Abstract

Tam and Zardecki recently developed a radiative transport solution for small angle scattering of a diverging laser beam in a particulate medium. Their solution is extended to the case of a converging laser beam focused in a particulate medium and also to the case of a diverging laser beam which propagates an arbitrary distance through clear air before entering the particulate medium.

© 1980 Optical Society of America

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References

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  1. W. G. Tam and A. Zardecki, “Laser beam propagation in particulate media,” J. Opt. Soc. Am. 69, 68–70 (1979).
    [Crossref]
  2. H. S. Snyder and W. T. Scott, “Multiple Scattering of Fast Charged Particles,” Phys. Rev. 76, 220–225 (1949).
    [Crossref]

1979 (1)

1949 (1)

H. S. Snyder and W. T. Scott, “Multiple Scattering of Fast Charged Particles,” Phys. Rev. 76, 220–225 (1949).
[Crossref]

Scott, W. T.

H. S. Snyder and W. T. Scott, “Multiple Scattering of Fast Charged Particles,” Phys. Rev. 76, 220–225 (1949).
[Crossref]

Snyder, H. S.

H. S. Snyder and W. T. Scott, “Multiple Scattering of Fast Charged Particles,” Phys. Rev. 76, 220–225 (1949).
[Crossref]

Tam, W. G.

Zardecki, A.

J. Opt. Soc. Am. (1)

Phys. Rev. (1)

H. S. Snyder and W. T. Scott, “Multiple Scattering of Fast Charged Particles,” Phys. Rev. 76, 220–225 (1949).
[Crossref]

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Figures (1)

FIG. 1
FIG. 1

Scattering geometry: (a) diverging beam; (b) converging beam.

Equations (10)

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I ( φ , r , 0 ) = δ ( r + φ z 0 ) exp ( b φ 2 ) .
I ( φ , r , z ) = π exp ( σ z ) b ( 2 π ) 4 d 2 ξ d 2 η exp ( Ω ( z ) ) × exp { [ ξ + η ( z z 0 ) ] 2 / 4 b } exp [ i ( ξ φ + η r ) ] ,
I ( φ , r , z ) = π exp ( σ z ) b ( 2 π ) 4 m = 0 σ m m ! I m ( φ , r , z ) ,
I 0 ( φ , r , z ) = 2 ( 2 π ) 3 b ( z z 0 ) 2 exp ( b r 2 ( z z 0 ) 2 ) δ ( φ r ( z z 0 ) )
I m ( φ , r , z ) = ( 2 π ) 2 z m 2 × exp ( φ 2 / 4 α ) 0 1 d z 1 0 1 d z m Δ 1 ( m ) × exp { α Δ ( m ) [ r z ( 1 4 a j = 1 m z j + z 0 4 b ) φ α ] } , ( 5 )
α = m / 4 a + 1 / 4 b ,
Δ ( m ) = 1 4 a 2 [ m j = 1 m ( 1 z j ) 2 ( j = 1 m ( 1 z j ) 2 ] + 1 4 a b [ m + j = 1 m ( 1 z j ) 2 2 j = 1 m ( 1 z j ) + m z 0 2 2 m z 0 + 2 z 0 j = 1 m ( 1 z j ) ] ,
z j = z j / z j = 0 to m .
θ = r / z .
I m ( φ , θ , z ) = ( 2 π ) 2 z m 2 exp ( φ 2 4 α ) J m ( φ , θ ) ,