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Coherent Backscattering from Multiple Scattering Systems - KOPS ...

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5 Experiments<br />

E sample<br />

(t) / E half−space<br />

(t)<br />

1<br />

0.8<br />

0.6<br />

0.4<br />

0.2<br />

0<br />

free space distribution<br />

half−space distribution<br />

−9.5 −9 −8.5 −8 −7.5 −7 −6.5 −6<br />

log 10<br />

(t [sec])<br />

Figure 5.2: Comparison of the albedos. The time-dependent albedos A(t) of an<br />

absorbing cut-out of the free space (solid line) and respectively a half-space (dashed<br />

line) at the position of the sample are similar enough to replace the latter by the former<br />

in the calculation of the total albedo. Calculations for teflon sample with radius R =<br />

20 mm, thickness L = 50 mm, diffusion coefficient D = 16500 m2 /s, absorption time<br />

τ = 3.3 ns, transport mean free path l ∗ = 220 µm, refractive index n = 1.35.<br />

be equal to one, meaning that the incoming light power P in is equal to the scattered power<br />

P out of a lossless sample. The albedo can therefore also be defined as<br />

A = P with losses<br />

P lossless<br />

=<br />

∫ ∫<br />

∫ surface ∫<br />

surface<br />

j with losses (⃗r ⊥ , t) d⃗r ⊥ dt<br />

j lossless (⃗r ⊥ , t) d⃗r ⊥ dt<br />

(5.3)<br />

In our multiple scattering samples, most of the loss of light energy is due to absorption. If it<br />

were the only reason for energy loss, the albedo could be easily calculated as<br />

A(τ) = 1 − 2(l∗ +z 0 )<br />

e− √<br />

Dτ<br />

(5.4)<br />

2(l ∗ +z<br />

√ 0 )<br />

Dτ<br />

with diffusion coefficient D, absorption time τ, and penetration depth z 0 .<br />

However, as the samples are not infinitely large, one can not a priori exclude that losses at the<br />

side and rear boundaries have significant influence on the photon distribution in the sample.<br />

Unfortunately there is no expression for the photon density distribution in a 3-dimensionally<br />

confined absorbing medium. Therefore we approximate the losses in a sample with finite<br />

radius R and thickness L by comparing the energy inside the non-absorbing infinite half-<br />

46

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