Useful way to compensate for intrinsic birefringence caused by calcium fluoride in optical lithography systems

Zelong Zhou; Hongbo Shang; Yongxin Sui; Huaijiang Yang

doi:10.3788/COL201816.032201

Journals >Chinese Optics Letters >Volume 16 >Issue 3 >Page 032201 > Article

Chinese Optics Letters
Vol. 16, Issue 3, 032201 (2018)

Useful way to compensate for intrinsic birefringence caused by calcium fluoride in optical lithography systems

Zelong Zhou^1,2,*, Hongbo Shang¹, Yongxin Sui¹, and Huaijiang Yang¹

Author Affiliations

¹Engineering Researcher Center of Extreme Precision Optics, Changchun Institute of Optics, Fine Mechanics and Physics, Chinese Academy of Sciences, Changchun 130033, China

²University of Chinese Academy of Sciences, Beijing 100049, China

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DOI: 10.3788/COL201816.032201 Cite this Article Set citation alerts

Zelong Zhou, Hongbo Shang, Yongxin Sui, Huaijiang Yang, "Useful way to compensate for intrinsic birefringence caused by calcium fluoride in optical lithography systems," Chin. Opt. Lett. 16, 032201 (2018) Copy Citation Text

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Fig. 1. OZP terms arranged according to the M-symmetry properties.

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Fig. 2. Flow chart of PSO rotation.

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Fig. 3. Layout of a patent lithographic lens.

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Fig. 4. Convergence curve of PSO rotation.

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Fig. 5. Retardation pupil maps (a) before and (b) after PSO rotation.

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Fig. 6. OZP coefficients of retardation before and after PSO rotation.

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$j$	$O Z_{j}$	$j$	$O Z_{j}$	$j$	$O Z_{j}$
1	$(\begin{matrix} 1 & 0 \\ 0 & - 1 \end{matrix})$	4	$\sqrt{3} (2 ρ^{2} - 1) (\begin{matrix} 1 & 0 \\ 0 & - 1 \end{matrix})$	7	$\sqrt{8} (3 ρ^{3} - 2 ρ) (\begin{matrix} \cos θ & \sin θ \\ \sin θ & - \cos θ \end{matrix})$
$- 1$	$(\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix})$	$- 4$	$\sqrt{3} (2 ρ^{2} - 1) (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix})$	$- 7$	$\sqrt{8} (3 ρ^{3} - 2 ρ) (\begin{matrix} \cos θ & - \sin θ \\ - \sin θ & - \cos θ \end{matrix})$
2	$2 ρ (\begin{matrix} \cos θ & \sin θ \\ \sin θ & - \cos θ \end{matrix})$	5	$\sqrt{6} ρ^{2} (\begin{matrix} \cos 2 θ & \sin 2 θ \\ \sin 2 θ & - \cos 2 θ \end{matrix})$	8	$\sqrt{8} (3 ρ^{3} - 2 ρ) (\begin{matrix} \sin θ & - \cos θ \\ - \cos θ & - \sin θ \end{matrix})$
$- 2$	$2 ρ (\begin{matrix} \cos θ & - \sin θ \\ - \sin θ & - \cos θ \end{matrix})$	$- 5$	$\sqrt{6} ρ^{2} (\begin{matrix} \cos 2 θ & - \sin 2 θ \\ - \sin 2 θ & - \cos 2 θ \end{matrix})$	$- 8$	$\sqrt{8} (3 ρ^{3} - 2 ρ) (\begin{matrix} \sin θ & \cos θ \\ \cos θ & - \sin θ \end{matrix})$
3	$2 ρ (\begin{matrix} \sin θ & - \cos θ \\ - \cos θ & - \sin θ \end{matrix})$	6	$\sqrt{6} ρ^{2} (\begin{matrix} \sin 2 θ & - \cos 2 θ \\ - \cos 2 θ & - \sin 2 θ \end{matrix})$	9	$\sqrt{5} (6 ρ^{4} - 6 ρ^{2} + 1) (\begin{matrix} 1 & 0 \\ 0 & - 1 \end{matrix})$
$- 3$	$2 ρ (\begin{matrix} \sin θ & \cos θ \\ \cos θ & - \sin θ \end{matrix})$	$- 6$	$\sqrt{6} ρ^{2} (\begin{matrix} \sin 2 θ & \cos 2 θ \\ \cos 2 θ & - \sin 2 θ \end{matrix})$	$- 9$	$\sqrt{5} (6 ρ^{4} - 6 ρ^{2} + 1) (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix})$