Abstract

The implementation of gradient-index materials into optical systems requires that a complete set of manufacturing tolerances be calculated. To evaluate systems where the gradient may be tilted, decentered, or displaced, modifications to present ray-trace procedures are developed. Examples of the effect of such manufacturing errors are given.

© 1976 Optical Society of America

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References

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  1. M. Rimmer, Appl. Opt. 9, 533 (1970).
    [Crossref] [PubMed]
  2. D. T. Moore, J. Opt. Soc. Am. 65, 451 (1975).
    [Crossref]

1975 (1)

1970 (1)

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Figures (7)

FIG. 1
FIG. 1

Coordinate system for general ray trace. θ is the tilt of gradiant relative to the optical axis of lens in xy plane, h is the displacement of the gradient relative to polar tangent plane, and k is the decentration of gradient.

FIG. 2
FIG. 2

Method of ray-trace vertification. Rays A and B are equivalent. The path of ray A is calculated by the equations of this paper and the path of ray B by those of the previous work.2

FIG. 3
FIG. 3

Example of the effect of tilts of a radical gradient in a single element. Each number represents the numbered rays which intersected the image plane at that point; (a) is the untilted system. (b) is a tilt of the gradient of 0.5°, (c) of 1.0° and (d) of 5.0°. The diameter of the diffraction spot is 5.0 μm.

FIG. 4
FIG. 4

Example of the effect of decentration of a radial gradient in a single element. (a) the centered system, (b) a decentration of 0.0125 cm, (c) of 0.0625 cm, and (d) of 0.125 cm. The diameter of the diffraction spot is 5.0 μm.

FIG. 5
FIG. 5

Schematic of the effect of the removal of material from an axial gradient material. In case (a) no material is removed and the resulting surface is tangent to the polar tangent plane of the designed system. If additional material is removed then the actual polar tangent plane and that used in the design do not coincide.

FIG. 6
FIG. 6

Ray-displacement plane for various amounts of gradient displacement. (a) corresponds to 0.0 cm displacement, (b) to 0.01 cm, (c) to 0.05 cm, and (d) 0.1 cm.

FIG. 7
FIG. 7

Schematic for ray-trace procedure in a tilted and decentered gradient-index lens. The gradient is rotationaly symmetric in the primed coordinate system. The tilt angle is 0.10 rad and the decentration is +0.20 cm.

Tables (6)

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TABLE I Radial gradient singlet.

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TABLE II Peak to peak deviation for f/4 radial gradient singlet.

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TABLE III Axial gradient corrector plate for f/2.5.

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TABLE IV Parameters of lens with tilted and decentered gradient.

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TABLE V Polynomial expansions of ray-trace equations for tilted and decentered gradient.

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TABLE VI Ray-trace data for tilt decenter gradient.

Equations (28)

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N ( x , ξ ) = n = 0 m = 0 N n m x m ξ n ,
N ( x , ξ ) = n = 0 m = 0 N n m x m ξ n .
x = ( x - h ) cos θ + ( Y - k ) sin θ , Y = - ( x - h ) sin θ + ( Y - k ) cos θ .
N = n = 0 m = 0 { [ - ( x - h ) sin θ + ( Y - k ) cos θ ] 2 + Z 2 } n × { ( x - h ) cos θ + ( Y - k ) sin θ } m N n m .
( N x Y ˙ - N Y ) ( 1 + Y ˙ 2 + Z ˙ 2 ) + N Y ¨ = 0 , ( N x Z ˙ - N Z ) ( 1 + Y ˙ 2 + Z ˙ 2 ) + N Z ¨ = 0.
Y = j = 1 A j x j - 1 ,             Z = j = 1 B j x j - 1 .
N X = j = 1 G j x j - 1 , N Y = j = 1 E j x j - 1 , N Z = j = 1 F j x j - 1 , N = j = 1 D j x j - 1 .
A k + 2 = - 1 k ( k + 1 ) D 1 ( m = 1 k m A m + 1 n = 1 k + 1 - m G k + 2 - m - n H n - m = 1 k E m H k + 1 - m + m = 1 k - 1 m ( m + 1 ) A m + 2 D k + 1 - m ) , B k + 2 = - 1 k ( k + 1 ) D 1 ( m = 1 k m B m + 1 n = 1 k + 1 - m G k + 2 - m - n H n - m = 1 k F m H k + 1 - m + m = 1 k - 1 m ( m + 1 ) B m + 2 D k + 1 - m ) .
N = n = 0 N ξ n m = 0 M ζ m N n m .
ζ = j = 1 d 1 , j x j - 1 ,
d 1 , 1 = - h cos θ + A 1 sin θ - k sin θ , d 1 , 2 = cos θ + A 2 sin θ , d 1 , n = A n sin θ             for n 3.
ξ = j = 1 c 1 , j x j - 1 ,
c 1 , j = I = 1 j ( A l A j + 1 - l + B l B j + 1 - l ) ,
A 1 = ( A 1 - k ) cos θ + h sin θ , A 2 = A 2 cos θ - sin θ , A n = A n cos θ ,             n 3.
ζ m = j = 1 d m , j x j - 1 , ξ n = j = 1 c n , j x j - 1 .
d m , j = i = 1 j d 1 , i d m - 1 , j + 1 - i ,
c n , j = i = 1 j c 1 , i c n - 1 , j + 1 - i .
D j = n = 0 N i = 1 j m = 0 M N n m d m , i c n , j + 1 - i .
G j = - 2 sin θ l = 1 j A j + 1 - l n = 1 N i = 1 l m = 0 M n N n m d m , i c n - 1 , l + 1 - i + cos θ n = 0 N i = 1 j m = 1 M m N n m d m - 1 , i c n , j + 1 - i ,
E j = 2 cos θ l = 1 j A j + 1 - l n = 1 N i = 1 l m = 0 M n N n m d m , i c n - 1 , l + 1 - i + sin θ n = 0 N i = 1 j m = 1 M m N n m d m - 1 , i c n , j + 1 - i ,
F j = 2 l = 1 j B j + 1 - l n = 1 N i = 1 l m = 0 M n N n m d m , i c n - 1 , l + 1 - i .
N ( x , ξ ) = N 00 + N 02 ξ 2 + N 03 ξ 3 .
N = 1.605509 - 0.040 x + 0.029481 x 2 + 0.001779 x 3 .
0.9928769 i ¯ - 0.1191448 j ¯ ,
0.9992035 i ¯ - 0.0399044 j ¯ .
0.9957504 i ¯ - 0.0920930 j ¯ .
0.9995183 i ¯ + 0.03103549 j ¯ ,
0.9874026 i ¯ - 0.1582281 j ¯ .