Abstract

An algebraic formula is derived for the fifth-order distortion of a GRIN-rod lens. By measuring distortion, one can then determine the coefficients of the profile formula.

© 1983 Optical Society of America

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References

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  1. N. Yamamoto, K. Iga, Appl. Opt. 19, 1101 (1980).
    [CrossRef] [PubMed]
  2. E. W. Marchand, Gradient Index Optics (Academic, New York, 1978).
  3. K. Iga, K. Kokomori, T. Sakayori, Appl. Phys. Lett. 26, 573 (1975).
    [CrossRef]
  4. W. Streifer, K. B. Paxton, Appl. Opt. 10, 769 (1971).
    [CrossRef] [PubMed]
  5. J. Rees, Appl. Opt. 21, 1009 (1982).
    [CrossRef] [PubMed]

1982 (1)

1980 (1)

1975 (1)

K. Iga, K. Kokomori, T. Sakayori, Appl. Phys. Lett. 26, 573 (1975).
[CrossRef]

1971 (1)

Iga, K.

N. Yamamoto, K. Iga, Appl. Opt. 19, 1101 (1980).
[CrossRef] [PubMed]

K. Iga, K. Kokomori, T. Sakayori, Appl. Phys. Lett. 26, 573 (1975).
[CrossRef]

Kokomori, K.

K. Iga, K. Kokomori, T. Sakayori, Appl. Phys. Lett. 26, 573 (1975).
[CrossRef]

Marchand, E. W.

E. W. Marchand, Gradient Index Optics (Academic, New York, 1978).

Paxton, K. B.

Rees, J.

Sakayori, T.

K. Iga, K. Kokomori, T. Sakayori, Appl. Phys. Lett. 26, 573 (1975).
[CrossRef]

Streifer, W.

Yamamoto, N.

Appl. Opt. (3)

Appl. Phys. Lett. (1)

K. Iga, K. Kokomori, T. Sakayori, Appl. Phys. Lett. 26, 573 (1975).
[CrossRef]

Other (1)

E. W. Marchand, Gradient Index Optics (Academic, New York, 1978).

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Figures (1)

Fig. 1
Fig. 1

Gradient-index rod and meridional ray.

Tables (1)

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Table I Computation of Distortion

Equations (39)

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n 2 = N 0 2 ( 1 Y 2 + h 4 Y 4 + h 6 Y 6 ) , Y = g r , r 2 = x 2 + y 2 ,
t 0 = B g , c 0 = cos t 0 , s 0 = sin t 0 ,
f = 1 / N 0 g s 0 , f 1 = c 0 f ,
m = c 0 C / f , 1 / m = c 0 A / f , ( c 0 A / f ) ( c 0 C / f ) = 1.
D = ( y ˆ m h ) / m h .
y ¨ = ( n n / y ) / l 2 , n = f ( r ) ,
n 0 = N 0 , l = N 0 cos θ 0 .
y ¨ = ( N 0 g / l ) 2 y ( 1 + 2 h 4 Y 2 + 3 h 6 Y 4 ) .
Y + Y = 2 h 4 Y 3 + 3 h 6 Y 5 , Y = g y , t = z ( N 0 g / l ) ,
Y = b 0 sin t , b 0 = sin θ 0 .
Y = q / N 0 , q = n sin θ ,
Y + Y = 2 h 4 b 0 3 sin 3 t .
Y + Y = f ( t ) ,
Y p = 0 t f ( x ) sin ( t x ) d x ,
Y = b 0 s + ( 3 / 4 ) h 4 b 0 3 ( s c t s 3 / 3 ) , c = cos t , s = sin t ,
Y = b 0 s + N 1 ( s c t s 3 / 3 ) + N 2 [ 7 s 3 15 s + c t ( 11 + 4 c 2 ) + 2 s t 2 ] + N 3 [ ( 2 / 5 ) s 5 + s 3 3 ( s c t ) ]
Y = q / N 0 = b 0 c + N 1 s ( t c s ) + N 2 [ 17 c s 2 + s t ( 12 s 2 19 ) + 2 c t 2 ] + N 3 [ c s 2 ( 3 + 2 s 2 ) 3 s t ] ,
N 1 = ( 3 / 4 ) h 4 b 0 3 , N 2 = ( 9 / 64 ) h 4 2 b 0 5 N 3 = ( 1 / 16 ) ( h 6 ( 1 / 2 ) h 4 2 ) b 0 5 .
y ˆ = y + C tan θ ,
sin θ = q , tan θ = q / ( 1 q 2 ) 1 / 2 ,
g y ˆ = Y + C g [ q + ( 1 / 2 ) q 3 + ( 3 / 8 ) q 5 ] , q = N 0 Y
c = cos t , s = sin t , t = B ( N 0 g / l ) .
l = N 0 cos θ 0 , N 0 / l = 1 / ( 1 b 0 2 ) 1 / 2 .
t = ( B g ) [ 1 + 1 2 b 0 2 + ( 3 / 8 ) b 0 4 ]
t 0 = B g , t 2 = 1 2 t 0 b 0 2 , t 4 = ( 3 / 8 ) t 0 b 0 4
c = cos ( t 0 + t 2 + t 4 ) , s = sin ( t 0 + t 2 + t 4 ) .
c = c 0 ( 1 1 2 t 2 2 ) s 0 ( t 2 + t 4 ) , s = s 0 ( 1 1 2 t 2 2 ) + c 0 ( t 2 + t 4 ) , c 0 = cos t 0 , s 0 = sin t 0 .
g m h = b 0 ( s 0 + N 0 g C c 0 ) .
D = b 0 2 s + c C 1 ( K 0 + h 4 K 1 + h 4 2 K 2 + h 6 K 3 ) C 1 = C g N 0 , b 0 = sin θ 0 , t = B g , c = cos t , s = sin t .
K 0 = A 0 + C 1 [ B 0 + 1 2 N 0 2 c 3 + ( 3 / 8 ) N 0 4 b 0 2 c 5 ] , K 1 = ( 3 / 4 ) { A 1 + C 1 B 1 [ 1 ( 3 / 2 ) ( b 0 N 0 ) 2 ] } , K 2 = ( b 0 2 / 32 ) [ A 3 + C 1 B 3 ( 9 / 2 ) ( A 2 + C 1 B 2 ) ] , K 3 = ( b 0 2 / 16 ) ( A 3 + C 1 B 3 ) ,
A 0 = ( 1 / 8 ) t [ 4 c + b 0 2 ( 3 c s t ) ] , A 1 = s c t s 3 / 3 + s t 2 ( t c s ) , t 2 = 1 2 t b 0 2 , A 2 = 7 s 3 15 s + c t ( 11 + 4 c 2 ) , A 3 = ( 2 / 5 ) s 5 + s 3 3 s + 3 c t ,
B 0 = ( 1 / 8 ) t [ 4 s + b 0 2 ( 3 s + c t ) ] , B 1 = s t c s 2 + t 2 ( c t s + 3 s 3 ) , B 2 = 17 c s 2 + t ( 12 s 3 19 s ) + 2 c t 2 , B 3 = c s 2 ( 3 + 2 s 2 ) 3 s t .
a K 1 + a 2 K 2 + b K 3 = K ¯ 0 , a = h 4 , b = h 6
E = ( a K 1 + a 2 K 2 + b K 3 K ¯ 0 ) 2 ,
E / a = E / b = 0 ,
( K 1 + 2 a K 2 ) ( a K 1 + a 2 K 2 + b K 3 K ¯ 0 ) = 0 , [ K 3 ( a K 1 + a 2 K 2 + b K 3 K ¯ 0 ) ] = 0 .
a K 1 2 + 3 a 2 K 1 K 2 + b K 1 K 3 + 2 a 3 K 2 2 + 2 a b K 2 K 3 + 2 a K 0 K 2 = K ¯ 0 K 1 , a K 1 K 3 + a 2 K 2 K 3 + b K 3 2 = K ¯ 0 K 3 .
h 4 = a , h b = b .
B = C = 10 , N 0 = 1.525 , g = 2 π / 67 , h 4 = 2 / 3 0.67 , h 6 = 17 / 45 0.38.

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