## Abstract

The second-order focusing properties of a grating spectrograph with fixed entrance slit and a circular image field are investigated analytically. It is shown that the Rowland mount is the only circular field spectrograph mount which allows perfect tangential focusing (to second order) over a predetermined wavelength range, while two classes of mounts exist, one of which is nontrivial, for which-astigmatism is corrected exactly on a circular arc.

© 1990 Optical Society of America

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### Equations (36)

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(1)
$${({x}^{\prime}-{x}_{0})}^{2}+{({y}^{\prime}-{y}_{0})}^{2}={{R}^{\prime}}^{2},$$
(2)
$${F}_{20}=\frac{{\text{cos}}^{2}\alpha}{2r}+\frac{{\text{cos}}^{2}\beta}{2{r}^{\prime}}-{a}_{20}(\text{cos}\alpha +\text{cos}\beta )+\frac{m\mathrm{\lambda}}{{\mathrm{\lambda}}_{0}}{H}_{20}=0,$$
(3)
$$x(y,z)=\sum _{i=0}^{\infty}\sum _{j=0}^{\infty}{a}_{ij}{y}^{i}{z}^{j}.$$
(4)
$${r}^{\prime}(\beta )=\frac{{\text{cos}}^{2}\beta}{A+B\hspace{0.17em}\text{cos}\beta +C\hspace{0.17em}\text{sin}\beta},$$
(5)
$$A=B\hspace{0.17em}\text{cos}\alpha +C\hspace{0.17em}\text{sin}\alpha -\frac{{\text{cos}}^{2}\alpha}{r},$$
(7)
$$C=-\frac{2{dH}_{20}}{{\mathrm{\lambda}}_{0}}.$$
(8)
$${r}^{\prime}(\beta )={x}_{0}\hspace{0.17em}\text{cos}\beta +{y}_{0}\hspace{0.17em}\text{sin}\beta \pm \sqrt{{[{x}_{0}\hspace{0.17em}\text{cos}\beta +{y}_{0}\hspace{0.17em}\text{sin}\beta ]}^{2}+{{R}^{\prime}}^{2}-{r}_{0}^{2}},$$
(9)
$$\begin{array}{r}({K}_{0}^{2}-{L}_{0}^{2})+2({K}_{0}{K}_{1}-{L}_{0}{L}_{1})\hspace{0.17em}\text{sin}\beta \\ +({K}_{1}^{2}+2{K}_{0}{K}_{2}-{K}_{0}^{2}-{L}_{1}^{2}-2{L}_{0}{L}_{2})\hspace{0.17em}{\text{sin}}^{2}\beta \\ +(2{K}_{0}{K}_{3}+2{K}_{1}{K}_{2}-2{K}_{0}{K}_{1}-2{L}_{0}{L}_{3}-2{L}_{1}{L}_{2})\hspace{0.17em}{\text{sin}}^{3}\beta \\ +({K}_{2}^{2}+2{K}_{1}{K}_{3}-{K}_{1}^{2}-2{K}_{0}{K}_{2}-{L}_{2}^{2}-2{L}_{0}{L}_{4}-2{L}_{1}{L}_{3})\hspace{0.17em}{\text{sin}}^{4}\beta \\ +(2{K}_{2}{K}_{3}-2{K}_{0}{K}_{3}-2{K}_{1}{K}_{2}-2{L}_{1}{L}_{4}-2{L}_{2}{L}_{3})\hspace{0.17em}{\text{sin}}^{5}\beta \\ +({K}_{3}^{2}-{K}_{2}^{2}-2{K}_{1}{K}_{2}-{L}_{3}^{2}-2{L}_{2}{L}_{4})\hspace{0.17em}{\text{sin}}^{6}\beta \\ -(2{K}_{2}{K}_{3}+2{L}_{3}{L}_{4})\hspace{0.17em}{\text{sin}}^{7}\beta -({K}_{3}^{2}+{L}_{4}^{2})\hspace{0.17em}{\text{sin}}^{8}\beta =0,\end{array}$$
(10)
$${K}_{0}=2QAB+2{x}_{0}A,$$
(11)
$${K}_{1}=2QBC+2{y}_{0}B+2{x}_{0}C,$$
(12)
$${K}_{2}=-2{x}_{0}A,$$
(13)
$${K}_{3}=-2{y}_{0}B-2{x}_{0}C,$$
(14)
$${L}_{0}=1-2{x}_{0}B-{QA}^{2}-{QB}^{2},$$
(15)
$${L}_{1}=-2{y}_{0}A-2QAC,$$
(16)
$${L}_{2}=-2+4{x}_{0}B-2{y}_{0}C+{QB}^{2}-{QC}^{2},$$
(17)
$${L}_{3}=2{y}_{0}A,$$
(18)
$${L}_{4}=1-2{x}_{0}B+2{y}_{0}C,$$
(19)
$$Q={{R}^{\prime}}^{2}-{r}_{0}^{2}.$$
(20)
$$\sum _{n=0}^{8}{c}_{n}\hspace{0.17em}{\text{sin}}^{n}\beta =0,$$
(21)
$${K}_{0}={K}_{1}={K}_{2}={K}_{3}={L}_{0}={L}_{1}={L}_{2}={L}_{3}={L}_{4}=0.$$
(22)
$${F}_{02}=\frac{1}{2r}+\frac{1}{2{r}^{\prime}}-{a}_{02}(\text{cos}\alpha +\text{cos}\beta )+\frac{m\mathrm{\lambda}}{{\mathrm{\lambda}}_{0}}{H}_{02}=0,$$
(23)
$${r}^{\prime}(\beta )=\frac{1}{D+E\hspace{0.17em}\text{cos}\beta +F\hspace{0.17em}\text{sin}\beta},$$
(24)
$$D=E\hspace{0.17em}\text{cos}\alpha +F\hspace{0.17em}\text{sin}\alpha -\frac{1}{r},$$
(26)
$$F=-\frac{2{dH}_{02}}{{\mathrm{\lambda}}_{0}}.$$
(27)
$$\begin{array}{l}({M}_{0}^{2}-{N}_{0}^{2})+2({M}_{0}{M}_{1}-{N}_{0}{N}_{1})\hspace{0.17em}\text{sin}\beta \\ +\hspace{0.17em}({M}_{1}^{2}-{M}_{0}^{2}-{N}_{1}^{2}-2{N}_{0}{N}_{2})\hspace{0.17em}{\text{sin}}^{2}\beta \\ -\hspace{0.17em}(2{M}_{0}{M}_{1}+2{N}_{1}{N}_{2})\hspace{0.17em}{\text{sin}}^{3}\beta -({M}_{1}^{2}+{N}_{2}^{2})\hspace{0.17em}{\text{sin}}^{4}\beta =0,\end{array}$$
(28)
$${M}_{0}=2QDE+2{x}_{0}D,$$
(29)
$${M}_{1}=2QEF+2{y}_{0}E+2{x}_{0}F,$$
(30)
$${N}_{0}=1-2{x}_{0}E-{QD}^{2}-{QE}^{2},$$
(31)
$${N}_{1}=-2{y}_{0}D-2QDF,$$
(32)
$${N}_{2}=2{x}_{0}E-2{y}_{0}F+{QE}^{2}-{QF}^{2}.$$
(33)
$${M}_{0}={M}_{1}={N}_{0}={N}_{1}={N}_{2}=0.$$
(34)
$${R}^{\prime}={r}_{0}=\frac{1}{4}\sqrt{\frac{1}{{a}_{02}^{2}}+\frac{{\mathrm{\lambda}}_{0}^{2}}{{({dH}_{02})}^{2}}},$$
(35)
$$({x}_{0},{y}_{0})=\left(\frac{1}{4{a}_{02}},\frac{-{\mathrm{\lambda}}_{0}}{4{dH}_{02}}\right).$$
(36)
$$r=\frac{1}{2}{\left[{a}_{02}\hspace{0.17em}\text{cos}\alpha -\frac{{dH}_{02}}{{\mathrm{\lambda}}_{0}}\hspace{0.17em}\text{sin}\alpha \right]}^{-1}.$$