## Abstract

The possibility of converting the continuous refractive-index profile into a two-index solution is shown. The technique of conversion that is developed makes it possible to take dispersion into account. This ensures that it is possible to achieve a good agreement between theory and practice in a broad spectral range. Several gradient-index filters have been developed and produced, and measuring results are presented.

© 1992 Optical Society of America

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### Equations (15)

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(1)
$$\mathbf{M}=\left[\begin{array}{ll}\text{cos}\hspace{0.17em}\mathrm{\Phi}\hfill & i\hspace{0.17em}\text{sin}\hspace{0.17em}\mathrm{\Phi}/n(x)\hfill \\ i\hspace{0.17em}n(x)\text{sin}\hspace{0.17em}\mathrm{\Phi}\hfill & \text{cos}\hspace{0.17em}\mathrm{\Phi}\hfill \end{array}\right],$$
(2)
$$\mathrm{\Phi}=2\mathrm{\pi}/\mathrm{\lambda}\hspace{0.17em}OT=2\mathrm{\pi}/\mathrm{\lambda}\hspace{0.17em}n(x)t(x),$$
(3)
$$t(x)=O{T}_{\text{tot}}/[(N-1)n(x)],$$
(4)
$$OT=n(x)t(x)\ll \mathrm{\lambda},$$
(5)
$$\mathbf{M}=\left[\begin{array}{ll}1\hfill & i\hspace{0.17em}2\mathrm{\pi}/\mathrm{\lambda}\hspace{0.17em}t(x)\hfill \\ i\hspace{0.17em}2\mathrm{\pi}/\mathrm{\lambda}\hspace{0.17em}{n}^{2}(x)t(x)\hfill & 1\hfill \end{array}\right].$$
(6)
$$\mathbf{M}=\left[\begin{array}{ll}1\hfill & i\hspace{0.17em}2\mathrm{\pi}/\mathrm{\lambda}[{t}_{H}(x)+{t}_{L}(x)]\hfill \\ i\hspace{0.17em}2\mathrm{\pi}/\mathrm{\lambda}[{{n}_{H}}^{2}{t}_{H}(x)+{{n}_{L}}^{2}{t}_{L}(x)]\hfill & 1\hfill \end{array}\right].$$
(7)
$${t}_{H}(x)=\frac{{n}^{2}(x)-{{n}_{L}}^{2}}{{{n}_{H}}^{2}-{{n}_{L}}^{2}}t(x),$$
(8)
$${t}_{L}(x)=t(x)-{t}_{H}(x).$$
(9)
$${t}_{H}(x)=\frac{{n}^{2}(x)-{{n}_{L}}^{2}({\mathrm{\lambda}}_{c})}{{{n}_{H}}^{2}({\mathrm{\lambda}}_{c})-{{n}_{L}}^{2}({\mathrm{\lambda}}_{c})}t(x),$$
(10)
$${t}_{L}(x)=t(x)-{t}_{H}(x).$$
(11)
$$n(x,\mathrm{\lambda})=\{{{n}_{H}}^{2}(\mathrm{\lambda})[{t}_{H}(x)/t(x)]+{{n}_{L}}^{2}(\mathrm{\lambda}){[{t}_{L}(x)/t(x)\}}^{1/2}.$$
(12)
$$n(x,\mathrm{\lambda})={[A(x){{n}_{H}}^{2}(\mathrm{\lambda})+[1-A(x)]{{n}_{L}}^{2}(\mathrm{\lambda})]}^{1/2},$$
(13)
$$A(x)=\frac{{n}^{2}(x)-{{n}_{L}}^{2}({\mathrm{\lambda}}_{c})}{{{n}_{H}}^{2}({\mathrm{\lambda}}_{c})-{{n}_{L}}^{2}({\mathrm{\lambda}}_{c})}.$$
(14)
$$\frac{{\mathrm{\lambda}}^{*}}{\mathrm{\lambda}}=\frac{n({\mathrm{\lambda}}_{0})}{n(\mathrm{\lambda})}.$$
(15)
$$\frac{{\mathrm{\lambda}}^{*}}{\mathrm{\lambda}}=\frac{{\mathrm{\alpha}}_{HL}}{1+{\mathrm{\alpha}}_{HL}}\frac{{n}_{H}({\mathrm{\lambda}}_{0})}{{n}_{H}(\mathrm{\lambda})}+\frac{1}{1+{\mathrm{\alpha}}_{HL}}\frac{{n}_{L}({\mathrm{\lambda}}_{0})}{{n}_{L}(\mathrm{\lambda})},$$