Abstract

We describe how optical system defects (tilt/jitter, decenter, and despace) propagate through an arbitrary paraxial optical system that can be described by an ABCD ray transfer matrix. A pedagogical example is given that demonstrates the effect of alignment errors on a typical optical system.

© 1988 Optical Society of America

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References

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  1. H. T. Yura, S. G. Hanson, J. Opt. Soc. Am. A 4, 1931 (1987).
    [CrossRef]
  2. C. A. Massey, A. E. Siegman, Appl. Opt. 8, 875 (1968).
  3. R. Kingslake, Lens Design Fundamentals (Academic, New York, 1978), Chap. 2.
  4. A. E. Siegman, Lasers (Oxford U. Press, New York, 1986), Chap. 5.
  5. S. A. Collins, J. Opt. Soc. Am. 60, 1168 (1970).
    [CrossRef]

1987 (1)

1970 (1)

1968 (1)

C. A. Massey, A. E. Siegman, Appl. Opt. 8, 875 (1968).

Collins, S. A.

Hanson, S. G.

Kingslake, R.

R. Kingslake, Lens Design Fundamentals (Academic, New York, 1978), Chap. 2.

Massey, C. A.

C. A. Massey, A. E. Siegman, Appl. Opt. 8, 875 (1968).

Siegman, A. E.

C. A. Massey, A. E. Siegman, Appl. Opt. 8, 875 (1968).

A. E. Siegman, Lasers (Oxford U. Press, New York, 1986), Chap. 5.

Yura, H. T.

Appl. Opt. (1)

C. A. Massey, A. E. Siegman, Appl. Opt. 8, 875 (1968).

J. Opt. Soc. Am. (1)

J. Opt. Soc. Am. A (1)

Other (2)

R. Kingslake, Lens Design Fundamentals (Academic, New York, 1978), Chap. 2.

A. E. Siegman, Lasers (Oxford U. Press, New York, 1986), Chap. 5.

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Figures (2)

Fig. 1
Fig. 1

The example system without alignment errors.

Fig. 2
Fig. 2

The example system with alignment errors: s, despace; h, decenter; and θ, ray tilt.

Equations (12)

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θ j = θ j + δ θ j = [ 0 θ j ] + [ 0 δ θ j ] .
θ j = [ h j θ j ] + [ δ h j δ θ j ] ,
r f + = A j + B j ( r j ) + h j ,
T j = [ h j θ j ] ,
t = j = 1 n ( A j h j + B j θ j ) ,
t = j = 1 n ( C j h j + D j θ j ) ,
y 7 = [ 0 f 3 / m m ϕ 3 0 ] y 1 ,
y 7 = M 3 M 2 S M 1 y 1 + M 3 M 2 [ h 0 ] + M 3 [ 0 θ ] ,
M i = [ 0 f i ϕ i 0 ] , i = 1 , 2 , 3
S = [ 1 s 0 1 ] ,
u ( r 2 ) = i k m 2 π f 3 exp [ i k ϕ 2 f 3 h ( i ˆ · r 2 ) ] × d 2 r 1 u i ( r 1 ) exp ( i k m ϕ 1 ϕ 2 f 3 s r 1 2 / 2 f 3 ) × exp [ i k m ( f 3 θ i ˆ + r 2 ) · r 1 ] ,
y + d y = ( M + d M ) y + M ( y + d y ) .

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