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[J]. Chinese Optics Letters, 2018, 16(3): 032201 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	$\sqrt{5} (6 ρ^{4} - 6 ρ^{2} + 1) (\begin{matrix} 1 & 0 \\ 0 & - 1 \end{matrix})$
$- 3$	$2 ρ (\begin{matrix} \sin θ & \cos θ \\ \cos θ & - \sin θ \end{matrix})$	$- 6$	$\sqrt{6} ρ^{2} (\begin{matrix} \sin 2 θ & \cos 2 θ \\ \cos 2 θ & - \sin 2 θ \end{matrix})$	$- 9$	$\sqrt{5} (6 ρ^{4} - 6 ρ^{2} + 1) (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix})$