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  1. R. C. Spencer, G. Hyde, J. Opt. Soc. Amer. 54, 1388A (1965).
  2. R. C. Spencer, paper presented at Colloque d’Optique, Quebec1967 [see Appl. Opt. 7, 757 (1968) for a report of this meeting].
  3. R. C. Spencer, G. Hyde, IEEE Trans. AP-16, 317 (1968).
    [CrossRef]

1968 (1)

R. C. Spencer, G. Hyde, IEEE Trans. AP-16, 317 (1968).
[CrossRef]

1965 (1)

R. C. Spencer, G. Hyde, J. Opt. Soc. Amer. 54, 1388A (1965).

Hyde, G.

R. C. Spencer, G. Hyde, IEEE Trans. AP-16, 317 (1968).
[CrossRef]

R. C. Spencer, G. Hyde, J. Opt. Soc. Amer. 54, 1388A (1965).

Spencer, R. C.

R. C. Spencer, G. Hyde, IEEE Trans. AP-16, 317 (1968).
[CrossRef]

R. C. Spencer, G. Hyde, J. Opt. Soc. Amer. 54, 1388A (1965).

R. C. Spencer, paper presented at Colloque d’Optique, Quebec1967 [see Appl. Opt. 7, 757 (1968) for a report of this meeting].

IEEE Trans. (1)

R. C. Spencer, G. Hyde, IEEE Trans. AP-16, 317 (1968).
[CrossRef]

J. Opt. Soc. Amer. (1)

R. C. Spencer, G. Hyde, J. Opt. Soc. Amer. 54, 1388A (1965).

Other (1)

R. C. Spencer, paper presented at Colloque d’Optique, Quebec1967 [see Appl. Opt. 7, 757 (1968) for a report of this meeting].

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Figures (2)

Fig. 1
Fig. 1

Reflection from a spherical mirror.

Fig. 2
Fig. 2

Caustic region—circle of least confusion CC′.

Tables (1)

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Table I Properties of Circle of Least Confusion

Equations (22)

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x = h 3 ,
4 z 2 = 1 + 3 h 2 - 4 h 6 ,
4 r 2 = 1 + 3 h 2 .
C C = 2 x = 2 h 3 ,
C 1 C 1 = 2 x 1 = 2 h 1 3 ,
C C = 2 x = 2 h 2 .
4 O C ¯ 1 2 = 1 + 3 h 1 2 ,
4 O F ¯ 1 2 = sec 2 α = 1 + t 2 .
h 1 2 = t 2 / 3.
C 1 C 1 = 2 t 3 / ( 27 ) 1 2 .
h = σ t / 2
z sin 2 α + x cos 2 α = h .
z 2 sin 2 2 α = h 2 - 2 h x cos 2 α + x 2 cos 2 2 α .
( 1 + 3 h 2 ) ( sin 2 2 α ) / 4 = h 2 - 2 h x cos 2 α + x 2 .
1 + 3 h 2 = 1 + t 2 - 2 x t ( 1 - t 2 ) ( 1 + t 2 ) 1 2 + x 2 t 2 ( 1 + t 2 ) 2 .
3 σ 2 = 4 - σ 3 ( 1 - t 2 ) ( 1 + t 2 ) 1 2 + σ 0 t 2 ( 1 + t 2 ) 2 / 16.
( σ - 1 ) ( 2 + σ ) 2 = σ 3 [ 1 - ( 1 - t 2 ) ( 1 + t 2 ) 1 2 ] + σ 6 [ t 2 ( 1 + t 2 ) 2 / 16 ] = σ 3 T 1 ( t ) + σ 6 T 2 ( t ) .
= ( σ - 1 ) = σ 3 ( T 1 + σ 3 T 2 ) ( 2 + σ ) 2 .
C 0 = C 1 + C 2 2 + , C 0 = ( T 1 + T 2 ) , C 1 = 3 ( 3 - T 1 - 2 T 2 ) , C 2 = 3 ( 2 - T 1 - 5 T 2 ) .
= C 0 / ( C 1 + C 2 ) .
Δ σ = ( 2 + σ a ) 2 3 σ a [ ( 2 + σ a ) - σ a ( T 1 + 2 σ a 3 T 2 ) ] .
T 1 = 1 - ( 1 - 2 h 2 ) ( 1 - h 2 ) 3 / 2 ,             T 2 = h 2 16 ( 1 - h 2 ) 3 .

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