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Optimal selection of aspheric surfaces in optical design

Akira Yabe

Open Access Open Access

Abstract

Optical systems frequently use aspheric surfaces to improve performance. Typically the designer uses a combination of experience and trail and error to decide which surfaces to make aspheric, then the shape of the aspheric surface is optimized as part of the optical system. In this paper, a method of optimally choosing which surfaces are to be made aspheric is developed. This is implemented by allowing the surface number of the asphere to become a variable in the optimization. Imaginary surfaces are used as an intermediary to make the problem continuous for the optimization. The method of implementation and design examples are given.

©2005 Optical Society of America

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Figures (13)
Fig. 1.
Fig. 1. Thin glass layer after the sphere on the rear side of the glass.

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Fig. 2.
Fig. 2. Thin glass layer before the sphere on the front side of the glass.

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Fig. 3.
Fig. 3. Thin air layer before the sphere on the rear side of the glass.

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Fig. 4.
Fig. 4. Thin air layer after the sphere on the front side of the glass.

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Fig. 5.
Fig. 5. Starting point of the examples.

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Fig. 6.
Fig. 6. Merit function values for the different position of aspherics.

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Fig. 7.
Fig. 7. Real surface number for each solution with 1 aspherics.

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Fig. 8.
Fig. 8. Merit function value of each solution with 1 aspherics.

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Fig. 9.
Fig. 9. Solution for FNO=1.8 and the tangent of the field of view =0.28.

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Fig. 10.
Fig. 10. Real surface numbers for each solution with 4 aspherics.

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Fig. 11.
Fig. 11. Merit function value of each solution with 4 aspherics.

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Fig. 12.
Fig. 12. Solution for FNO=1.5 and the tangent of the field of view =0.33.

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Fig. 13.
Fig. 13. Solutions for ArF microlithography lens

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Tables (4)

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Table 1. Lens data of Fig. 1

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Table 2. Lens data of Fig. 2

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Table 3. Lens data of Fig. 3

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Table 4. Lens data of Fig. 4

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Equations (14)

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

C = b C k + a C k + 1
N = b N k + a N k + 1 ′
z 1 ( x , y ) = az ( x , y )
z 2 ( x , y ) = − bz ( x , y )
z 1 ( x , y ) = − az ( x , y )
z 2 ( x , y ) = bz ( x , y )
z 1 ( x , y ) − z 2 ( x , y ) = z ( x , y )
z 1 ( x , y ) − z 2 ( x , y ) = − z ( x , y )
a ( 0 ) = 0
a ( 1 ) = 1
a ( r ) = r
a ′ ( 0 ) = 0
a ′ ( 1 ) = 0
a ( r ) = ( 1 − cos ⁡ ( π r ) ) 2
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