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
εo
Ei
µo
εt
r
k = k i zˆ
Et
Hi
Er
Hr
µo
r
k = kt zˆ
Ht
r
k = − k i 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
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Scattering of Electromagnetic Waves from Objects
r r
r r
Incident plane wave: E i (r )
Scattered wave: E s (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: E i (r )
εo
a
ε1
λ 2
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
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Rayleigh Scattering: Basic Mechanism
r r
Incident plane wave: E i (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
Incident plane wave: E i (r )
z
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 E i
From homework (3), the z-directed dipole moment p induced in a sphere in the
presence of E-field E is:
⎛ ε −ε ⎞
p = 4π ε o a 3 ⎜⎜ 1 o ⎟⎟ E
⎝ ε1 + 2 εo ⎠
In the present case, the dipole moment phasor p induced in the sphere is therefore:
⎛ ε −ε ⎞
p = 4π ε o a 3 ⎜⎜ 1 o ⎟⎟ E i
⎝ ε1 + 2 εo ⎠
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
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Rayleigh Scattering: Scattered Radiation
Hertzian Dipole
z
Induced Dipole
z
+ + ++ +
q
d
a
ε1
- - -- -
-q
⎛ ε1 − εo ⎞
⎟⎟ E i
⎝ ε1 + 2 εo ⎠
3
Dipole moment = p = 4π ε o a ⎜⎜
r r
j ηo k Id
Eff (r ) = θˆ
sin(θ ) e − j k r
4π r
Dipole moment = p = qd
The far-field scattered radiation is:
Current = I = j ω q
r
r
j η o k ( jω p )
E s − ff (r ) = θˆ
sin(θ ) e − j k r
4π r
⇒ Id = j ω qd = j ω p
Therefore:
= − θˆ
r r
j ηo k ( jω p )
Eff (r ) = θˆ
sin(θ ) e − j k r
4π r
k 2 a3 ⎛ ε 1 − ε o ⎞
⎟ E i sin(θ ) e − j k r
⎜
r ⎜⎝ ε 1 + 2ε o ⎟⎠
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Rayleigh Scattering: Total Scattered Power
z
+ + ++ +
a
ε1
- - -- -
2π π
r
r 2
E s − ff (r )
0 0
2ηo
Ps = ∫ ∫
r 2 sin(θ ) dθ dφ
2
=
4π 4 6 ⎛ ε 1 − ε o ⎞
⎟⎟ E i 2
k a ⎜⎜
3ηo
⎝ ε 1 + 2ε o ⎠
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
4
Rayleigh Scattering: Scattering Cross-Section
Total scattered power Ps from a dielectric sphere is:
2π π
r
r 2
E s − ff (r )
0 0
2ηo
Ps = ∫ ∫
r 2 sin(θ ) dθ dφ =
2
4π 4 6 ⎛ ε 1 − ε o ⎞
2
⎟⎟ E i
k a ⎜⎜
3ηo
⎝ ε 1 + 2ε o ⎠
The incident power per unit area was just the
Poynting vector of the incident wave:
r r 2
E i (r )
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
σs = r r
E i (r )
2
σs is also the ratio of the total scattered power
2ηo
to the power per unit area of the incident wave
at the location of the scatterer
For the dielectric sphere:
Ps
σs = r r
E i (r )
2
2ηo
=
8π 4 6 ⎛ ε 1 − ε o ⎞
k a ⎜⎜
⎟⎟
3
⎝ ε 1 + 2ε o ⎠
2
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Rayleigh Scattering: Why is the Sky Blue
The scattering cross-section of a dielectric sphere is:
σs =
8π 4 6 ⎛ ε 1 − ε o ⎞
⎟⎟
k a ⎜⎜
3
⎝ ε 1 + 2ε o ⎠
2
The scattered power is inversely proportional to the fourth power of the wavelength:
σs ∝ k4 ∝
1
λ4
Shorter wavelengths are scattered more than longer wavelengths in the Rayleigh limit
Sun
Why is the sky blue?
Molecules/atoms in the atmosphere
Rayleigh scatter the sunlight
Earth
Sky appears blue since
the shorter wavelengths
are scattered more (and
violet is absorbed in the
upper atmosphere)
Sun is actually
white – all
wavelengths
are present
Sun appears yellow/orange
since shorter wavelengths
have been scattered out in
the direct line-of-sight
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
5
Example: Radar Range Equation
Power per unit area at the location of the target:
St arg et =
Pin
4π r 2
G (θ , φ )
If the scattering cross-section of the target is σs
then the total scattered power Ps is:
Ps = σ s St arg et = σ s
Pin
4π r
2
G (θ , φ )
If the target scatters isotropically (equally in
all directions) then the power Pout received by
the matched antenna is:
Pout = η p
Pin
= ηp
Pout
Transmitting-receiving
antenna (matched)
r
⇒
Ps
4π r 2
Pin
A(θ , φ )
(4π r 2 )2
σ s G (θ , φ ) A(θ , φ )
σ G 2 (θ , φ ) ⎛⎜ λ2
Pout
= ηp s
2 ⎜ 4π
Pin
⎝
4π r 2
(
)
Radar range
equation
2
2
⎞
⎟ = η p σ s G (θ , φ ) λ
3 4
⎟
(4π ) r
⎠
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
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