Abstract

Analysis is made of the effect of omitting the compensating plate in a corner-reflector interferometer. It is concluded that a noncompensated instrument has disadvantages when compared to a compensated one, but when properly adjusted it may still be useable with a photoelectric detector. The flux is integrated over rectangular field stops.

© 1957 Optical Society of America

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References

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  1. E. R. Peck and S. W. Obetz, J. Opt. Soc. Am. 43, 505 (1953).
    [Crossref] [PubMed]
  2. E. R. Peck, J. Opt. Soc. Am. 46, 380(A) (1956).
  3. E. R. Peck, J. Opt. Soc. Am. 38, 1015 (1948).
    [Crossref] [PubMed]
  4. E. R. Peck, J. Opt. Soc. Am. 45, 931 (1955).
    [Crossref]

1956 (1)

E. R. Peck, J. Opt. Soc. Am. 46, 380(A) (1956).

1955 (1)

1953 (1)

1948 (1)

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Figures (5)

Fig. 1
Fig. 1

Corner-reflector interferometer. P1, P2: corner reflectors. D: dividing plate. C: location of compensating plate which is here omitted.

Fig. 2
Fig. 2

Diagram of conventions used in the analysis. C1, C2: optical centers of the corner reflectors. N′, F: second nodal point and focal plane of ideal projection lens of unit focal length.

Fig. 3
Fig. 3

Unit propagation vectors for refraction of light through the dividing plate.

Fig. 4
Fig. 4

Greatest spread in order of interference over a circular field stop vs carriage position.

Fig. 5
Fig. 5

Interference fringes and rectangular field stop. ΔNa, ΔNb: order differences between the optic axis and the x and y intercepts of the field stop.

Equations (17)

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t ( μ p · p - p · p p · p ) .
δ = 2 t { [ μ 2 - 1 + ( p · p ) 2 ] 1 2 - p · p } .
δ = k 1 + k 2 x + k 4 y 2 + k 5 x 2 + k 8 x 3 + k 9 x y 2
k 1 = 2 t ( σ - cos κ ) , k 2 = 2 t ( - 1 + σ - 1 cos κ ) sin κ , k 4 = t ( 1 - σ - 1 cos κ ) cos κ , k 5 = t [ cos κ - σ - 1 cos 2 κ + σ - 3 ( μ 2 - 1 ) sin 2 κ ] , k 8 = t [ 1 - 2 σ - 1 cos κ + σ - 3 cos 3 κ - σ - 5 ( μ 2 - 1 ) sin 2 κ cos κ ] sin κ , k 9 = t ( 1 - 2 σ - 1 cos κ + σ - 3 cos 3 κ ) sin κ .
δ = a 1 + a 2 x + a 4 y 2 + a 5 x 2 + a 8 x 3 + a 9 x y 2 ,
a 1 = k 1 + 2 ζ a 5 = k 5 - ζ a 2 = k 2 + 2 ξ a 8 = k 8 - ξ a 4 = k 4 - ζ a 9 = k 9 - ξ .
ξ = - 1 2 k 2 = t ( 1 - σ - 1 cos κ ) sin κ .
N = a 1 λ - 1 + a 4 λ - 1 y 2 + a 5 λ - 1 x 2 .
( k 5 - k 4 ) λ - 1 θ r 2 .
Φ = B A 2 λ a 4 a 5 - 1 2 C p - 1 [ 1 + p cos ( 2 π N 0 - ϕ ) ]
Δ N a = a 5 λ - 1 a 2 f - 2 ;             Δ N b = a 4 λ - 1 b 2 f - 2 .
C = [ ( F 1 a 2 + F 2 a 2 ) ( F 1 b 2 + F 2 b 2 ) ] 1 2 ,
p = 1 4 C Δ N a Δ N b - 1 2 ,
ϕ = - sin - 1 a 4 a 4 - 1 F 1 a F 2 b + a 5 a 5 - 1 F 1 b F 2 a C = - tan - 1 a 4 a 4 - 1 F 1 a F 2 b + a 5 a 5 - 1 F 1 b F 2 a F 1 a F 1 b - a 4 a 5 a 4 a 5 - 1 F 2 a F 2 b .
u a = 2 Δ N a 1 2 ;             u b = 2 Δ N b 1 2 .
ζ 0 = ( a 2 k 5 + b 2 k 4 ) ( a 2 + b 2 ) - 1 .
ϕ = 1 3 π ( a 2 + b 2 ) f - 2 2 ( ζ - ζ 0 ) λ - 1 .