## Abstract

We investigated the ability of aberration-corrected concave holographic gratings used in the Rowland mount at normal diffraction to provide high spectral resolution in the far-ultraviolet region. By assuming that astigmatism and spherical aberration are geometrically corrected by an ellipsoid, we show that holography can be used to correct the remaining prominent second-type coma. Stigmatic sources require a laser wavelength that is too far in the ultraviolet for current recording technology. However, at 3336 Å a simple compact symmetric mount, which involves two spherical mirrors, can generate aberrated wave fronts that can be used to record a coma-corrected holographic grating. When compared with the equivalent equally spaced straight-groove grating, which requires a modified ellipsoid substrate, holography cancels the additional asymmetrical term of deformation that permits the use of a simpler surface for the substrate. Some areas of potential difficulty in the holographic mounting are briefly analyzed.

© 1991 Optical Society of America

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### Equations (16)

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(1)
$$F=\phantom{\rule{0.2em}{0ex}}\u3008\mathit{\text{AP}}\u3009+\u3008\mathit{\text{PB}}\u3009+\mathit{\text{nm}}\phantom{\rule{0em}{0ex}}\lambda \phantom{\rule{0.1em}{0ex}},$$
(2)
$$n\phantom{\rule{0em}{0ex}}{\lambda}_{0}=(\u3008\mathit{\text{CP}}\u3009-\u3008\mathit{\text{DP}}\u3009)-(\u3008\mathit{\text{CO}}\u3009-\u3008\mathit{\text{DO}}\u3009)\phantom{\rule{0.1em}{0ex}},$$
(3)
$$x\phantom{\rule{0.1em}{0ex}}(y,z)={a}_{20}{y}^{2}+{a}_{02}{z}^{2}+{a}_{30}{y}^{3}+{a}_{12}y{z}^{2}+{a}_{40}{y}^{4}+{a}_{22}{y}^{2}{z}^{2}+{a}_{04}{z}^{4}\phantom{\rule{0.1em}{0ex}}.$$
(4)
$$F(y,z)={F}_{00}+y{F}_{10}+1\phantom{\rule{0em}{0ex}}/\phantom{\rule{0em}{0ex}}2{y}^{2}{F}_{20}+1\phantom{\rule{0em}{0ex}}/\phantom{\rule{0em}{0ex}}2{z}^{2}{F}_{02}+1\phantom{\rule{0em}{0ex}}/\phantom{\rule{0em}{0ex}}2{y}^{3}{f}_{30}+1\phantom{\rule{0em}{0ex}}/\phantom{\rule{0em}{0ex}}2y{z}^{2}{F}_{12}+1\phantom{\rule{0em}{0ex}}/\phantom{\rule{0em}{0ex}}8{y}^{4}{F}_{40}+1\phantom{\rule{0em}{0ex}}/\phantom{\rule{0em}{0ex}}4{y}^{2}{z}^{2}{F}_{22}+1\phantom{\rule{0em}{0ex}}/\phantom{\rule{0em}{0ex}}8{z}^{4}{F}_{04}\phantom{\rule{0.1em}{0ex}}.$$
(5)
$${F}_{i,j}={M}_{i,j}+\frac{m\lambda}{{\lambda}_{0}}{H}_{i,j}\phantom{\rule{0.1em}{0ex}}.$$
(6)
$$\sigma ={\lambda}_{0}\phantom{\rule{0em}{0ex}}/\phantom{\rule{0em}{0ex}}(sin\phantom{\rule{0em}{0ex}}\delta -sin\phantom{\rule{0em}{0ex}}\gamma )\phantom{\rule{0.1em}{0ex}}.$$
(7)
$${H}_{02}=\frac{1}{Rcos\phantom{\rule{0em}{0ex}}\gamma}-\frac{1}{Rcos\phantom{\rule{0em}{0ex}}\delta}-\frac{1}{\rho}(cos\phantom{\rule{0em}{0ex}}\gamma -cos\phantom{\rule{0em}{0ex}}\delta )=0\phantom{\rule{0.1em}{0ex}},$$
(8)
$${F}_{12}={M}_{12}+\frac{m\phantom{\rule{0em}{0ex}}\lambda}{{\lambda}_{0}}{H}_{12}=0\phantom{\rule{0.1em}{0ex}},$$
(9)
$$\sigma =\frac{{\lambda}_{0}}{2sin\phantom{\rule{0em}{0ex}}(|\phantom{\rule{0em}{0ex}}\delta \phantom{\rule{0em}{0ex}}|)}\phantom{\rule{0.1em}{0ex}},$$
(10)
$$\begin{array}{l}{M}_{12}=\left[\frac{1}{R\phantom{\rule{0.2em}{0ex}}cos\phantom{\rule{0.2em}{0ex}}\alpha}-\frac{cos\phantom{\rule{0.2em}{0ex}}\alpha}{\rho}\right]\frac{sin\phantom{\rule{0.2em}{0ex}}\alpha}{R\phantom{\rule{0.2em}{0ex}}cos\phantom{\rule{0.2em}{0ex}}\alpha}=4.53\times {10}^{-8}\phantom{\rule{0.2em}{0ex}}({\text{mm}}^{-2}),\hfill \\ {H}_{12}=\frac{-2\phantom{\rule{0.2em}{0ex}}sin\phantom{\rule{0.2em}{0ex}}\delta}{R\phantom{\rule{0.2em}{0ex}}cos\phantom{\rule{0.2em}{0ex}}\delta}\left(\frac{1}{R\phantom{\rule{0.2em}{0ex}}cos\phantom{\rule{0.2em}{0ex}}\delta}-\frac{cos\phantom{\rule{0.2em}{0ex}}\delta}{\rho}\right)\phantom{\rule{0.1em}{0ex}}.\hfill \end{array}$$
(11)
$${cos}^{2}\delta =\frac{m\phantom{\rule{0em}{0ex}}\lambda}{\sigma R}\frac{1}{{M}_{12}R+m\lambda \phantom{\rule{0em}{0ex}}/\phantom{\rule{0em}{0ex}}\sigma \rho}$$
(12)
$${H}_{12}=\frac{{S}_{c}sin\phantom{\rule{0.2em}{0ex}}\gamma}{{r}_{c}}-\frac{{S}_{d}sin\phantom{\rule{0em}{0ex}}\delta}{{r}_{d}}+{C}_{2}cos\phantom{\rule{0.2em}{0ex}}\gamma -{D}_{2}cos\phantom{\rule{0em}{0ex}}\delta \phantom{\rule{0.1em}{0ex}},$$
(13)
$$\begin{array}{cc}{S}_{c}=\frac{{r}_{c}}{{{r}_{c}}^{\prime 2}}-\frac{cos\phantom{\rule{0.2em}{0ex}}\gamma}{\rho},& {S}_{d}=\frac{{r}_{d}}{{{r}_{d}}^{\prime 2}}-\frac{cos\phantom{\rule{0.2em}{0ex}}\gamma}{\rho}\end{array}\phantom{\rule{0.1em}{0ex}}.$$
(14)
$$\begin{array}{l}{C}_{2}=sin{\eta}_{c}sec{\eta}_{c}\left[\frac{{S}_{c1}}{{p}_{c}}-\frac{{S}_{c2}}{{q}_{c}\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}{r}_{c}}\right]\phantom{\rule{0.2em}{0ex}}\left(\frac{{q}_{c}}{{r}_{c}}-1\right)\phantom{\rule{0.2em}{0ex}}{\left(1\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}\frac{{q}_{c}}{{{r}_{c}}^{\prime}}\right)}^{2},\\ {D}_{2}=sin{\eta}_{d}sec{\eta}_{d}\left[\frac{{S}_{d1}}{{p}_{d}}-\frac{{S}_{d2}}{{q}_{d}-{r}_{d}}\right]\phantom{\rule{0.2em}{0ex}}\left(\frac{{q}_{d}}{{r}_{d}}-1\right)\phantom{\rule{0.2em}{0ex}}{\left(1-\frac{{q}_{d}}{{{r}_{d}}^{\prime}}\right)}^{2}\phantom{\rule{0.1em}{0ex}},\end{array}$$
(15)
$$\begin{array}{ll}{S}_{c1}=\frac{1}{{p}_{c}}-\frac{cos\phantom{\rule{0.2em}{0ex}}{\eta}_{c}}{{R}_{c}},\hfill & {S}_{c2}=\frac{{q}_{c}\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}{r}_{c}}{{({q}_{c}\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}{{r}_{c}}^{\prime})}^{2}}-\frac{cos\phantom{\rule{0.2em}{0ex}}{\eta}_{c}}{{R}_{c}},\hfill \\ {S}_{d1}=\frac{1}{{p}_{d}}-\frac{cos\phantom{\rule{0.2em}{0ex}}{\eta}_{d}}{{R}_{d}},\hfill & {S}_{d2}=\frac{{q}_{d}\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}{r}_{d}}{{({q}_{d}\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}{{r}_{d}}^{\prime})}^{2}}-\frac{cos\phantom{\rule{0.2em}{0ex}}{\eta}_{d}}{{R}_{d}}\phantom{\rule{0.1em}{0ex}}.\hfill \end{array}$$
(16)
$$\begin{array}{cc}{F}_{12}={M}_{12}+\frac{m\lambda}{{\lambda}_{0}}{H}_{12}=0;& \text{hence}\phantom{\rule{0.2em}{0ex}}{H}_{12}=-1.56\times {10}^{-7}({\text{mm}}^{-2})\phantom{\rule{0.1em}{0ex}}.\end{array}$$