## Abstract

The diffraction of Hermite–Gaussian beams by *N* equally spaced slits in a planar screen (a lamellar finite grating) at the scalar diffraction regime is analyzed. We start from the Rayleigh–Sommerfeld theory for two-dimensional problems with Dirichlet conditions and Kirchhoff’s approximation. The theory is presented for beams at oblique incidence; however, in the numerical simulation mainly normal incidence is considered. For Gaussian beams the ratio of the intensity diffracted at normal direction *E* at minimum and maximum transmitted power is studied in this paper as a function of the beam width *L*. Also, for Gaussian beams, the angular positions ${\theta}_{min}$ of the first minimum of the diffraction patterns as a function of the *L*-spot diameter is analyzed. Two methods to determine *L* for Gaussian beams are proposed. For Hermite–Gaussian beams an interesting diffraction property previously presented for one slit [J. Opt. Soc. Am. A **12**, 2440 (1995)] is generalized to finite gratings, namely, $\tau =\mathrm{\lambda}(E/N),$ where *τ* is the transmission coefficient and λ is the wavelength of the incident radiation. We study this property in detail and analyze its validity conditions in terms of the optogeometrical parameters of the system. This property might be useful in determining the total energy transmitted by a lamellar grating from knowledge of the intensity diffracted in the normal direction.

© 2001 Optical Society of America

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### Equations (25)

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(1)
$$E({x}_{0},{y}_{0})=\frac{i}{2}{\int}_{-\infty}^{\infty}E(x,0)\frac{\partial}{\partial {y}_{0}}{{H}_{0}}^{1}(\mathit{kr})\mathrm{d}x,$$
(2)
$$E({x}_{0},{y}_{0})=\frac{i}{2}\sum _{j=1}^{N}{\int}_{{a}_{j}}^{{a}_{j}+l}{E}^{i}(x,0)\frac{\partial}{\partial {y}_{0}}{{H}_{0}}^{1}(\mathit{kr})\mathrm{d}x,$$
(3)
$${{H}_{0}}^{1}(\mathit{kr})\approx \sqrt{2/\pi \mathit{kr}}exp(i\pi /4)exp(\mathit{ikr}).$$
(4)
$$\frac{\partial}{\partial {y}_{0}}{{H}_{0}}^{1}(\mathit{kr})\approx -i\sqrt{2k/\pi}exp(-i\pi /4)\times \frac{exp[\mathit{ik}({r}_{0}-xsin\theta )]}{{{r}_{0}}^{1/2}}cos\theta ,$$
(5)
$$E({x}_{0},{y}_{0})=f(\theta )exp({\mathit{ikr}}_{0})/\sqrt{{r}_{0}},$$
(6)
$$f(\theta )=\sqrt{k}exp(-i\pi /4)cos\theta {\widehat{E}}^{i}(ksin\theta ,0),$$
(7)
$${\widehat{E}}^{i}(\alpha ,0)=\frac{1}{\sqrt{2\pi}}\sum _{j=1}^{N}{\int}_{{a}_{j}}^{{a}_{j}+l}{E}^{i}(x,0)exp(-i\alpha x)\mathrm{d}x.$$
(8)
$$I(\theta )={k}^{2}{cos}^{2}\theta |{\widehat{E}}^{i}(ksin\theta ,0){|}^{2},$$
(9)
$${E}^{i}(x,y)=\frac{1}{\sqrt{2\pi}}{\int}_{-k}^{k}A(\alpha )exp[i(\alpha x-\beta y)]\mathrm{d}\alpha ,$$
(10)
$${\widehat{E}}^{i}(ksin\theta ,0)=\frac{l}{2\pi}{\int}_{-k}^{k}A(\alpha )exp[i(\alpha -ksin\theta )l/2]\times exp[i(N-1)(\alpha -ksin\theta )(d+l)/2]\times \frac{sin[(\alpha -ksin\theta )l/2]}{(\alpha -ksin\theta )l/2}\times \frac{sin[N(\alpha -ksin\theta )(d+l)/2]}{sin[(\alpha -ksin\theta )(d+l)/2]}\mathrm{d}\alpha .$$
(11)
$${I}_{0}=-\mathrm{Im}{\int}_{-\infty}^{\infty}{E}^{i*}\frac{\partial {E}^{i}}{\partial y}\mathrm{d}x={\int}_{-k}^{k}\beta |A(\alpha ){|}^{2}\mathrm{d}\alpha ,$$
(12)
$${E}^{i}(x,y=0)={H}_{m}\left[\frac{2}{L}(x-b)\right]exp\left[-\frac{2(x-b{)}^{2}}{{L}^{2}}\right],$$
(13)
$$A(\alpha )=\frac{L}{2}(i{)}^{m}{H}_{m}\left[-\frac{L}{2}{q}_{1}({\theta}_{i})\right]{q}_{2}({\theta}_{i})exp[-i\alpha b]\times exp[-{q}_{1}({\theta}_{i}{)}^{2}{L}^{2}/8],$$
(14)
$${q}_{1}({\theta}_{i})=\alpha cos{\theta}_{i}-\beta sin{\theta}_{i}$$
(15)
$${q}_{2}({\theta}_{i})=cos{\theta}_{i}+(\alpha /\beta )sin{\theta}_{i}.$$
(16)
$${A}_{d}(\alpha )=A(\alpha )exp(\alpha ).$$
(17)
$$K={E}_{min}/{E}_{max}$$
(18)
$${\theta}_{min}=\frac{\mathrm{\lambda}}{2D}\left[1+\frac{0.3709}{(L/l{)}^{2}}\right].$$
(19)
$$E=N\frac{\tau}{\mathrm{\lambda}},$$
(20)
$${\widehat{E}}^{i}(\alpha ,0)=\frac{1}{2\pi}\sum _{j=1}^{N}{\int}_{-k}^{k}A({\alpha}^{\prime})\mathrm{d}{\alpha}^{\prime}\times {\int}_{{a}_{j}}^{{a}_{j}+l}exp[i({\alpha}^{\prime}-\alpha )x]\mathrm{d}x,$$
(21)
$${\int}_{{a}_{j}}^{{a}_{j+l}}exp[i({\alpha}^{\prime}-\alpha )x]\mathrm{d}x$$
(22)
$$=lexp[({\alpha}^{\prime}-\alpha )({a}_{j}+l/2)]\times \frac{sin[({\alpha}^{\prime}-\alpha )l/2]}{({\alpha}^{\prime}-\alpha )l/2},$$
(23)
$${\widehat{E}}^{i}(\alpha ,0)=\frac{l}{2\pi}{\int}_{-k}^{k}A({\alpha}^{\prime})\frac{sin[({\alpha}^{\prime}-\alpha )l/2]}{({\alpha}^{\prime}-\alpha )l/2}\times exp[i({\alpha}^{\prime}-\alpha )l/2]\mathrm{d}{\alpha}^{\prime}\times \sum _{j=1}^{N}exp[i({\alpha}^{\prime}-\alpha ){a}_{j}].$$
(24)
$$\sum _{j=1}^{N}exp[i({\alpha}^{\prime}-\alpha ){a}_{j}]$$
(25)
$$=\sum _{j=1}^{N}exp[i({\alpha}^{\prime}-\alpha )(l+d)(j-1)]=\frac{1-exp[\mathit{iN}({\alpha}^{\prime}-\alpha )(l+d)]}{1-exp[i({\alpha}^{\prime}-\alpha )(l+d)]}.$$