High-precision measurement technique of isoplanatic angle

Xiwen Qiang; Min Wu; Fei Zong; Shengwei Zhai; Yuehong Hu; Shuanglian Feng; Junwei Zhao; Jinyong Chang

doi:10.11884/HPLPB202133.210215

Journals >High Power Laser and Particle Beams >Volume 33 >Issue 8 >Page 081008 > Article

High Power Laser and Particle Beams
Vol. 33, Issue 8, 081008 (2021)

High-precision measurement technique of isoplanatic angle

Xiwen Qiang, Min Wu, Fei Zong, Shengwei Zhai..., Yuehong Hu, Shuanglian Feng, Junwei Zhao and Jinyong Chang|Show fewer author(s)

Author Affiliations

The 27th Research Institute of China Electronics Technology Group Corporation, Zhengzhou 450047, China

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DOI: 10.11884/HPLPB202133.210215 Cite this Article

Xiwen Qiang, Min Wu, Fei Zong, Shengwei Zhai, Yuehong Hu, Shuanglian Feng, Junwei Zhao, Jinyong Chang. High-precision measurement technique of isoplanatic angle[J]. High Power Laser and Particle Beams, 2021, 33(8): 081008 Copy Citation Text

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Abstract

Isoplanatic angle is an important parameter in atmospheric optics field, and it is important for astronomical site optics selection, and evaluation of adaptive compensation for effects of atmospheric turbulence. Past measurement techniques of isoplanatic angle in the world is analyzed, and fundamental measurement principle of isoplanatic angle is introduced. A three annuli mirror was designed and its usability and precision has been verified. Primary specifications was put forward and a prototype has been developed for high precision measurement of isoplanatic angles. The results show that the prototype an measure isoplanatic angle day and night.

Keywords

adaptive optics atmospheric optics isoplanatic angle measurement technique tri-annuli apodizing mask

$\sigma _{{\rm{angle}}}^2 = {\left( {\frac{\theta }{{{\theta _0}}}} \right)^{5/3}}$

(1)

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${\theta _0} = {\left[ {2.91{k^2}{{\sec }^{8/3}}\varphi \int_0^L {C_{{n}}^2({\textit{z}} ){{\textit{z}} ^{5/3}}{{\rm{d}}}{\textit{z}} } } \right]^{ - 3/5}}$

(2)

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$\sigma _{\rm{s}}^2\left( \varphi \right) = 4{(2{\text{π}} )^4}0.033{\sec ^{8/3}}\varphi {k^2}{A^{ - 2}}\int_0^\infty {C_{{n}}^2({\textit{z}} )} W({\textit{z}} ){{\rm{d}}}{\textit{z}} $

(3)

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$W({\textit{z}} ) = c{{\textit{z}} ^{5/3}}$

(4)

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${\theta _0}(\varphi ) = 12.9{A^{ - 6/5}}{c^{3/5}}{\left[ {\sigma _{\rm{s}}^2(\varphi )} \right]^{ - 3/5}}$

(5)

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$W({\textit{z}} ) = {\int_0^\infty {\left| {\int {{\rm{d}}\rho \rho {{\rm{J}}_0}(L\rho )P(L\rho )} } \right|} ^2}{L^{ - 8/3}}{\sin ^2}\Bigg[\frac{{{L^2}{\textit{z}} }}{{2k}}\Bigg]{\rm{d}}L$

(6)

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${{\rm{J}}_{{n}}}(x) = {(x/2)^2}{\sum\limits_{k = 0}^\infty {\frac{{{{( - 1)}^k}}}{{k!(n + k)!}}\left( {\frac{x}{2}} \right)} ^{2k}}$

(7)

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$P(L\rho ) = \left\{

\begin{array}{l} 1, R_{1} ⩽ ρ ⩽ R_{2}, R_{3} ⩽ ρ ⩽ R_{4}, R_{5} ⩽ ρ ⩽ R_{6} \\ 0, 0 ⩽ ρ ⩽ R_{1}, R_{2} ⩽ ρ ⩽ R_{3}, R_{4} ⩽ ρ ⩽ R_{5} \end{array}

\right.$

(8)

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${{C}}_{{n}}^2({\textit{z}} ) = 8.16 \times {10^{ - 54}}{{\textit{z}} ^{10}}\exp \Bigg(\frac{{ - {\textit{z}} }}{{1000}}\Bigg) + 3.02 \times {10^{ - 17}}\exp \Bigg(\frac{{ - {\textit{z}} }}{{1500}}\Bigg) + 1.90 \times {10^{ - 15}}\exp \Bigg(\frac{{ - {\textit{z}} }}{{100}}\Bigg)$

(9)

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