## Abstract

The effect of light absorption by sample in the analysis of Maker
fringe data for estimating a second-order nonlinear coefficient has
been studied experimentally. Two theories, one by Jerphagnon and
Kurtz that neglects the absorption effect and one by Herman and Hayden
that takes into account the absorption effect, were compared with the
experimental results. It was found that Jerphagnon and Kurtz’s
formula was unable to predict correctly not only the magnitude but also
the incident angle dependence or the sample thickness dependence of the
second harmonic signal generated by the sample with strong absorption,
whereas the theory by Herman and Hayden was able to make those
predictions fairly well. It was also found that the error in the
estimated nonlinear coefficient when one uses Jerphagnon and Kurtz’s
formula could be as large as 2–4 times the true value, depending on
sample thickness.

© 1998 Optical Society of America

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### Equations (4)

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(1)
$$\begin{array}{l}P_{2\mathrm{\omega}}{}^{\mathrm{\gamma}\to p}=\frac{128{\mathrm{\pi}}^{3}}{\mathit{cA}}\frac{|t_{\mathit{af}}{}^{1\mathrm{\gamma}}{|}^{4}|t_{\mathit{fs}}{}^{2p}{|}^{2}|t_{\mathit{sa}}{}^{2p}{|}^{2}}{n_{2}{}^{2}c_{2}{}^{2}}P_{\mathrm{\omega}}{}^{2}{\left(\frac{2\mathrm{\pi}L}{\mathrm{\lambda}}\right)}^{2}{{d}_{\mathrm{eff}}}^{2}\times exp\left[-2\left({\mathrm{\delta}}_{1}+{\mathrm{\delta}}_{2}\right)\right]\frac{{sin}^{2}\mathrm{\psi}+{sinh}^{2}\mathrm{\chi}}{{\mathrm{\psi}}^{2}+{\mathrm{\chi}}^{2}},\\ \phantom{\rule{2.9em}{0ex}}\mathrm{\psi}=\left(2\mathrm{\pi}L/\mathrm{\lambda}\right)\left({n}_{1}{c}_{1}-{n}_{2}{c}_{2}\right),\\ \phantom{\rule{2.6em}{0ex}}{\mathrm{\delta}}_{1}=\left(2\mathrm{\pi}L/\mathrm{\kappa}\right)\left({n}_{1}{\mathrm{\chi}}_{1}/{c}_{1}\right),{\mathrm{\delta}}_{2}=\left(2\mathrm{\pi}L/\mathrm{\kappa}\right)\left({n}_{2}{\mathrm{\chi}}_{2}/{c}_{2}\right),\\ \phantom{\rule{2.9em}{0ex}}\mathrm{\chi}={\mathrm{\delta}}_{1}-{\mathrm{\delta}}_{2}=\left(2\mathrm{\pi}L/\mathrm{\lambda}\right)\left({n}_{1}{\mathrm{\kappa}}_{1}/{c}_{1}-{n}_{2}{\mathrm{\kappa}}_{2}/{c}_{2}\right),\\ \phantom{\rule{1.5em}{0ex}}t_{\mathit{af}}{}^{1\mathrm{\gamma}}=\left\{\begin{array}{cc}\frac{2cos\left(\mathrm{\theta}\right)}{{c}_{1}+{n}_{1}cos\left(\mathrm{\theta}\right)},& \mathrm{\gamma}=p,\\ \frac{2cos\left(\mathrm{\theta}\right)}{cos\left(\mathrm{\theta}\right)+{n}_{1}{c}_{1}},& \mathrm{\gamma}=s\end{array}\right.,\\ \phantom{\rule{1.5em}{0ex}}t_{\mathit{fs}}{}^{2p}=\frac{2{n}_{2}{c}_{2}}{{n}_{2s}{c}_{2}+{n}_{2}{c}_{2s}},t_{\mathit{sa}}{}^{2p}=\frac{2{n}_{2s}{c}_{2s}}{{n}_{2s}cos\left(\mathrm{\theta}\right)+{c}_{2s}},\end{array}$$
(2)
$${d}_{\mathrm{eff}}=\left\{\begin{array}{ll}2{d}_{31}{c}_{1}{s}_{1}{c}_{2}+{d}_{31}c_{1}{}^{2}{s}_{2}+{d}_{33}s_{1}{}^{2}{s}_{2},& \mathrm{\gamma}=p\\ {d}_{31}{s}_{2},& \mathrm{\gamma}=s\end{array}\right..$$
(3)
$$\begin{array}{l}{P}_{2\mathrm{\omega}}=\frac{512{\mathrm{\pi}}^{3}}{c{\mathrm{\omega}}^{2}}{d}^{2}{\left(t_{\mathrm{\omega}}{}^{\mathrm{\gamma}}\right)}^{4}T_{2\mathrm{\omega}}{}^{\mathrm{\gamma}}{p}^{2}\left(\mathrm{\theta}\right)P_{\mathrm{\omega}}{}^{2}\left[1/{\left(n_{1}{}^{2}-n_{2}{}^{2}\right)}^{2}\right]{sin}^{2}\mathrm{\psi},\\ \phantom{\rule{1em}{0ex}}\mathrm{\psi}=\left(2\mathrm{\pi}L/\mathrm{\lambda}\right)\left({n}_{1}{c}_{1}-{n}_{2}{c}_{2}\right),\\ \phantom{\rule{0.2em}{0ex}}t_{\mathrm{\omega}}{}^{\mathrm{\gamma}}=\left\{\begin{array}{cc}\frac{2cos\mathrm{\theta}}{{n}_{1}{c}_{1}+cos\mathrm{\theta}},& \mathrm{\gamma}=s\\ \phantom{\rule{0.5em}{0ex}}\frac{2cos\mathrm{\theta}}{{n}_{1}cos\mathrm{\theta}+{c}_{1}},& \mathrm{\gamma}=p\end{array}\right.,\\ p\left(\mathrm{\theta}\right)={p}_{1}{p}_{2}\mathrm{projection\; factor},\\ T_{2\mathrm{\omega}}{}^{\mathrm{\gamma}}=\left\{\begin{array}{l}\frac{2{n}_{2}{c}_{2}cos\left(\mathrm{\theta}+{n}_{1}{c}_{1}\right)\left({n}_{1}{c}_{1}+{n}_{2}{c}_{2}\right)}{{n}_{2}{c}_{2}+cos{\mathrm{\theta}}^{3}},\mathrm{\gamma}=s\\ \frac{2{n}_{2}{c}_{2}\left({n}_{1}cos\mathrm{\theta}+{c}_{1}\right)\left({n}_{2}{c}_{1}+{n}_{1}{c}_{2}\right)}{{\left({c}_{2}+{n}_{2}cos\mathrm{\theta}\right)}^{3}},\mathrm{\gamma}=p\end{array}\right..\end{array}$$
(4)
$$\mathrm{\Delta}n=\frac{\mathrm{\lambda}}{{D}_{\mathrm{eff}}\mathrm{\pi}}{sin}^{-1}\left(\sqrt{T}\right),{D}_{\mathrm{eff}}=\frac{\mathit{nD}}{{\left({n}^{2}-{sin}^{2}\mathrm{\theta}\right)}^{1/2}},$$