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Lecture 34

Electromagnetic

In this lecture you will learn:

• Scattering of electromagnetic from objects

• Why the is

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 : 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

Assumption:

• Assume that the radius is much smaller than the 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 “

• 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 are scattered more than longer wavelengths in the Rayleigh limit

Why is the sky blue? Sun is actually – all Molecules/atoms in the atmosphere wavelengths Rayleigh scatter the 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

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