## Abstract

We show that for any rotationally symmetric apodizer on the full aperture, there is a family of apodizers on the annular aperture with the same functional Strehl ratio vs defocus (*W*_{20}) and vs spherical aberration (*W*_{40}). However, in the latter case, the coefficients *W*_{20} and *W*_{40} are reduced by the factors (1 − *ɛ*^{2}) and (1 − *ɛ*^{2})^{2}, respectively, where *ɛ* is the central obscuration ratio. We indicate that the best focal plane is shifted from *W*_{20} = −*W*_{40} to *W*_{20} = −(1 + *ɛ*^{2})*W*_{40}. These general results allow us to design and to compare novel apodizers on annular apertures which reduce the influence of *W*_{20} and *W*_{40}. The Strehl ratios of a novel family of apodizers are discussed to illustrate our general results.

© 1988 Optical Society of America

Full Article |

PDF Article
### Equations (27)

Equations on this page are rendered with MathJax. Learn more.

(1)
$$p(r,{W}_{20},{W}_{40})=2\pi \hspace{0.17em}{\int}_{0}^{\mathrm{\Omega}}\tilde{p}(\rho )\hspace{0.17em}\text{exp}\{i2\pi [{W}_{20}{(\rho /\mathrm{\Omega})}^{2}+{W}_{40}{(\rho /\mathrm{\Omega})}^{4}]\}\times {J}_{0}(2\pi r\rho )\rho d\rho .$$
(2)
$$S({W}_{20},{W}_{40})=\mid p(r=0,{W}_{20},{W}_{40}){\mid}^{2}/\mid p(r=0,{W}_{20}=0,{W}_{40}=0){\mid}^{2}.$$
(3)
$$\begin{array}{l}\mid q({W}_{20}+{W}_{40},{W}_{40}){\mid}^{2}=\mid p(r=0,{W}_{20},{W}_{40}){\mid}^{2}\\ =\mid \pi {\mathrm{\Omega}}^{2}{\int}_{0}^{\mathrm{\Omega}}\tilde{p}(\rho )\hspace{0.17em}\text{exp}\{i2\pi \{{W}_{20}{(\rho /\mathrm{\Omega})}^{2}+{W}_{40}{(\rho /\mathrm{\Omega})}^{4}]\}d[{(\rho /\mathrm{\Omega})}^{2}]{\mid}^{2}.\end{array}$$
(4)
$$\zeta ={(\rho /\mathrm{\Omega})}^{2}-0.5,\mathrm{\hspace{0.17em}\u200a\u200a}\mathrm{\hspace{0.17em}\u200a\u200a}\mathrm{\hspace{0.17em}\u200a\u200a}\tilde{q}(\zeta )=\tilde{p}(\rho ),$$
(5)
$$\mid q({W}_{20}+{W}_{40},{W}_{40}){\mid}^{2}=\mid \pi {\mathrm{\Omega}}^{2}{\int}_{-0.5}^{0.5}\tilde{q}(\zeta )\times \text{exp}\{i2\pi [{W}_{40}{\zeta}^{2}+({W}_{40}+{W}_{20})\zeta ]\}d\zeta {\mid}^{2}.$$
(6)
$$\rho =\mathrm{\Omega}{[(1-{\varepsilon}^{2})\zeta +0.5(1+{\varepsilon}^{2})]}^{1/2},\mathrm{\hspace{0.17em}\u200a\u200a}\mathrm{\hspace{0.17em}\u200a\u200a}\mathrm{\hspace{0.17em}\u200a\u200a}\tilde{A}(\rho ,\varepsilon )=\tilde{q}(\zeta ).$$
(7)
$$A(r,\varepsilon )=2\pi {\int}_{\varepsilon \mathrm{\Omega}}^{\mathrm{\Omega}}\tilde{A}(\rho ,\varepsilon )\hspace{0.17em}\text{exp}\{i2\pi [{W}_{20}{(\rho /\mathrm{\Omega})}^{2}+{W}_{40}{(\rho /\mathrm{\Omega})}^{4}]\}\times {J}_{0}(2\pi r\rho )\rho d\rho .$$
(8)
$$\mid {q}_{A}({W}_{20}+{W}_{40},{W}_{40}){\mid}^{2}=\mid \pi {\mathrm{\Omega}}^{2}{\int}_{\u220a\mathrm{\Omega}}^{\mathrm{\Omega}}\tilde{A}(\rho ,\varepsilon )\times \text{exp}\{i2\pi [{W}_{20}{(\rho /\mathrm{\Omega})}^{2}+{W}_{40}{(\rho /\mathrm{\Omega})}^{4}]\}d[{(\rho /\mathrm{\Omega})}^{2}]{\mid}^{2}.$$
(9)
$$\mid {q}_{A}({W}_{20}+{W}_{40},{W}_{40}){\mid}^{2}=\mid \pi {\mathrm{\Omega}}^{2}(1-{\varepsilon}^{2})\hspace{0.17em}{\int}_{-0.5}^{0.5}\tilde{q}(\zeta )\times \text{exp}[i2\pi {(1-{\varepsilon}^{2})}^{2}{W}_{40}{\zeta}^{2}]\times \text{exp}\{i2\pi (1-{\varepsilon}^{2})\times [{W}_{20}+{W}_{40}(1+{\varepsilon}^{2})]\zeta \}d\zeta {\mid}^{2}.$$
(10)
$$\mid {q}_{A}({W}_{20}+{W}_{40},{W}_{40}){\mid}^{2}={(1-{\varepsilon}^{2})}^{2}\mid q\{(1-{\varepsilon}^{2})\times [{W}_{20}+(1+{\varepsilon}^{2}){W}_{40}],\times {(1-{\varepsilon}^{2})}^{2}{W}_{40}\}{\mid}^{2}.$$
(11)
$$\begin{array}{l}\mid {S}_{A}({W}_{20},{W}_{40}){\mid}^{2}=\mid {q}_{A}({W}_{20}+{W}_{40},{W}_{40}){\mid}^{2}/\mid {q}_{A}(0,0){\mid}^{2}\\ =\mid q\{(1-{\varepsilon}^{2})[{W}_{20}+(1+{\varepsilon}^{2}){W}_{40}],\times {(1-{\varepsilon}^{2})}^{2}{W}_{40}\}{\mid}^{2}/\mid q(0,0){\mid}^{2}.\end{array}$$
(12)
$${S}_{A}({W}_{20})=S[(1-{\varepsilon}^{2}){W}_{20}].$$
(13)
$${W}_{20}=-(1+{\varepsilon}^{2}){W}_{40}.$$
(14)
$${S}_{A}({W}_{20},{W}_{40})=S[(1-{\varepsilon}^{2}){W}_{20},\mathrm{\hspace{0.17em}\u200a\u200a}\mathrm{\hspace{0.17em}\u200a\u200a}\mathrm{\hspace{0.17em}\u200a\u200a}{(1-{\varepsilon}^{2})}^{2}{W}_{40}].$$
(15)
$$\begin{array}{l}T=\pi {\int}_{0}^{\mathrm{\Omega}}\mid \tilde{p}(\rho ){\mid}^{2}\rho d\rho /\pi {\mathrm{\Omega}}^{2}\\ ={\int}_{-0.5}^{0.5}\mid \tilde{q}(\zeta ){\mid}^{2}d\zeta ;\end{array}$$
(16)
$$\begin{array}{l}{T}_{A}=\pi {\int}_{\varepsilon \mathrm{\Omega}}^{\mathrm{\Omega}}\mid A(\rho ,\varepsilon ){\mid}^{2}\rho d\rho /\pi {\mathrm{\Omega}}^{2}\\ =(1-{\varepsilon}^{2})\hspace{0.17em}{\int}_{-0.5}^{0.5}\mid \tilde{q}(\zeta ){\mid}^{2}d\zeta .\end{array}$$
(17)
$${T}_{A}=(1-{\varepsilon}^{2})T.$$
(18)
$$\tilde{q}(\zeta )=\text{exp}(-2\pi \alpha {\zeta}^{2}).$$
(19)
$$\tilde{A}(\rho ,\varepsilon )=\text{exp}(-\pi \alpha /2)\xb7\text{exp}(-2\pi \alpha [{(\rho /\mathrm{\Omega})}^{2}-{\varepsilon}^{2}]{(1-{\varepsilon}^{2})}^{-1}\times \{[{(\rho /\mathrm{\Omega})}^{2}-{\varepsilon}^{2}]{(1-{\varepsilon}^{2})}^{-1}-1\}).$$
(20)
$$\tilde{p}(\rho )=\text{exp}(-\pi \alpha /2)\hspace{0.17em}\text{exp}\{-2\pi \alpha {(\rho /\mathrm{\Omega})}^{2}[{(\rho /\mathrm{\Omega})}^{2}-1]\}.$$
(21)
$$\rho =\mathrm{\Omega}\sqrt{1+{\varepsilon}^{2}}/\sqrt{2}.$$
(22)
$${W}_{20}=-\{[(1-{\varepsilon}^{8})/4-(1-{\varepsilon}^{4})(1-{\varepsilon}^{6})/6(1-{\varepsilon}^{2})]/[(1-{\varepsilon}^{6})/3-{(1-{\varepsilon}^{4})}^{2}/4(1-{\varepsilon}^{2})]\}{W}_{40}.$$
(23)
$$1-{\varepsilon}^{8}=(1+{\varepsilon}^{4})(1-{\varepsilon}^{4}),$$
(24)
$$\begin{array}{l}1-{\varepsilon}^{6}=(1-{\varepsilon}^{2})(1+{\varepsilon}^{2}+{\varepsilon}^{4})\\ =(1-{\varepsilon}^{4})(1+{\varepsilon}^{2}+{\varepsilon}^{4})/(1+{\varepsilon}^{2}),\end{array}$$
(25)
$${W}_{20}=-\{[3(1-{\varepsilon}^{4})(1+{\varepsilon}^{4})-2(1-{\varepsilon}^{4})(1+{\varepsilon}^{2}+{\varepsilon}^{4})]/[4(1-{\varepsilon}^{4})(1+{\varepsilon}^{2}+{\varepsilon}^{4})/(1+{\varepsilon}^{2})-3(1-{\varepsilon}^{4})(1+{\varepsilon}^{2})]\}{W}_{40},$$
(26)
$${W}_{20}=-\{[3(1+{\varepsilon}^{4})-2(1+{\varepsilon}^{2}+{\varepsilon}^{4})]/[4(1+{\varepsilon}^{2}+{\varepsilon}^{4})-3{(1+{\varepsilon}^{2})}^{2}]\}(1+{\varepsilon}^{2}){W}_{40}.$$
(27)
$${W}_{20}=-(1+{\varepsilon}^{2}){W}_{40}.$$