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Light Scattering When Light Passes Through a Medium Some of It Is Directed Away from Its Direction of Travel

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Light Scattering When Light Passes Through a Medium Some of It Is Directed Away from Its Direction of Travel Light Scattering When light passes through a medium some of it is directed away from its direction of travel. Any photons that are diverted from their direction of propagation are scattered. In the atmosphere both scattering and absorption occur and their combination is known as EXTINCTION. If we have a medium containing a number of scattering elements we can define a scattering coefficient, ks, analogous to an absorption coefficient ka=nσ where n is the number of absorbers per unit volume and σ is the absorption cross section. The intensity of light of wavelength λ, after passing through a medium of thickness L, I(λ), may be given by I(λ)=I0(λ)exp(-ksL), where I0(λ) is the Intensity of light entering the medium and ksLis the optical depth, τ. When both scattering and absorption occur the overall extinction can be defined as ke=ks+ka. And the intensity can be found again using the Beer-Lambert law: I(λ)=I0(λ)exp(-keL). Climate and Energy Lecture 4 2007 Rayleigh Scattering When the scattering elements are small compared to the wavelength of light being scattered then the treatment is simpler than in other cases. The analysis was first performed by Lord Rayleigh (1871) to explain why the sky is blue. As the incident light’s e-m field varies the molecules and small particles are continuously redistributing their charges. The field causes the charges to oscillate at the frequency of the radiation and these charges then re-radiate an e-m field at the same frequency as the forcing field. However the emitted radiation will not be propagated in the same direction as the forcing field and will not necessarily have the same polarization. When equal positive and negative charges are displaced in opposite directions the charge distribution is called a dipole. The dipole moment p is the product of the quantity of displaced charge and the displaced distance. The vector p points in the direction of the charge displacement. If the incident field is E0exp(iωt), the magnitude of the magnitude of the polarisation is p = αtE0exp(iωt), where α is the polarisability of the particle, ω is the angular velocity of the particle and t is time Climate and Energy Lecture 4 2007 Rayleigh Scattering 2 Light interaction with gas molecules and small particles can be modelled using dipoles. Let r be the unit vector in the direction from the dipole to the observation point and p the vector in the direction of the electric field or dipole moment. The scattered field at a distance r from a dipole is: E = ωαtE0exp(iωt), o θ ω 2 r E = exp iωt − r × ( p× r) r 2 e c r c P r el Where c is the speed of light and r is large enough to be in the far field of the dipole. The triple product is a vector lying in the plane containing the range vector (NOT r) and the dipole vector p. The scattered electric field only has components in this scattered plane. The vector in the direction of propogation of the incident light and the vector from the dipole to the observation point define a plane. The component of the polarisation vector in this plane is αE0lrl and perpendicular αE0rrr and el and er are the unit field vectors in these directions. Climate and Energy Lecture 4 2007 Rayleigh Scattering 3 If the incident electric field is (E0rer+ E0lel)exp(iωt), then the scattered field components at P are 2 αω r αω 2 r E (r) = exp iωt − E e r 2 0r r El(r) = expiωt − E0lcosθ()er × r c r c c2r c The intensity of the scattered light assuming that the incident light is unpolarised is 2 4 α ω 2 I = (1+ cos θ )I0 c4r 2 If the scatterer is a sphere of radius a and dielectric constant K, its polarisability is given as [(K-1)(K+2)]a3. So we can write: a6ω 4 m2 −1 I = 1+ cos2 θ I 4 2 2 ( ) 0 c r m + 2 Climate and Energy Lecture 4 2007 Rayleigh Scattering 4 • The angular distribution of the scattered intensity varies from 2 in the forward and backward directions to 1 at 90°, where the l component vanishes and the scattered radiation is plane polarised transverse to the scattering plane. • The scattering intensity and energy removed from the incident beam is proportional to the sixth power of the radius (or for nearly spherical particles to the square of the volume). • The scattered intensity and energy removed from the incident beam is proportional to ω4 or to λ-4. • The intensity of Rayleigh scattering is proportional to the sixth power of the radius. The absorption, however, is proportional to the volume. As a result scattering falls off more rapidly than absorption, or put another way the single scattering albedo decreases with decreasing particle size. Climate and Energy Lecture 4 2007 Rayleigh Scattering 5 The sky is blue as blue light is scattered out of the direct solar beam much more efficiently than longer wavelengths. The horizon appears whiter than the zenith because the increased path leads to increased scattering. During sunrise and sunset blue and yellow light is efficiently removed from the direct beam over the very long atmospheric paths at the high solar zenith angles leading to spectacular red skies. Climate and Energy Lecture 4 2007 The Direction of Scattering Climate and Energy Lecture 4 2007 Mie Scattering 1 If the scatterers are large compared to the wavelength of light then geometric optics provide a good approximation. However, in most atmospheric situations aerosols are neither large nor small enough to be treated simply. The size parameter, α = 2πa/λ, is between 1 and 20. This problem was first solved by Gustav Mie (1908) by determining the wave vector in spherical coordinates for e-m waves, specified by Maxwell’s equations. In general particles absorb as well as scatter and Mie gave solutions for the absorption, scattering and extinction cross sections as a function of the scattering angle. A function for the extinction efficiency factor, Qext, for spheres of refractive index n in a medium of unity refractive index can be derived using Mie theory which describe the efficiency with which light is scattered as a function of the size parameter. Climate and Energy Lecture 4 2007 Mie Scattering 2 Qext is independent of particle size and can be related to the extinction cross 2 section, Cext by Qext=Cext/πr , where r is the particle radius. Similar relationships can be formed for Qscat and Qabs, the scattering and absorption coefficients. The refractive index, n, is given as a complex number where the real part is the scattering component and the imaginary component is the absorptive component. We can define the single scattering albedo, ω0, as the fraction of light scattered by a particle to the total extinction caused by it: ω0 = Qscat/Qext = Qscat/(Qscat+ Qabs) = Cscat/Cext = Cscat/(Cscat + Cabs). Climate and Energy Lecture 4 2007 Mie Scattering 3 Climate and Energy Lecture 4 2007 Ensemble scattering, absorption and extinction coefficients So far we have only considered a single particle interacting with radiation in isolation. What we want to know and have to measure is how a bulk air parcel, containing many aerosol particles interacts. If the aerosol number concentration is very large then the scattering is very complex as radiation undergoes many interactions in a volume. However, in practice the atmosphere, even at its most polluted, has a relatively small number concentration. Even, in highly polluted environments, where the number concentration may reach 106 particles cm-3 the particles only occupy approximately 10-6 of the volume. This means that we can approximate the problem by only considering single scattering. Given this we can assume that the total scattered intensity is the sum of the intensities scattered by each of the particles. Climate and Energy Lecture 4 2007 Ensemble scattering, absorption and extinction coefficients 2 Now the ensemble scattering, absorption and extinction coefficients (ks, ka, and ke) are functions of particle diameter (Dp), refractive index (n), and wavelength of incident light (λ). If we assume that the entire particle population has the same refractive index, then max Dp πD2 k λ = p Q n,α N(D )dD e () ∫ ext () p p 0 4 Where N(Dp) is the number concentration of particles of diameter Dp. From here on we assume, unless otherwise stated that ke is dependent on wavelength. Similar expressions can be written for the scattering and absorption coefficients. And ke = ks + ka. So at a given wavelength, the intensity, I, at a distance z through a scattering and absorbing layer can be related to the initial intensity, I0 by the Beer Lambert Law: I=I0exp(-kez) = I0exp(-τ) Where τ is the optical depth of the aerosol layer. Climate and Energy Lecture 4 2007 Ensemble scattering, absorption and extinction coefficients 3 Often these coefficients are expressed in terms of the aerosol mass distribution function M(Dp): 3 πDp M ()D = ρ N()D p p 6 p The result is: Dmax p 3 k = Q n,α M (D )dD e ∫ ext () p p 0 2ρ p Dp Which can be written: max Dp k = E D ,n,λ M (D )dD e ∫ ext ()p p p 0 Where Eext(Dp, n, λ) is the mass extinction coefficient. Likewise Escat and Eabs: 3 3 Eabs ()Dp ,n,λ = Qabs ()n,α E (D ,n,λ)= Q ()n,α 2ρ D scat p scat p p 2ρ p Dp Climate and Energy Lecture 4 2007 Ensemble scattering, absorption and extinction coefficients 4 The figure shows how Escat and Eabs vary as a function of particle size over a range of refractive indices.
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