## Abstract

The FDTD method has been successfully used for many electromagnetic problems, but its application to laser material processing has been limited because even a several-millimeter domain requires a prohibitively large number of grids. In this article, we present a novel FDTD method for simulating large-scale laser beam absorption problems, especially for metals, by enlarging laser wavelength while maintaining the material’s reflection characteristics. For validation purposes, the proposed method has been tested with in-house FDTD codes to simulate *p*-, *s*-, and circularly polarized 1.06 μm irradiation on Fe and Sn targets, and the simulation results are in good agreement with theoretical predictions.

© 2013 Optical Society of America

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### Equations (12)

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(1)
$${R}_{s}=\frac{{\left[n-\mathrm{cos}{\theta}_{i}\right]}^{2}+{\kappa}^{2}}{{\left[n+\mathrm{cos}{\theta}_{i}\right]}^{2}+{\kappa}^{2}},$$
(2)
$${R}_{p}=\frac{{\left[n-\frac{1}{\mathrm{cos}{\theta}_{i}}\right]}^{2}+{\kappa}^{2}}{{\left[n+\frac{1}{\mathrm{cos}{\theta}_{i}}\right]}^{2}+{\kappa}^{2}},$$
(3)
$$\tilde{n}=n+i\kappa ,$$
(4)
$${n}^{2}=\frac{{c}^{2}}{2}\left(\sqrt{{\left(\mu \epsilon \right)}^{2}+{\left(\frac{\mu \sigma \lambda}{2\pi c}\right)}^{2}}+\mu \epsilon \right),$$
(5)
$${\kappa}^{2}=\frac{{c}^{2}}{2}\left(\sqrt{{\left(\mu \epsilon \right)}^{2}+{\left(\frac{\mu \sigma \lambda}{2\pi c}\right)}^{2}}-\mu \epsilon \right),$$
(6)
$$\begin{array}{ccc}\mu \epsilon ={a}_{1}& \text{and}& \frac{\mu \sigma \lambda}{2\pi c}={a}_{2}\end{array}.$$
(7)
$${n}^{2}={\mu}_{r}\left(\frac{\sqrt{{\epsilon}_{1}^{2}+{\epsilon}_{2}^{2}}+{\epsilon}_{1}}{2}\right),$$
(8)
$${\kappa}^{2}={\mu}_{r}\left(\frac{\sqrt{{\epsilon}_{1}^{2}+{\epsilon}_{2}^{2}}-{\epsilon}_{1}}{2}\right),$$
(9)
$${\tilde{\epsilon}}_{r}={\epsilon}_{1}+i{\epsilon}_{2}=\left({\epsilon}_{\infty}-\frac{{\omega}_{p}^{2}}{{\omega}^{2}+{\gamma}_{p}^{2}}\right)+i\left(\frac{{\omega}_{{}_{p}}^{2}{\gamma}_{p}}{{\omega}^{3}+\omega {\gamma}_{p}^{2}}\right),$$
(10)
$$\omega =\frac{2\pi c}{\lambda}.$$
(11)
$$\text{error}=\frac{\sqrt{{\displaystyle \underset{0}{\overset{\pi /2}{\int}}{\left[{R}_{s,p}^{*}(\theta )-{R}_{s,p}(\theta )\right]}^{2}d\theta}}}{\sqrt{{\displaystyle \underset{0}{\overset{\pi /2}{\int}}{\left[{R}_{s,p}(\theta )\right]}^{2}d\theta}}}.$$
(12)
$${R}_{s,p}^{*}=\frac{{\displaystyle {\int}_{0}^{\Re}{R}_{s,p}({\mathrm{sin}}^{-1}\left(r/\Re \right)){E}_{0}{e}^{-{(r/{r}_{0})}^{2}}dr}}{{\displaystyle {\int}_{0}^{\Re}{E}_{0}{e}^{-{(r/{r}_{0})}^{2}}dr}},$$