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Enhanced depth of field integral imaging with sensor resolution constraints

Raúl Martínez-Cuenca, Genaro Saavedra, Manuel Martínez-Corral, and Bahram Javidi

Open Access Open Access

Abstract

One of the main challenges in integral imaging is to overcome the limited depth of field. Although it is widely assumed that such limitation is mainly imposed by diffraction due to lenslet imaging, we show that the most restricting factor is the pixelated structure of the sensor (CCD). In this context, we demonstrate that by proper reduction of the fill factor of pickup microlenses, the depth of field can be substantially improved with no deterioration of lateral resolution.

©2004 Optical Society of America

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Supplementary Material (2)

Media 1: GIF (333 KB)     
Media 2: GIF (632 KB)     

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Figures (6)
Fig. 1.
Fig. 1. Scheme, not to scale, of the capture setup of a 3D IIS. The field lens collects the rays from the outermost microlenses, the camera lens projects the images onto the CCD.

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Fig. 2.
Fig. 2. (a) Grey-scale representation of Eq. (4). Any cross-sections correspond to the spot produced by an object point at a depth z. White lines delimit the back-projected pixel size. The effect of defocus is much more appreciable for positive values of z; (b) Spot diameter for different values of the fill factor. The black thick line is used to mark the back-projected pixel size.

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Fig. 3.
Fig. 3. (a) Synthetic object; (b) 2D elemental images captured from 49 views; (c) Enlarged view of the central image. The object was placed at z=0 and the fill factor was set at ϕ/p=1.0

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Fig. 4.
Fig. 4. (a) 2D elemental images of the object captured from 49 different views; (b) Enlarged view of the central image. The object was placed at z=0 and the fill factor was set at ϕ/p=0.5.

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Fig. 5.
Fig. 5. Central elemental image as the object is displaced from z=0 to z=67.5 mm. Left-hand image corresponds to ϕ/p=0.5. Right-hand one to ϕ/p=1.0 (Video file of 0.33 Mb).

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Fig. 6.
Fig. 6. Evolution of the reconstructed image as the object is displace from z=0 to z=67.5 mm. Left-hand image corresponds to ϕ/p=0.5. Right-hand one to ϕ/p=1.0 (Video file of 0.63 Mb).

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Equations (5)

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H ( x ' ; x , z ) = ∣ ∑ m exp { i π ∣ m p − x ∣ 2 λ ( a − z ) } ∫ P z ( x o ) exp { − i 2 π x o x ' + [ M z ( m p − x ) − m p ] λ g d 2 x o } ∣ 2 ,
P z ( x o ) = p ( x o ) exp { i π z λ a ( a − z ) ∣ x o ∣ 2 } .
H ( x ' , x , z ) = ∣ P ˜ z ( x ' λ g ) ∣ 2 ⊗ ∑ m δ { x ' − [ m p ( 1 − M z ) − M z x ] } .
H o r z = ∣ ∫ 0 ϕ 2 p ( r o ) exp { i π λ z a ( a − z ) r o 2 } J o ( 2 π r r o λ g ) r o d r o ∣ 2 ,
∫ 0 D 2 H o r z r dr = 0.84 ∫ 0 ∞ H o r z r dr .
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