Lecture 34 Electromagnetic Scattering Scattering Of

Lecture 34 Electromagnetic Scattering Scattering Of

Lecture 34 Electromagnetic Scattering In this lecture you will learn: • Scattering of electromagnetic waves from objects • Rayleigh Scattering • Why the sky is blue • Radar range equation ECE 303 – Fall 2007 – Farhan Rana – Cornell University Scattering of Electromagnetic Waves from a Plane Interface x ε µ Ei ε µ o o r t o k = ki zˆ Et Hi r k = kt zˆ Er Hr Ht r k = −ki zˆ z=0 z Incident, transmitted, and reflected waves are all plane waves The reflected and transmitted waves can also be called the scattered waves ECE 303 – Fall 2007 – Farhan Rana – Cornell University 1 Scattering of Electromagnetic Waves from Objects r Incident plane wave: E rr r r i () Scattered wave: Es (r ) εo ε1 λ phase fronts Questions: • How does one find the scattered field? • How much power from the incident field goes into the scattered field? • In which direction(s) does the scattered power go? ECE 303 – Fall 2007 – Farhan Rana – Cornell University Scattering of Electromagnetic Waves From Spherical Particles Assumption: • Assume that the particle radius is much smaller than the wavelength of the incident wave, i.e.: ka << 1 - When the above condition holds the scattering is called “Rayleigh Scattering” - When the above condition does not hold the scattering is called “Mie Scattering” • When ka << 1, the particle sees a uniform E-field that is slowly oscillating in time r r Incident plane wave: Ei ()r ε o a ε1 λ 2 ECE 303 – Fall 2007 – Farhan Rana – Cornell University 2 Rayleigh Scattering: Basic Mechanism r r Incident plane wave: Ei ()r εo ++ + ++ a ε1 - - - - - λ 2 One way to understand scattering is as follows: i) The incident E-field induces a time-varying dipole moment in the sphere (recall the electrostatics problem of a dielectric sphere in a uniform E-field from homework 3) ii) The time-varying dipole radiates like a Hertzian dipole, and this radiation is the scattered radiation ECE 303 – Fall 2007 – Farhan Rana – Cornell University Rayleigh Scattering: Induced Dipole r r z Incident plane wave: Ei ()r εo ++ + ++ a ε1 - - - - - λ 2 Suppose the z-directed E-field phasor for the incident plane wave at the location of r r the particle is: E()r = 0 = zˆ Ei From homework (3), the z-directed dipole moment p induced in a sphere in the presence of E-field E is: 3 ⎛ ε1 − εo ⎞ p = 4π εoa ⎜ ⎟ E ⎝ ε1 + 2εo ⎠ In the present case, the dipole moment phasor p induced in the sphere is therefore: 3 ⎛ ε1 − εo ⎞ p = 4π εoa ⎜ ⎟ Ei ⎝ ε1 + 2εo ⎠ ECE 303 – Fall 2007 – Farhan Rana – Cornell University 3 Rayleigh Scattering: Scattered Radiation Hertzian Dipole Induced Dipole z z ++ q + ++ d a ε1 - - -q - - - ⎛ ε − ε ⎞ Dipole moment = p = 4π ε a3 ⎜ 1 o ⎟ E r jη k Id o ⎜ ⎟ i E ()rr = θˆ o sin()θ e− j k r ⎝ ε1 + 2εo ⎠ ff 4π r Dipole moment = p = qd The far-field scattered radiation is: Current = I = j ω q r r j ηo k ( jω p) − j k r ⇒ Id = j ω qd = j ω p E ()r = θˆ sin()θ e s−ff 4π r Therefore: k 2 a3 ⎛ ε − ε ⎞ = −θˆ ⎜ 1 o ⎟ E sin()θ e− j k r r j η k ()jω p r ⎜ ε + 2ε ⎟ i E ()rr = θˆ o sin()θ e− j k r ⎝ 1 o ⎠ ff 4π r ECE 303 – Fall 2007 – Farhan Rana – Cornell University Rayleigh Scattering: Total Scattered Power z ++ + ++ a ε1 - - - - - r 2 2ππ r Es−ff ()r 2 Ps = ∫∫ r sin()θ dθ dφ 00 2ηo 2 4π 4 6⎛ ε1 − εo ⎞ 2 = k a ⎜ ⎟ Ei 3ηo ⎝ ε1 + 2εo ⎠ ECE 303 – Fall 2007 – Farhan Rana – Cornell University 4 Rayleigh Scattering: Scattering Cross-Section Total scattered power Ps from a dielectric sphere is: r r 2 2 2ππ E ()r s−ff 2 4π 4 6 ⎛ ε1 − εo ⎞ 2 Ps = ∫∫ r sin()θ dθ dφ = k a ⎜ ⎟ Ei 00 2ηo 3ηo ⎝ ε1 + 2εo ⎠ r r 2 The incident power per unit area was just the Ei ()r Poynting vector of the incident wave: 2ηo The scattering cross-section σs of a scatterer is defined as the area of a plane oriented perpendicular to the direction of incident wave that would intercept the same total incident power as the power Ps that the scatterer radiates Ps σ is also the ratio of the total scattered power σ s = s r r 2 to the power per unit area of the incident wave Ei ()r 2ηo at the location of the scatterer For the dielectric sphere: 2 P 8π ⎛ ε − ε ⎞ σ = s = k 4 a6 ⎜ 1 o ⎟ s r r 2 3 ⎜ ε + 2ε ⎟ Ei ()r 2ηo ⎝ 1 o ⎠ ECE 303 – Fall 2007 – Farhan Rana – Cornell University Rayleigh Scattering: Why is the Sky Blue The scattering cross-section of a dielectric sphere is: 2 8π 4 6⎛ ε1 − εo ⎞ σ s = k a ⎜ ⎟ 3 ⎝ ε1 + 2εo ⎠ The scattered power is inversely proportional to the fourth power of the wavelength: 4 1 σ s ∝ k ∝ λ4 Shorter wavelengths are scattered more than longer wavelengths in the Rayleigh limit Why is the sky blue? Sun Sun is actually white – all Molecules/atoms in the atmosphere wavelengths Rayleigh scatter the sunlight are present Sun appears yellow/orange since shorter wavelengths have been scattered out in Sky appears blue since the direct line-of-sight the shorter wavelengths are scattered more (and violet is absorbed in the Earth upper atmosphere) ECE 303 – Fall 2007 – Farhan Rana – Cornell University 5 Example: Radar Range Equation Power per unit area at the location of the target: Pin St arget = G()θ ,φ 4π r 2 If the scattering cross-section of the target is σs then the total scattered power Ps is: r Pin Ps = σ s St arget = σ s 2 G()θ ,φ 4π r If the target scatters isotropically (equally in all directions) then the power Pout received by the matched antenna is: Ps Pout = ηp A()θ ,φ 4π r 2 Pin Pin = ηp σ s G()()θ ,φ A θ ,φ Radar range 2 2 Pout ()4π r equation P σ G2()θ ,φ ⎛ λ2 ⎞ σ G2()θ ,φ λ2 Transmitting-receiving ⇒ out = η s ⎜ ⎟ = η s antenna (matched) p 2 ⎜ ⎟ p 3 4 Pin (4π r 2 ) ⎝ 4π ⎠ ()4π r ECE 303 – Fall 2007 – Farhan Rana – Cornell University ECE 303 – Fall 2007 – Farhan Rana – Cornell University 6.

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