Abstract

We present a method for single-image passive ranging and three-dimensional (3D) imaging in incoherent light based on chiral phase coding. A chiral linear phase variation across the aperture of an optical system results in a frequency response with a characteristic pattern of fringes such that the spatial period and inclination of the pattern depend on the focusing error. From this dependency, the absolute focusing error and, hence, the distance to the object can be found. In the experiments a resolution of 1.4μm is achieved with a 20mm aperture lens in a 4mm interval at a distance of 140mm from the lens. A resolution of 0.7mm is obtained at a distance of 11m with the range finder employing two 25.4mm spherical mirrors spaced apart by 140mm. We also demonstrate 3D imaging of weakly textured objects.

© 2011 Optical Society of America

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References

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R. J. Pieper and T. -C. Poon, J. Mod. Opt. 37, 2055 (1990).
[CrossRef]

Appl. Opt. (2)

J. Mod. Opt. (1)

R. J. Pieper and T. -C. Poon, J. Mod. Opt. 37, 2055 (1990).
[CrossRef]

J. Opt. A. (1)

Z. Zalevsky and A. Zlotnik, J. Opt. A. 10, 064014 (2008).
[CrossRef]

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Opt. Lett. (1)

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Figures (3)

Fig. 1
Fig. 1

(a) Chiral optical mask with two rectangular openings: 1, flat optical window; 2, wedge prism. Optical response of the mask: (b) in-focus and (c) out-of-focus MTFs.

Fig. 2
Fig. 2

(a) Ranging at 140 mm with a single-aperture range finder using a circular chiral mask, 1. (b) Ranging at 11 m with a two-aperture range finder: 2, distant object; 3, 4, spherical mirrors, one mirror is tilted out of plane of the arrangement; and 5, camera. The inset in (a) shows a data portion acquired with a 5 μm step.

Fig. 3
Fig. 3

Passive 3D imaging with a single- aperture range finder: (a) an original weakly textured scene and (b) the corresponding depth map; (c) a prism on a weakly textured background and (d) its depth map.

Equations (8)

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φ = π D 2 4 λ ( 1 f 1 z o 1 z i ) ,
H ( ω x , ω y , φ ) = H 0 ( ω x , ω y , φ ) cos Ψ ( + ) exp ( i Ψ ( ) ) ,
Ψ ( ± ) = ( A ω x + B ω y ) / 2 + ξ φ ω x ( d ± c )
H 0 ( ω x , ω y , φ ) = sin [ ξ φ ω x ( a | ω x | ) ] ξ φ ω x a sin [ η φ ω y ( b | ω y | ) ] η φ ω y b
Λ = 2 π { [ A / 2 + ξ φ ( c + d ) ] 2 + B 2 / 4 } 1 / 2 ,
α = arctan [ A / B + 2 ξ φ ( c + d ) / B ] ,
S ( φ ) = | I ˜ ( ω x , ω y ) | cos ( 2 Ψ ( + ) ) d ω x d ω y .
C ( u , v ) = I ( x + u / 2 , y + v / 2 ) I ( x u / 2 , y v / 2 ) d x d y .

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