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Resonance in scattering and absorption from large noble gas clusters

J. Zweiback, T. Ditmire, and M.D Perry

Open Access Open Access

Abstract

Light scattering in large noble gas clusters irradiated by intense laser pulses was studied and compared to absorption measurements. The scattering signal shows the presence of a peak, when the pulse width was varied, similar to one previously reported in absorption measurements. The peak of the scattering, however, occurs at a longer pulse width than for absorption. This result disagrees with a simple simulation and may be due to propagation or non-linear effects not included in the model.

©2000 Optical Society of America

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Figures (6)
Fig. 1.
Fig. 1. Emission pattern for scattered laser light. The target was argon at 600 psi. This fit is a cos2(θ). The intensity was 1.3×1016 W/cm2. The pulse width was 800 fs.

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Fig. 2.
Fig. 2. Scattering (bottom) and absorption (top) in xenon as a function of pulse width. The input energy was 6.4 mJ which corresponds to an intensity of 2.4×1017 W/cm2 for a 50 fs pulse. Data for 50 psi (triangles), 100 psi (circles), and 200 psi (squares) backing pressures are shown. The lines represent model calculations for 50 psi (dotted), 100 psi (dashed), and 200 psi (solid)

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Fig. 3.
Fig. 3. Calculated scattering (dashed) and absorption (solid) cross sections from Mie theory. The index of refraction was constant for all radii.

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Fig. 4.
Fig. 4. Calculated dielectric constant for an expanding plasma sphere. Plot shows both real (dashed) and imaginary (solid) components.

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Fig. 5.
Fig. 5. Calculated cross-sections for an expanding plasma sphere. Plot shows both absorption (solid) and scattering (dashed) cross-sections. Calculation was done for a 300 eV plasma

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Fig. 6.
Fig. 6. Calculated scattering (dashed) and heating (solid) for a 100 Å xenon cluster. The pulse width was 450 fs and the intensity was 4.1×1015.

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Equations (2)

Equations on this page are rendered with MathJax. Learn more.

σ sca = 8 π 3 k 4 r 6 ∣ ∊ − 1 ∊ + 2 ∣ 2 , σ abs = 4 π k r 3 Im { ∊ − 1 ∊ + 2 } .
∊ = 1 − n 0 n crit ( r 0 r ) 3 ( 1 + i ν ω ) − 1
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