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Numerical reconstruction of digital holograms with variable viewing angles

Lingfeng Yu, Yingfei An, and Lilong Cai

Open Access Open Access

Abstract

Here we describe a new method for numerically reconstructing an object with variable viewing angles from its hologram(s) within the Fresnel domain. The proposed algorithm can render the real image of the original object not only with different focal lengths but also with changed viewing angles. Some representative simulation results and demonstrations are presented to verify the effectiveness of the algorithm.

©2002 Optical Society of America

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Supplementary Material (3)

Media 1: GIF (145 KB)     
Media 2: GIF (1584 KB)     
Media 3: GIF (1607 KB)     

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Figures (7)
Fig. 1.
Fig. 1. Reconstruction with changed viewing angles.

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Fig. 2.
Fig. 2. (a) Recording and reconstructing a hologram; (b) Texture of the object.

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Fig. 3.
Fig. 3. (a) Magnitude; (b) phase of the object wave on the hologram.

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Fig. 4.
Fig. 4. (GIF, 145KB) A demo of reconstructing the real object with fixed li =1.2m but with changed viewing angles from -75° to 75°.

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Fig. 5.
Fig. 5. (a) Texture of layer 1 at 1.0m; (b) Texture of the layer 2 at 1.2m.

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Fig. 6.
Fig. 6. (a) Magnitude; (b) phase of the object wave on the hologram.

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Fig. 7.
Fig. 7. (a) (GIF, 1.54MB) A demo of reconstructing the real object with fixed viewing angle θ = 45° but with changed focal length between 1.0m and 1.2m; (b) (GIF, 1.56MB) A demonstration with both the viewing angle and the focal length adjusted.

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Equations (10)

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E ( x o , y o , z o ) = iE 0 λ ∫∫ τ x y exp ( iky sin θ ) exp [ ikr ( x , y , x o , y o ) ] r ( x , y , x o , y o ) χ ( x , y , x o , y o ) dxdy
r = ( z o − y sin θ ) 2 + ( x o − x ) 2 + ( y o − y cos θ ) 2
E ( x o , y o , z o ) = exp ( ikr o ) ∫∫ τ x y exp [ ik 2 r o ( x 2 + y 2 ) ]
× exp { − ik r o [ x o x + y o y cos θ + ( z o − r o ) y sin θ ] } dxdy
ik 2 r o ( x 2 + y 2 ) ≈ ik 2 z o ( x 2 + y 2 )
ξ = x o λr o
η = 1 λ r o [ y o cos θ + ( z o − r o ) sin θ ]
E ( ξ , η , z o ) = exp ( ikr o ) ∫ τ x y exp [ ik 2 z o ( x 2 + y 2 ) ] exp [ − i 2 π ( ξx + ηy ) ] dxdy
Δ f x o = 1 δx o ≅ Np λz o , and Δ f y o = 1 δy o ≅ Np cos θ λz o
P = α z o Δ f x o · α z o Δ f y o ≅ N 2 cos θ
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