## Abstract

It is now well known that a system of interference fringes photographed on a photographic plate has very interesting properties when used as a diffraction grating. This paper considers two cases. One is when the photographic plate is a plane surface; the aberration properties of this are worked out as a function of wavelength. There are three positions for which the spherical aberration is zero. Of these, one is of little interest as this is simply zero-order position. Another case considered was that of a photographic plate in the form of a concave spherical surface. In this case, there are three positions at which the spherical aberration is zero. The nature of variation of spherical aberration between these zero-aberration positions is presented in the form of curves computed from the theory of these gratings.

© 1971 Optical Society of America

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### Equations (38)

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(1)
$$\text{tan}\alpha =m\hspace{0.17em}\text{tan}\beta .$$
(2)
$$\text{sin}\alpha +\text{sin}\beta ={\mathrm{\lambda}}_{0}/\sigma ,$$
(3)
$$1/\sigma ={{\mathrm{\lambda}}_{0}}^{-1}[\text{sin}\beta +m\hspace{0.17em}\text{tan}\beta {(1+{m}^{2}\hspace{0.17em}{\text{tan}}^{2}\beta )}^{-{\scriptstyle \frac{1}{2}}}].$$
(4)
$$\text{sin}\alpha +\text{sin}\varphi =\mathrm{\lambda}/\sigma .$$
(5)
$$\text{sin}\varphi =(\mathrm{\lambda}/{\mathrm{\lambda}}_{0})\hspace{0.17em}\text{sin}\beta +[(\mathrm{\lambda}/{\mathrm{\lambda}}_{0})-1]m\hspace{0.17em}\text{tan}\beta \hspace{0.17em}{(1+{m}^{2}\hspace{0.17em}{\text{tan}}^{2}\beta )}^{-{\scriptstyle \frac{1}{2}}}.$$
(6)
$$\varphi =(\mathrm{\lambda}/{\mathrm{\lambda}}_{0})\beta +[(\mathrm{\lambda}/{\mathrm{\lambda}}_{0})-1]m\beta ,$$
(7)
$$1/{l}_{E}=[(m+1)\mathrm{\lambda}/{\mathrm{\lambda}}_{0}-m](1/{l}_{B}).$$
(8)
$$\text{tan}{\varphi}_{0}=[(m+1)\mathrm{\lambda}/{\mathrm{\lambda}}_{0}-m]\hspace{0.17em}\text{tan}\beta .$$
(9)
$$A{A}^{\prime}=({\varphi}_{0}-\varphi ).$$
(10)
$$\begin{array}{ll}\hfill \text{sin}\varphi & =[(\mathrm{\lambda}/{\mathrm{\lambda}}_{0})-1]\hspace{0.17em}\text{sin}\alpha +(\mathrm{\lambda}/{\mathrm{\lambda}}_{0})k\hspace{0.17em}\text{tan}\alpha {(1+{k}^{2}\hspace{0.17em}{\text{tan}}^{2}\alpha )}^{-{\scriptstyle \frac{1}{2}}},\\ \hfill \text{tan}{\varphi}_{0}& =[(1+k)\mathrm{\lambda}/{\mathrm{\lambda}}_{0}-1]\hspace{0.17em}\text{tan}\alpha ,\end{array}$$
(11)
$$\text{or}\mathrm{\hspace{0.17em}\u200a\u200a}\mathrm{\hspace{0.17em}\u200a\u200a}\mathrm{\hspace{0.17em}\u200a\u200a}\begin{array}{l}\mathrm{\lambda}/{\mathrm{\lambda}}_{0}=(2m-1)/(m+1)\\ \mathrm{\lambda}/{\mathrm{\lambda}}_{0}=(2-k)/(1+k).\end{array}$$
(12)
$$\text{sin}\varphi =\left(\frac{2m-1}{m+1}\right)\hspace{0.17em}\text{tan}\beta {(1+{\text{tan}}^{2}\beta )}^{-{\scriptstyle \frac{1}{2}}}+\left(\frac{m-2}{m+1}\right)m\hspace{0.17em}\text{tan}\beta {(1+{m}^{2}\hspace{0.17em}{\text{tan}}^{2}\beta )}^{-{\scriptstyle \frac{1}{2}}}$$
(13)
$$\text{sin}{\varphi}_{0}=(m-1)\hspace{0.17em}\text{tan}\beta {[1+{(m-1)}^{2}\hspace{0.17em}{\text{tan}}^{2}\beta ]}^{-{\scriptstyle \frac{1}{2}}}.$$
(14)
$$\text{sin}{\varphi}_{0}=(m-1)\hspace{0.17em}\text{tan}\beta -\frac{{(m-1)}^{3}}{2}{\text{tan}}^{3}\beta +{\scriptstyle \frac{3}{8}}{(m-1)}^{5}\hspace{0.17em}{\text{tan}}^{5}\beta \cdots $$
(15)
$$\text{sin}\varphi =(m-1)\hspace{0.17em}\text{tan}\beta -\frac{{(m-1)}^{3}}{2}{\text{tan}}^{3}\beta +{\scriptstyle \frac{3}{8}}({m}^{5}-3{m}^{4}+3{m}^{3}-3{m}^{2}+3m-1)\hspace{0.17em}{\text{tan}}^{5}\beta +\cdots .$$
(16)
$$\begin{array}{l}\text{sin}{\varphi}_{0}=(1-k)\hspace{0.17em}\text{tan}\alpha -\frac{{(1-k)}^{3}}{2}{\text{tan}}^{3}\alpha +{\scriptstyle \frac{3}{8}}{(1-k)}^{5}\hspace{0.17em}{\text{tan}}^{5}\alpha \cdots \\ \text{sin}\varphi =(1-k)\hspace{0.17em}\text{tan}\alpha -\frac{{(1-k)}^{3}}{2}{\text{tan}}^{3}\alpha +{\scriptstyle \frac{3}{8}}(1-3k+3{k}^{2}-3{k}^{3}+3{k}^{4}-{k}^{5})\hspace{0.17em}{\text{tan}}^{5}\alpha +\cdots .\end{array}$$
(17)
$$\text{and}\mathrm{\hspace{0.17em}\u200a\u200a}\mathrm{\hspace{0.17em}\u200a\u200a}\mathrm{\hspace{0.17em}\u200a\u200a}\begin{array}{l}\text{sin}\alpha +\text{sin}\beta ={\mathrm{\lambda}}_{0}/\sigma \\ \text{sin}\beta +\text{sin}\varphi =\mathrm{\lambda}/\sigma ,\end{array}$$
(18)
$$(\text{sin}\alpha /\text{sin}\beta )=\text{constant}=m(\text{say}).$$
(19)
$$1/\sigma =(m+1)\hspace{0.17em}\text{sin}\beta /{\mathrm{\lambda}}_{0}$$
(20)
$$\text{sin}\varphi =[(m+1)\mathrm{\lambda}/{\mathrm{\lambda}}_{0}-1]\hspace{0.17em}\text{sin}\beta .$$
(21)
$${L}^{\prime}=(r\hspace{0.17em}\text{sin}\varphi )/\text{sin}(\alpha +\beta -\varphi ).$$
(22)
$${l}^{\prime}=r\varphi /(\alpha +\beta -\varphi ).$$
(24)
$$\varphi =[(m+1)\mathrm{\lambda}/{\mathrm{\lambda}}_{0}-1]\beta .$$
(25)
$${l}^{\prime}/r=[(m+1)\mathrm{\lambda}/{\mathrm{\lambda}}_{0}-1]/\{(m+1)-[(m+1)\mathrm{\lambda}/{\mathrm{\lambda}}_{0}-1]\}=A.$$
(26)
$${l}^{\prime}/r=\text{sin}{\varphi}_{0}/\text{sin}(\alpha +\beta -{\varphi}_{0}).$$
(27)
$$\text{tan}{\varphi}_{0}=A\hspace{0.17em}\text{sin}\hspace{0.17em}(\alpha +\beta )/[1+A\hspace{0.17em}\text{cos}(\alpha +\beta )].$$
(28)
$$\text{sin}\alpha +\text{sin}\beta ={\mathrm{\lambda}}_{0}/\sigma $$
(29)
$$\text{sin}\beta =[{\mathrm{\lambda}}_{0}/(m+1)](1/\sigma )$$
(30)
$$2\hspace{0.17em}\text{sin}\beta =[2{\mathrm{\lambda}}_{0}/(m+1)](1/\sigma ).$$
(31)
$$\text{sin}\alpha =[m{\mathrm{\lambda}}_{0}/(m+1)](1/\sigma )$$
(32)
$$2\hspace{0.17em}\text{sin}\alpha =[2m{\mathrm{\lambda}}_{0}/(m+1)](1/\sigma ).$$
(33)
$$\text{sin}\beta ={\mathrm{\lambda}}_{0}/\sigma .$$
(34)
$$\text{sin}\alpha =m{\mathrm{\lambda}}_{0}/\sigma .$$
(35)
$$\text{sin}\alpha +\text{sin}\beta =(m+1){\mathrm{\lambda}}_{0}/\sigma .$$
(36)
$$\text{sin}\alpha ={\mathrm{\lambda}}_{0}/\sigma .$$
(37)
$$\text{sin}\beta =({\mathrm{\lambda}}_{0}/m)\hspace{0.17em}(1/\sigma );$$
(38)
$$\text{sin}\alpha +\text{sin}\beta =\{[(m+1){\mathrm{\lambda}}_{0}]/m\}\hspace{0.17em}(1/\sigma ).$$