Abstract

This paper describes two methods for obtaining three-dimensional holographic images containing superimposed constant-range contours for accurate cross-section tracing or contour mapping. The first and simplest method requires two illuminating sources with identical wavelength but at slightly different positions. This method has the drawback of requiring the hologram to be made at right-angles to the illumination direction. This results in shadowed areas on the object. The second method requires an illuminating source containing two wavelengths. This method also requires precise positioning of two reference sources, one for each wavelength. The positions of the reference sources may be optimized so that contours will appear over a maximum object angle. Experimental results that prove the principle are shown.

© 1967 Optical Society of America

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References

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  1. E. N. Leith and J. Upatnieks, J. Opt. Soc. Am. 52, 1123 (1962); J. Opt. Soc. Am. 53, 1377 (1963); J. Opt. Soc. Am. 54, 1295 (1964).
    [Crossref]
  2. K. A. Haines and B. P. Hildebrand, Phys. Letters 19, 10 (1965); Phys. Letters 20, 422 (1966).
    [Crossref]
  3. E. N. Leith, J. Upatnieks, and K. A. Haines, J. Opt. Soc. Am. 55, 981 (1965).
    [Crossref]
  4. R. W. Meier, J. Opt. Soc. Am. 55, 987 (1965).
    [Crossref]
  5. K. A. Haines and B. P. Hildebrand, Appl. Opt. 5, 595 (1966).
    [Crossref] [PubMed]

1966 (1)

1965 (3)

1962 (1)

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Figures (14)

Fig. 1
Fig. 1

Geometry for the hologram recording process.

Fig. 2
Fig. 2

Geometry for the reconstruction process.

Fig. 3
Fig. 3

Loci of interference for the two-source hologram.

Fig. 4
Fig. 4

Illustration of lateral image-point displacement in the two-frequency hologram.

Fig. 5
Fig. 5

Loci of interference lines for the two-frequency hologram.

Fig. 6
Fig. 6

Experimental arrangement for recording contour holograms using the two-source method.

Fig. 7
Fig. 7

Photograph of the true image of a contour hologram made by the two-source method.

Fig. 8
Fig. 8

Experimental arrangement for the demonstration of a contour hologram using the two-frequency method.

Fig. 9
Fig. 9

Photograph of the true image of a contour hologram made by the two-frequency method—object, a sphere.

Fig. 10
Fig. 10

Photograph of the true image of a contour hologram made by the two-frequency method—object, a coin.

Fig. 11
Fig. 11

Photograph of the coin without superimposed contours.

Fig. 12
Fig. 12

Photograph of the true image of a contour hologram made by the two-frequency method—object, a coin with the reference sources positioned for optimum object angle.

Fig. 13
Fig. 13

Geometry for calculation of image-point displacement.

Fig. 14
Fig. 14

Graph of angular image displacement as a function of object angle.

Equations (41)

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y b = ± b r b ( y 1 r 1 - y 2 r 2 y a b r a )
x b = ± b r b ( x 1 r 1 - x 2 r 2 x a b r a )
r b = ± 1 b ( 1 r 1 - 1 r 2 1 b r a ) - 1 ,
Lateral magnification M L = x b x 1 = y b y 1 = ± b r b r 1 .
Radial magnification M r = r b r 1 = ± b r b 2 r 1 2 = 1 b M L 2 .
Angular magnification M α = α b α 1 = b sin α 1 sin α b ,
θ = k 1 ( r 0 + r 1 - r 2 ) + k 2 ( r a - r b ) ,
s 1 = K s exp ( i k 1 r 01 ) s 2 = K s exp ( i k 1 r 02 ) .
I = 2 K 2 s 2 ( 1 + cos k 1 [ r 01 - r 02 ] ) .
k 1 ( r 01 - r 02 ) = 2 n π
r 01 - r 02 = n λ 1 .
Δ h = λ 1 / 2 sin ( Δ γ / 2 ) ,
tan γ = x b 2 - x b 1 z b 2 - z b 1 .
tan γ = C 1 tan α 1 + P ,
C 1 = M L 2 - M L 1 ( M L 2 / M α 2 - M L 1 / M α 1 )
P = [ ( M L 2 M α 2 - M L 1 M α 1 ) sin α 1 ] - 1 × { x 2 r 2 ( M L 2 - M L 1 ) + x a r a ( M L 2 b 2 - M L 1 b 1 ) } .
x 22 r 22 = λ 12 λ 11 x 21 r 21 - Δ λ 1 λ 11 x 1 r 1 .
θ 2 - θ 1 + k 2 r Δ b ,
θ 2 - θ 1 + k 2 Δ r b = 2 n π
r 0 + r 1 = ( n λ 11 λ 12 / Δ λ 1 ) + C 2 ,
C 2 = ( λ 12 r 21 - λ 11 r 22 ) / Δ λ 1 .
Δ r 1 = λ 11 λ 12 / 2 Δ λ 1 .
R = ( z b 1 x b 2 - z b 2 x b 1 ) / ( x b 1 - x b 2 ) .
R = sin ( α b 1 - α b 2 ) ( cos α b 1 / r b 2 ) - ( cos α b 2 / r b 1 ) ,
α b = cos - 1 ( x b / r b ) .
x 21 / r 21 = ± x a / b 1 r a
x 22 / r 22 = ± x a / b 2 r a .
R = r ˆ a sin ( α b 1 - α b 2 ) ( b 2 - b 1 ) cos α 1
r ˆ a = r a / [ 1 ± r a ( b 1 b 2 b 2 - b 1 ) ( 1 r 21 - 1 r 22 ) ]
α b 1 = cos - 1 ( b 1 cos α 1 )
α b 2 = cos - 1 ( b 2 cos α 1 ) .
δ Δ ,
= γ - η
Δ = Δ r b sec ( γ + α b 2 ) , η = tan - 1 { x b 2 / ( z b 2 + R 0 ) } , γ = tan - 1 { x b 2 / ( z b 2 + R ) } .
D λ 11 / ( λ 12 - λ 11 D q ) ,
q = ( λ 12 / λ 11 r 21 ± 1 / r 22 ) .
Δ = Δ r b sec ( γ + α 1 ) , η = tan - 1 { cos α 1 sin α 1 + r ˆ a / D sin α 10 } , γ = tan - 1 { cos α 1 sin α 1 + r ˆ a / D sin α 1 } .
Δ r b = D ( Δ λ 1 - q λ 11 D ) / ( λ 12 - q λ 11 D ) .
δ = { ( γ - η ) / [ γ - ( π / 2 - α 1 ) ] } Δ r b ,
γ ν 1 / [ 1.5 + r ˆ a / D - 1 2 ( π / 2 - α 1 ) 2 ] , η ν 1 / [ 1.5 + r ˆ a / D - 1 2 ( π / 2 - α 1 ) ( π / 2 - α 10 ) ] .
( π / 2 - α 1 ) 4 - 2 ( π / 2 - α 1 ) 3 ( π / 2 - α 10 ) + 4 ( π / 2 - α 1 ) 2 × ( 1 + r ˆ a / D ) - ( π / 2 - α 1 ) [ 6 + 16 r ˆ a / D + 8 ( r ˆ a / D ) ] / ( π / 2 - α 10 ) + [ 3 + 8 r ˆ a / D + 4 ( r ˆ a / D ) 2 ] = 0.