Abstract

Sampled images of two-dimensional objects are obtained from holograms recorded by using achromatic-fringe interferometer arrangements in which the beam splitting is achieved by a hologram of a random phase-sampling pattern. The use of monochromatic but spatially incoherent source suppresses both speckle and moiré without reducing contrast of the fringes at the hologram plane. The principle was analyzed in the case of Fourier holograms and verified experimentally.

© 1974 Optical Society of America

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References

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    [CrossRef]

1973 (2)

Y. Tsunoda and Y. Takeda, J. Appl. Phys. 44, 2422 (1973).
[CrossRef]

M. Kato and Y. Okino, Appl. Opt. 12, 1199 (1973).
[CrossRef] [PubMed]

1972 (2)

1970 (2)

1969 (1)

1968 (2)

1967 (1)

1966 (1)

J. M. Burch, J. W. Gates, R. G. N. Hills, and L. H. Tanner, Nature 212, 1347 (1966).
[CrossRef]

Bryngdahl, O.

Burch, J. M.

J. M. Burch, J. W. Gates, R. G. N. Hills, and L. H. Tanner, Nature 212, 1347 (1966).
[CrossRef]

Burckhardt, C. B.

Firester, A. H.

Fox, E. C.

Gates, J. W.

J. M. Burch, J. W. Gates, R. G. N. Hills, and L. H. Tanner, Nature 212, 1347 (1966).
[CrossRef]

Gerritsen, H. J.

Hannan, W. J.

Hills, R. G. N.

J. M. Burch, J. W. Gates, R. G. N. Hills, and L. H. Tanner, Nature 212, 1347 (1966).
[CrossRef]

Kato, M.

Leith, E. N.

Lohmann, A.

Okino, Y.

Ramberg, E. G.

Stewart, W. C.

Suzuki, T.

Takeda, Y.

Y. Tsunoda and Y. Takeda, J. Appl. Phys. 44, 2422 (1973).
[CrossRef]

Y. Takeda, Japan. J. Appl. Phys. 11, 656 (1972).
[CrossRef]

Tanner, L. H.

J. M. Burch, J. W. Gates, R. G. N. Hills, and L. H. Tanner, Nature 212, 1347 (1966).
[CrossRef]

Tsunoda, Y.

Y. Tsunoda and Y. Takeda, J. Appl. Phys. 44, 2422 (1973).
[CrossRef]

Upatnieks, J.

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Figures (6)

Fig. 1
Fig. 1

Achromatic-fringe interferometer arrangement with a holographic beam splitter.

Fig. 2
Fig. 2

The optical system for recording the Fourier hologram of an array of sampling spots.

Fig. 3
Fig. 3

Reconstruction of a transparency. In recording the hologram a spatially incoherent extended source was used.

Fig. 4
Fig. 4

Reconstruction of the same transparency as in Fig. 3, but a point source was used.

Fig. 5
Fig. 5

Reconstruction of the same transparency as in Figs. 3 and 4. The phases 0 or π were distributed randomly to the sampling spots. The size of the hologram was about 3 mm × 3 mm with lens of 120 mm focal length.

Fig. 6
Fig. 6

The same result as in Fig. 5, but with 1.5 mm × 1.5 mm hologram.

Equations (19)

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u ( x ) = p = - N N δ ( x - p Δ ) exp ( i ϕ p ) + C δ ( x - l ) ,
t ( ξ ) = - u ( x ) exp ( i 2 π λ f x ξ ) d x = p = - N N exp ( i 2 π λ f p Δ ξ ) exp ( i ϕ p ) + C exp ( i 2 π λ f l ξ ) ,
T B ( ξ ) = t ( ξ ) t * ( ξ ) = C p = - N N exp ( i 2 π λ f ( p Δ - l ) ξ ) exp ( i ϕ p ) + C p = - N N exp ( - i 2 π λ f ( p Δ - l ) ξ ) exp ( - i ϕ p ) + | p = - N N exp ( i 2 π λ f p Δ ξ ) exp ( i ϕ p ) | 2 + C 2 .
T ( ξ ) = C p = - N N exp ( i 2 π λ f ( p Δ - l ) ξ ) exp ( i ϕ p ) .
- δ ( x s - x s ) exp ( i 2 π λ f x s ξ ) d x s = exp ( i 2 π λ f x s ξ ) .
u 0 ( x ) = - α α [ exp ( i 2 π λ f x s ξ ) T ( ξ ) ] exp ( i 2 π λ f x ξ ) d ξ , = C p - = N N exp ( i ϕ p ) × - α α exp ( i 2 π λ f ( x s + p Δ - l + x ) ξ ) d ξ ,
u R ( x ) = - α α C 2 exp ( i 2 π λ f x s ξ ) exp ( i 2 π λ f x ξ ) d ξ ,
u m ( ξ m ) = - [ T 0 ( x ) u 0 ( x ) ] exp ( i 2 π λ f x ξ m ) d x + - u R ( x ) exp ( i 2 π λ f x ξ m ) d x .
u 0 ( x ) C p = - N N δ ( x s + p Δ - l + x ) exp ( i ϕ p )
u R ( x ) C 2 δ ( x + x s ) .
u m ( ξ m ) C p = - N N - δ ( x s + p Δ - l + x ) × exp ( i ϕ p ) T 0 ( x ) exp ( i 2 π λ f x ξ m ) d x + C 2 - δ ( x + x s ) exp ( i 2 π λ f x ξ m ) d x = C p = - N N { [ T 0 ( l - p Δ - x s ) exp ( - i 2 π λ f x s ξ m ) ] × exp ( - i 2 π λ f ( p Δ - l ) ξ m ) exp ( i ϕ p ) } + C 2 exp ( - i 2 π λ f x s ξ m ) .
I ( ξ m ) = u m ( ξ m ) u * m ( ξ m ) C 3 p = - N N [ T 0 ( l - p Δ - x s ) × exp ( - i 2 π λ f ( p Δ - l ) ξ m ) exp ( i ϕ p ) ] + C 3 p = - N N [ T * 0 ( l - p Δ - x s ) × exp ( i 2 π λ f ( p Δ - l ) ξ m ) exp ( - i ϕ p ) ] + C 2 | p = - N N [ T 0 ( l - p Δ - x s ) × exp ( - i 2 π λ f ( p Δ - l ) ξ m ) exp ( i ϕ p ) ] | 2 + C 4 .
I 1 ( ξ m ) = C 3 p = - N N { [ - T 0 ( l - p Δ - x s ) d x s ] × exp ( - i 2 π λ f ( p Δ - l ) ξ m ) exp ( i ϕ p ) } .
2 = n Δ ,
d A 2.4 λ f / D H ,
Δ d A .
D H 2.4 λ f / Δ .
D L 4 a = 2 M Δ .
D H 4.8 λ f M / D L .