c}} 0&0&0\\ 0&0&0\\ {2.8}&{4.2}&{16.2} \end{array}{\rm{}}\begin{array}{*{20}{c}} 0&{2.3}&0\\ {3.2}&0&0\\ 0&0&0 \end{array}} \right) $$ (10) 将公式(1)、(6)、(7)、(10)
$({\rm{\theta}}=90^{ \circ },\;\phi=0^{ \circ })$ 代入公式(9),求得$ {{{d}}}_{{\rm{e}}{\rm{f}}{\rm{f}}}=-3.2\;{\rm{p}}{\rm{m}}/{\rm{V}} $ (输出3.43 μm)。 -
OPO的振荡阈值功率为[21]:
$${{{J}}_{{\rm{th}}}} = \frac{{2.25{{\tau }}}}{{{{\kappa }}{{{g}}_{{s}}}{{L}}_{{\rm{eff}}}^2}}{\left[ {\frac{{{L}}}{{2{{c\tau }}}}{\rm{ln}}33 + 2{{\alpha l}} + \ln \frac{1}{{\sqrt {{R}} }} + {\rm{ln}}2} \right]^2}$$ (11) 式中:
$ {\rm{\kappa }} $ 为增益系数,可表示为:$ {\rm{\kappa }}=\dfrac{2{{{\omega }}}_{{\rm{s}}}{{\rm{\omega }}}_{{\rm{i}}}{{{d}}}_{{\rm{e}}{\rm{f}}{\rm{f}}}^{2}}{{{{n}}}_{{\rm{s}}}{{{n}}}_{{\rm{i}}}{{{n}}}_{{\rm{p}}}{{{\varepsilon}}}_{0}{{{c}}}^{3}} $ ;$ {{{g}}}_{{\rm{s}}} $ 为模式耦合系数,可表示为:$ {{{g}}}_{{\rm{s}}}=\dfrac{{{{w}}}_{{\rm{p}}}^{2}}{{{{w}}}_{{\rm{p}}}^{2}+{{{w}}}_{{\rm{s}}}^{2}} $ ;$ {{{L}}}_{{\rm{e}}{\rm{f}}{\rm{f}}} $ 为参量有效增益长度,约等于晶体长度l;$ {{{w}}}_{{\rm{p}}} $ 为泵浦光腰斑半径;$ {{{w}}}_{{\rm{s}}} $ 为信号光腰斑半径,它们满足${\left(\dfrac{{\text{π}}}{2{{L}}{\rm{\lambda}}}\right)}^{2}{{{w}}}_{{\rm{s}}}^{6}+ {{{w}}}_{{\rm{s}}}^{2}-$ $\dfrac{{{{w}}}_{{\rm{p}}}^{2}}{2}=0 $ 。将其代入公式(11),可得:$$ \begin{split} {{{J}}_{{\rm{th}}}} =\; & \frac{{2.25{{{n}}_{{s}}}{{{n}}_{{i}}}{{{n}}_{{p}}}{{{\varepsilon }}_0}{{{c}}^3}\left( {{{w}}_{{p}}^2 + {{w}}_{{s}}^2} \right)}}{{2{{{\omega }}_{{s}}}{{{\omega }}_{{i}}}{{d}}_{{\rm{eff}}}^2{{w}}_{{p}}^2{{L}}_{{\rm{eff}}}^2}}{{\tau [}}\frac{{{L}}}{{2{{c\tau }}}}{\rm{ln}}33+\\ & 2{{\alpha l}} + \ln \frac{1}{{\sqrt {{R}} }} + {\rm{ln}}2]^2=\\ & \frac{{2.25{{{\lambda }}_{{s}}}{{{\lambda }}_{{i}}}{{{n}}_{{s}}}{{{n}}_{{i}}}{{{n}}_{{p}}}{{{\varepsilon }}_0}{{c}}}}{{2{{\left( {2{{{\text{π}}}}} \right)}^2}{{d}}_{{\rm{eff}}}^2{{L}}_{{\rm{eff}}}^2}} \cdot \frac{{\left( {{{w}}_{{p}}^2 + {{w}}_{{s}}^2} \right)}}{{{{w}}_{{p}}^2}} \cdot\\ & {{\tau }}[\frac{{{L}}}{{2{{c\tau }}}}{\rm{ln}}33 + 2{{\alpha l}} + \frac{1}{{\sqrt {{R}} }} + {\rm{ln}}2{]^2} \end{split} $$ (12) 各参数值如表3所示。
表 3 KTA-OPO与BGSe-OPO的具体参数
Table 3. Parameters of KTA-OPO and BGSe-OPO
Parameter Meaning KTA BGSe $ {{\tau }} $/ns Pulse width 10 10 $ {{{\lambda}}}_{{\rm{s}}} $/μm Wavelength of signal light 1.54 1.50 $ {{{\lambda}}}_{{\rm{i}}} $/μm Wavelength of idler light 3.43 3.64 $ {{{n}}}_{{\rm{s}}} $ Refractive index of signal light 1.806 2 2.485 4 $ {{{n}}}_{{\rm{i}}} $ Refractive index of idler light 1.773 0 2.461 1 $ {{{n}}}_{{\rm{p}}} $ Refractive index of pump light 1.816 7 2.512 8 $ {{{\varepsilon}}}_{0} $/F·m-1 Permittivity of vacuum $ 8.85\times {10}^{-12} $ $ 8.85\times {10}^{-12} $ c/m·s-1 Velocity of light $ 3\times {10}^{8} $ $ 3\times {10}^{8} $ $ {d}_{\rm{eff}} $/m·V-1 Effective nonlinear coefficient $ {\approx 3.2\times 10}^{-12} $ $ {\approx 11.9\times 10}^{-12} $ $ {L}_{\rm{eff}} $/mm Parametric effective gain length 20 15 $ {{L}} $/mm Optical length of OPO cavity 46 52.5 $ {{l}} $/mm Length of crystal 20 15 将表2中的数值代入公式(12),则有:
$$\begin{split} &{{{J}}_{{\rm{th}}}}{\rm{BGSe}} = 1.992\;8 \times {10^{11}} \cdot \frac{{\left( {{{w}}_{\rm{p}}^2 + {{w}}_{\rm{s}}^2} \right)}}{{{{w}}_{\rm{p}}^2}} \cdot \\ &{\left[ {\frac{{\rm{L}}}{{2{{c\tau }}}}{\rm{ln}}33 + 2{{\alpha l}} + \ln \frac{1}{{\sqrt {{R}} }} + {\rm{ln}}2} \right]^2}\left( {{\rm{W}}/{{\rm{m}}^2}} \right) \end{split} $$ (13) $$\begin{split} {{{J}}_{{\rm{th}}}}{\rm{KTA}} =\;& 5.676\;4{\rm{}} \times {10^{11}} \cdot \frac{{\left( {{{w}}_{\rm{p}}^2 + {{w}}_{\rm{s}}^2} \right)}}{{{{w}}_{\rm{p}}^2}} \cdot \left[\frac{{\rm{L}}}{{2{\rm{c\tau }}}}{\rm{ln}}33 +\right. \\ &\left. 2{{\alpha l}} + \ln \frac{1}{{\sqrt {{R}} }} + {\rm{ln}}2\right]^2\left( {{\rm{W}}/{{\rm{m}}^2}} \right) \end{split} $$ (14) 由公式(13)、(14)可得:
$$ \frac{{{{{I}}_{{\rm{th}}}}\left( {{\rm{BGSe}}} \right)}}{{{{{I}}_{{\rm{th}}}}\left( {{\rm{KTA}}} \right)}} = 0.351\;1 $$ (15) 因此,当光斑半径、吸收系数、反射率等相同的情况下,BGSe的OPO阈值为KTA阈值的35.11%。
Comparison of mid-infrared laser generated by optical parametric oscillation of BaGa4Se7 and KTiAsO4
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摘要: BaGa4Se7(硒镓钡,简称BGSe)与KTiOAsO4(砷酸氧钛钾,简称KTA)均可在1.06 μm激光泵浦下产生中红外激光。首先仿真计算出两种非线性晶体的相位匹配曲线,结果显示:切割角为(56.3°, 0°)的BGSe晶体在I类相位匹配条件下和切割角为(90°, 0°)的KTA在II-A类相位匹配条件下均可产生~3.5 μm的闲频光。然后理论计算出BGSe (56.3°, 0°, I类)的有效非线性系数为−11.9 pm/V,KTA(90°, 0°,II-A类)的有效非线性系数为−3.2 pm/V;在其他条件相同的情况下,15 mm长BGSe (56.3°, 0°, I类) 的OPO振荡阈值是20 mm长KTA (90°, 0°, II-A类) OPO振荡阈值的35.11%。最后通过实验验证BGSe (56.3°, 0°, I类, 15 mm) 的振荡阈值小于KTA(90°, 0°, II-A类, 20 mm),输出的中红外激光能量大于KTA。因此,BGSe是一种极具应用前景的中红外非线性晶体。Abstract: Both BaGa4Se7 (BGSe) and KTiAsO4 (KTA) can generate mid-infrared lasers pumped by 1.06 μm laser. Firstly, the phase matching curves of two kinds of non-linear crystals were simulated and calculated. The results show that BGSe with cutting angles of (56.3°, 0°) and KTA with cutting angle of (90°, 0°) can generate idle frequency light of ~3.5 micron under phase matching conditions of type I and type II-A, respectively. Then through theoretical calculation, the effective non-linear coefficients of BGSe (56.3°, 0°, type-I) is −11.9 pm/V, and that of KTA (90°, 0°, type II-A) is −3.2 pm/V. The OPO oscillation threshold of 15 mm long BGSe (56.3°, 0°, type-I) is 35.11% of that of 20 mm long KTA (90°, 0°, type II-A). Then, the experimental results show that the oscillation threshold of BGSe-OPO (56.3°, 0°, type-I, 15 mm) is smaller than that of KTA-OPO (90°, 0°, type-II-A, 20 mm). The output energy of BGSe (56.3°, 0°, type-I) is larger than that of KTA (90°, 0°, type II-A). Therefore, BGSe is a promising mid-infrared non-linear crystal.
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Key words:
- BaGa4Se7 /
- KTiOAsO4 /
- optical parametric oscillation /
- mid-infrared laser
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表 1 BGSe与KTA性能对比
Table 1. Comparison between BGSe and KTA
Name BGSe KTA Characteristic Biaxial, monoclinic, point group m[11] Uniaxial, point group mm2[13] Transmittance range 0.47-18µm[14] 0.35-5.3 µm [10] Damage threshold 557 MW/cm2[14](5 ns,1.064 µm,1 Hz) >600 MW/cm2 Nonzero tensor d11=24.3 pm/V, d13=20.4 pm/V [15] d33=16.2 pm/V, d31=2.8 pm/V, d32=4.2 pm/V, d31=2.8 pm/V
d24=3.2 pm/V, d15=2.3 pm/V[13],deff=4.47 pm/V[10]表 2 KTA晶体的Sellmeier系数
Table 2. Sellmeier coefficients of KTA
Ax 2.149 5 Ay 2.1308 Az 2.193 1 Bx 1.020 3 By 1.056 4 Bz 1.238 2 Cx 0.042 378 Cy 0.042 523 Cz 0.059 171 Dx 0.553 1 Dy 0.692 7 Dz 0.508 8 Ex 72.304 5 Ey 54.850 5 Ez 53.289 8 px 1.995 1 py 2.001 7 pz 1.892 0 qx 1.956 7 qy 1.726 1 qz 2.000 0 表 3 KTA-OPO与BGSe-OPO的具体参数
Table 3. Parameters of KTA-OPO and BGSe-OPO
Parameter Meaning KTA BGSe $ {{\tau }} $ /nsPulse width 10 10 $ {{{\lambda}}}_{{\rm{s}}} $ /μmWavelength of signal light 1.54 1.50 $ {{{\lambda}}}_{{\rm{i}}} $ /μmWavelength of idler light 3.43 3.64 $ {{{n}}}_{{\rm{s}}} $ Refractive index of signal light 1.806 2 2.485 4 $ {{{n}}}_{{\rm{i}}} $ Refractive index of idler light 1.773 0 2.461 1 $ {{{n}}}_{{\rm{p}}} $ Refractive index of pump light 1.816 7 2.512 8 $ {{{\varepsilon}}}_{0} $ /F·m-1Permittivity of vacuum $ 8.85\times {10}^{-12} $ $ 8.85\times {10}^{-12} $ c/m·s-1 Velocity of light $ 3\times {10}^{8} $ $ 3\times {10}^{8} $ $ {d}_{\rm{eff}} $ /m·V-1Effective nonlinear coefficient $ {\approx 3.2\times 10}^{-12} $ $ {\approx 11.9\times 10}^{-12} $ $ {L}_{\rm{eff}} $ /mmParametric effective gain length 20 15 $ {{L}} $ /mmOptical length of OPO cavity 46 52.5 $ {{l}} $ /mmLength of crystal 20 15 -
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