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Vs = cos Os cos 4J sx + cos Os sin 4J/iJ - sin Osz
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Rayleigh Scattering
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2.1 Rayleigh Scattering by a Small Particle
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In Rayleigh scattering, the particle size D is much less than wavelength A. In such a case an oscillatory dipole with moment 15 is induced inside the
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particle. The field radiated by the dipole is the scattered field. The far field radiated by a dipole 15 in the direction ks is
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_ k 2 e ikr E s = --4-ks x (k s x p)
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Let the electric field inside the particle be denoted by E int , where the subscript int denotes internal. The polarization per unit volume inside the particle is
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(E p -
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In Rayleigh scattering, the internal field is a constant vector inside the particle. The dipole moment of the particle is
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is the volume of the particle. Using (1.2.2) and (1.2.3) in (1.2.1)
E s = - - - ( Ep
k 2 e ikr
E) Va k s X k s X E int
A _ ]
k e = --(Ep
2 ikr
E) Va
aAa s E int ) + bs(b s . Eint}
Using (1.2.4), the scattering amplitude matrix can be determined by relating the internal field E int to the incident field E i . In Rayleigh scattering, the power absorbed by the particle is, from (1.1.26), (1.2.5)
2.2 Rayleigh Scattering by a Sphere
Consider a sphere of radius a field inside the particle is
A centered at the origin. Then the internal
E int =
+ 2E
The internal field Eint is parallel to the incident field E i . Then using (1.2.6) in (1.2.4) and Va = 41ra3 /3, r E s = e: fa [as(a s ' E i ) + bs(b s . E i )] (1.2.7) where (1.2.8) From (1.2.7)
E as = fa( as . E i )
E bs = fa(b s . Ed
2.2 Rayleigh Scattering by a Sphere
We can calculate the scattering amplitude matrix using the scattering plane orthonormal system. First let Eli = 1 and E2i = 0, thus E i = t. Then E1s = In and E 2s = 121. From (1.2.9a)-(1.2.9b), we obtain El s = fo(1s . L) = fo and E2s = 10(2 s . Ii) = O. Hence In = 10 and 121 = O. Next, let Eli = 0 and E2i = 1 so that Ei = 2i. Then E1s = h2 and E2s = 122. From (1.2.9a)(1.2.9b), E1s = fo(ls . 2i) = 0 and E2s = fo(2 s .2i) = I cos e. Therefore h2 = 0 and 122 = 10 cos e. Thus the scattering amplitude matrix assumes following simple form in the scattering plane system of coordinates:
In [ 121
h2] 122 -
[/0 0
Next, we use the orthonormal unit vectors of vertical and horizontal polarization,
EVs] [ Ehs
[Ivv Ihv
IVh] [EVi] Ihh Ehi
To get fvv and fhv, we let EVi = 1 and Ehi = 0 so that E i = Vi. Then from (1.2.9a)-(1.2.9b), Evs = lo(v s . Vi) and Ehs = lo(i/,s . Vi)' Thus,
fvv Ihv
lo(Vs . Vi)
fo [cos Os cos 0i cos( <Ps - <Pi) + sin Os sin Oil (1.2.12a)
= lo(h s . Vi) = - 10 COS Oi sin(4)s - 4>i) (1.2.12b) To get Ivh and fhh' we let E Vi = 0 and Ehi = 1 so that E i = hi. Then from (1.2.9a)-(1.2.9b), E vs = lo(vs . hi) and Ehs = lo(h s . hi) Thus,
fo cos Os sin(4)s - <Pi)
(1.2.13a) (1.2.13b)
Ihh = 10cos(4)s - 4>i)
To calculate the scattering cross section, suppose that the incident wave is horizontally polarized. The scattered power is determined by integration over all scattered angles of the sum of the scattered vertical and horizontal polarizations: ers =
= I/o l2 =
2 2 d4>s (11vh1 + I/hhl )
dOssinO s
21r d4>s [cos20ssin2(4)s-4>d+cos2(4>s-4>i)]
81r Ifol 2 3
81r k 4 a 6 \ Ep - E 1 3 Ep + 2E
If we compare the scattering cross section to the geometric cross section 2 1ra of a sphere, we obtain
s eT eTg
eT s 1ra 2
Ep -
+ 2E
2 E 1
Since ka 1, the scattering cross section is much less than the geometric cross section. It also follows from (1.2.15) that eTs/eTg = O(D4/>...4) , in agreement with (1.1.21).