Abstract

To evaluate the Earth’s surface with a sensor on a satellite, it is important for one to use a correction factor such as atmospheric turbulence. The spatial coherence of light influenced by turbulence was measured in the laboratory by use of a simulator equipped with a multiple reflection system. The turbulence was generated by changing the ambient temperatures from 32 to 48 °C. It was found that the spatial coherence of light that passed through the turbulent region decreased when the temperature increased. Double-slit interference fringes were analyzed by a photographic method. The visibility of the interference fringes is the degree of coherence.

© 1999 Optical Society of America

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References

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    [CrossRef]
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    [CrossRef] [PubMed]

1994 (1)

1978 (1)

1974 (1)

1972 (1)

1971 (1)

1970 (1)

1968 (1)

1966 (1)

Andrews, L. C.

Beran, M. J.

Fante, R. L.

Fried, D. L.

Gamo, H.

Ho, T. L.

Lutomirski, R. T.

Majumdar, A. K.

Miller, W. B.

Ricklin, J. C.

Yura, H. T.

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Figures (8)

Fig. 1
Fig. 1

Degree of coherence of a collimated Gaussian beam for L/ kl m 2 = 1, C n 2 k 2 Lη m -5/3 = 0.025, where η m = 5.91/l m . l m is the minimum scale of the turbulence. L/ ka 2 = 0 corresponds to an infinite plane wave; L/ ka 2 → ∞ corresponds to a spherical wave.1

Fig. 2
Fig. 2

Optical system of the multiple reflection simulator that was used to measure atmospheric turbulence.

Fig. 3
Fig. 3

Interference fringe obtained through double slits.

Fig. 4
Fig. 4

Average wind velocity within the turbulence chamber shown as a function of distance x.

Fig. 5
Fig. 5

Average wind velocity within the turbulence chamber shown as a function of height y with respect to distance z = 0.60 m.

Fig. 6
Fig. 6

Average air temperature within the turbulence chamber shown as a function of distance x.

Fig. 7
Fig. 7

Degree of coherence of the light influenced by turbulence from experimental results.

Fig. 8
Fig. 8

Dependence of degradation of degrees of coherence on wavelength and turbulence.

Equations (3)

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γx1, x2=Γˆx1, x2/Iˆx1Iˆx21/2,
γr12, Lexp-1.455Cn2k2Lr125/31-5/6L2/k2a4,
V=Imax-IminImax+Imin=2I1I21/2I1+I2 |γ12|,

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