Scattering Approach to Light in Matter: Radiation and Scattering of Plane Waves - Prof. Ch, Study notes of Optics

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Phys 531 Lecture 4 6 September 2005 Light in Matter: Scattering Approach Last time, talked about light in matter: Include charge, current terms in Maxwell eqs Try to only consider macroscopic effects (= polarization P) Result: wave equation similar to vacuum 0 →  c → c/n Generally, waves slower in medium Get n from microscopic model 1 Today: Take a different approach • Consider direct effect of microscopic charges on field Summarize: Each atom radiates a new wave total field = incident field + radiated field Call radiated = scattered Punch line: Incident and scattered field both travel at v = c, but total field looks like it travels slower 2 Outline: • Radiation • Scattering by dense medium • Scattering approach to index Next time: - Survey optical materials - Start considering boundaries between media 3 Radiation (Hecht 3.4) Want to consider sources explicitly: simplest source = radiating charge (Also, nice to know where light comes from!) Basic result: Accelerating charge emits EM wave Why? Hecht gives nice explanation see Figure 3.28 and discussion, pg. 59 4 Scattering (Hecht 4.2) Think about plane wave in matter atoms → oscillating dipoles → radiation Try to understand effect of radiated field I’ll give more mathematical derivation: Hecht gives more conceptual argument 9 Start: Plane wave incident on single atom: induce dipole p = 0χ1E with E = E0e i(k·r−ωt) and χ1 = “single-atom susceptability” = χ/N (N = density) Atom produces dipole field ≈ spherical wave centered at atom location 10 Draw wave fronts: incident field scattered field Fields add: Etot = Eincident + Escattered 11 Real medium has many atoms many Escat’s First model: scattering by glass cube • Glass: atoms randomly distributed • Assume cube size L  λ • Measure at distance d  L • Incident field: E = E0ẑe i(kx−ωt) take x = 0 at front face of cube 12 Setup: r L Einc z y x glass d 13 Need to add up scattered fields from each atom Will see that fields tend to cancel out, except when r is in front of medium 14 Assume each φj random 〈. . .〉 = average over possible values Then 〈f〉 = ∑ j 〈 eiφj 〉 = N 〈 eiφ 〉 for N = number of atoms = NL3 But 〈 eiφ 〉 = 1 2π ∫ 2π 0 eiφ dφ = 1 2πi ( ei2π − e0 ) = 0 So 〈f〉 = 0 = 〈E〉 19 But cancellation not perfect Irradiance I ∝ |E|2 so 〈I〉 ∝ 〈 |E|2 〉 6= 0 Get 〈 |f |2 〉 = 〈 ∑ j eiφj ∑ ` e−iφ` 〉 = ∑ j` 〈 ei(φj−φ`) 〉 If j 6= `, average is zero as before If j = `, average = 1 So 〈 |f |2 〉 = N 20 So rms scattered field at r ∝ N 1/2 Means scattered field per atom decreases like N−1/2 In a dense medium, scattering is suppressed. But still have Iscat ∝ N seems like what you would expect! Called Rayleigh scattering: why sky is blue Different from: - Case L . λ3, phases don’t cancel get large scattering amplitudes: superradiance - Forward scattering 21 Forward Scattering Consider scattered field in front of medium: r L Einc z y x d Difference: now xj and dj correlated 22 Set d = distance from r to back of cube if d  L, have dj ≈ d + L − xj Phase φj = k(dj + xj) ≈ k(d + L) = kx doesn’t depend on j Scattered fields don’t cancel out: ∑ j Ej(r) = ∑ j k2χ1E0 4πd ei[k(xj+dj)−ωt] = ∑ j k2χ1E0 4πd ei(kx−ωt) = N k2χ1 4πd E0e i(kx−ωt) 23 So Escat scales as N and I ∝ N 2! Forward scattering is strong Question: We get strong forward scattering because the phases φj from the different atoms are all the same. This happens because the two components of the phase, xj and dj, are correlated. But xj and dj are also correlated for backwards scattering... should we expect a strong backwards scattering effect? Hint: φj = k(dj + xj), and for backwards scattering, dj ≈ d − L + xj 24 For forward scattering, need φj ≈ constant, independent of j φj = k(dj + xj) ≈ k(d + L) + kρ2j 2d Estimate φj can vary by ∼1 radian Get forward scattering from atoms with ρj < ρmax = √ 2d k Defines volume Veff ≈ πρ 2 maxL = 2πLd k Atoms in Veff contribute to forward scattering 29 Put this volume into formula for scattered field: Escat = βEinc with β = k2χVeff 4πd = k2χ 4πd ( 2πLd k ) = kL 2 χ Question: How can d drop out of β? Surely, the further you are from the atoms, the weaker the scattered field should be! 30 Still missing one effect: most atoms at edge of volume, have dj > d+L−xj On average, contribute additional phase shift kd + 1 radkd ρmax 31 Won’t go through calculation (see Jackson §9.14) Result: extra factor of i Escat = i kL 2 χEinc Now go back to total field at r: Etot(r) = Einc(r) + Escat(r) = Einc(r) ( 1 + i 1 2 kLχ ) 32 We’ve implicitly assumed that scattering is weak (otherwise, scattered field would be re-scattered) So we are limited to kLχ  1 Then we can rewrite Etot Etot(r) = Einc(r)(1 + ikLχ/2) = Einc(r)e ikLχ/2 Since ex ≈ 1 + x for ||  1 33 So have Etot = E0e i[k(L+d)−ωt]eikLχ/2 Rearrage exponents: Etot(r) = E0e i(kd−ωt)eikL(1+χ/2) For χ  1, recognize 1 + χ 2 ≈ √ 1 + χ = n index of refraction So Etot(r) = E0e i(kd−ωt)einkL 34