![\"\begin{equation*}](\"https:\/\/pages.vassar.edu\/magnes\/wp-content\/ql-cache\/quicklatex.com-04c53280f2d54670343baae59001945f_l3.png\" "\\"Rendered")
<\/p>\n2.2.20 Nonlinear optics Robert W.Boyd<\/p>\n
This can be simplified to $I_{3} = I_{3}(max)\\frac{\\sin^{2}(\\frac{\\bigtriangleup kL}{2})}{(\\frac{\\bigtriangleup kL}{2})^{2}}$<\/p>\n
Fig 2.2.2 Nonlinear optics Robert W.Boyd<\/p>\n Link to mathematica code\u00a0https:\/\/vspace.vassar.edu\/tasanda\/sinc.nb<\/a><\/p>\n It shows the harmonic generation output as a function of the phase match\u00a0$\\bigtriangleup k$ when\u00a0$\\bigtriangleup k \\sim 0$<\/p>\n Link to mathematica code\u00a0https:\/\/vspace.vassar.edu\/tasanda\/Manipulation.nb<\/a><\/p>\n This model shows the effect of superposition of waves and explains why intensity is largest when $\\bigtriangleup k=0$<\/p>\n Deriving Efficiency<\/strong><\/p>\n Solving the general wave equations for the two frequencies $\\omega_{1}$(incident) and $\\omega_{2}$(Second harmonic) we obtain coupled-amplitude equations.<\/p>\n (2) <\/span> <\/span> (3) <\/span> <\/span> 2.6.10, 2.6.11 Nonlinear optics Robert W.Boyd<\/p>\n Where $\\bigtriangleup k = 2K_{1}-k_{2}$ is the wave vector mismatch. Integrating these equations gives us<\/p>\n $A_{1}=(\\frac{2\\pi I}{n_{1}c}) u_{1}e^{i\\phi_{1}}, \u00a0 \u00a0 \u00a0 A_{2}=(\\frac{2\\pi I}{n_{2}c})u_{2}e^{i\\phi_{2}$<\/p>\n 2.6.13, 2.6.14 Nonlinear optics Robert W.Boyd<\/p>\n The new field amplitudes $u_{1}$ and \u00a0$u_{2}$ are defined such that\u00a0$u_{1}(z)^{2} +\u00a0u_{2}(z)^{2} = 1$(conserved normalized quantity). Next we introduce a normalized distance parameter<\/p>\n $\\zeta=\\frac{z}{l}$ \u00a0 \u00a0 where \u00a0 \u00a0 \u00a0$l=(\\frac{n^{2}_{1}n_{2}c^{3}}{2\\pi I})^{\\frac{1}{2}}\\frac{1}{8\\pi \\omega_{1} d}$<\/p>\n 2.6.18, 2.6.19 Nonlinear optics Robert W.Boyd<\/p>\n $l$ is the distance over which the fields exchange energy.<\/p>\n $\\frac{du_{1}}{d\\zeta}=u_{1}u_{2}sin\\theta$ and $\\frac{du_{2}}{d\\zeta}=-u^{2}_{1}sin\\theta$.<\/p>\n 2.6.22, 2.6.23 Nonlinear optics Robert W.Boyd<\/p>\n If we assume $cos\\theta=0$ and \u00a0$sin\\theta=-1$ the\u00a0equations simplify to \u00a0$\\frac{du_{1}}{d\\zeta}=-u_{1}u_{2}$ \u00a0 and \u00a0 $\\frac{du_{2}}{d\\zeta}=u^{2}_{1}$<\/p>\n 2.6.31, 2.6.32 Nonlinear optics Robert W.Boyd<\/p>\n The second equation can be written as $\\frac{du_{2}}{d\\zeta}=1-u^{2}_{2}$(from conservation).<\/p>\n Hence $u_{2}=tanh(\\zeta+ \\zeta_{0})$. Initially we assume there is only the incident beam and no harmonic generation hence $u_{1}(0)=1, u_{2}(0)=0$. Therefore $u_{2}= tanh\\zeta$ and \u00a0$u_{1}=sech\\zeta$. Plotting these two equations below shows that incident waves are converted into the second harmonic.<\/p>\n![\"\"](\"http:\/\/pages.vassar.edu\/magnes\/files\/2012\/04\/Diagram-22.jpeg\")
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