Dr. S. Cruz-Pol
Quiz VIERNES
Wave Incidence
at Oblique angles
o  Uno de los problemas asignados en el
prontuario
o  Enviar contestaciones a preguntas en
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Sandra Cruz-Pol
ECE Dept. UPRM
Wave Incidence
I.  Normal incidence
Wave arrives at 90o from the surface
n
Standing waves
II.  Oblique incidence (lossless)
Wave arrives at an angle
n
Snell s Law and Critical angle
n
Two types: Parallel or Perpendicular
n
Brewster angle
Oblique incidence
First, we need to define:
o  Expression for uniform plane wave traveling in
any direction
x
Medium 1 : ε1 , µ1 Medium 2 : ε2, µ2
o  The Normal , an
o  Plane of incidence
o  Angle of incidence
θ
r
θt
y
θi
z
z=0
Oblique incidence
o  Uniform plane wave in general form

 

E (r , t ) = Eo cos( k ⋅ r − ωt ) = Re[Eo e j ( k ⋅r −ωt ) ]

r = xaˆ x + yaˆ y + zaˆ z the radius or position vector

k = k x aˆ x + k y aˆ y + k z aˆ z the wave number or propagation vector
k 2 = k x2 + k y2 + k z2 = ω 2 µε
o  For lossless unbounded media, k = β
Wave incidence at oblique angles
Oblique Incidence
x
Medium 2 : ε2, µ2
Medium 1 : ε1 , µ1
kr
kt
θr
θi
ki
θt
y
z
z=0
1
Dr. S. Cruz-Pol
Look at the direction of travel
Expression for Waves
x
Medium 2 : ε2, µ2
Medium 1 : ε1 , µ1
kr
Er = Ero cos( k rx x + k rz z − ωt )
kt
θr
θt
z
y
θi
kix
ki
Ei = Eio cos( kix x + kiz z − ωt )
θi
z=0
ki
E must be Continuous



Ei ( z = 0) + Er ( z = 0) = Et ( z = 0)
!i = ! r = !t = !
kix = krx = ktx = kx
kiy = kry = kty = ky
kix
So 700nm is not always red!!
kix = krx
β1 sin θi = β1 sin θ r
Critical angle, θc
k iz = β1 cosθ i
Snell Law
kix = ktx
o  Equating, we get
!1 sin " i = !2 sin " t
or
Also written as,
n1 sin ! i = n2 sin ! t
where, the index of refraction of a medium, ni ,
is defined as the ratio of the phase velocity in
free space (c) to the phase velocity in the
medium.
n = ε
1
sin θ c =
…All is reflected
o  When θt
the refracted
wave flows along the
surface and no energy is
transmitted into medium 2.
o  The value of the angle of
incidence corresponding to
this is called critical angle,
θc.
o  If θi > θc, the incident wave
is totally reflected.
k i = k ix2 + k iz2 = β1 = ω µ1ε 1
k ix = β1 sin θ i
From this we know that
frequency is a property of
the wave. So is color.
Give example of bathing suit.
=90o,
where
kiz
kiz
Tangential
Et = Eto cos( ktx x + ktz z − ωt )
=
n2
sin θ t [θ t = 90 o ]
n1
n2
ε
= r2
n1
ε r1
(for µ1 = µ 2 )
Example
sin ! c =
sin 42 o =
!r 2
sin 90 o
!r1
4
(1) = .67
9
sin 50 o = .77 = .67 (sin??o )
r1
Fiber optics
o  Light can be guided with total
reflections through thin dielectric
rods made of glass or transparent
plastic, known as optical fibers.
o  The only power lost is due to
reflections at the input and output
ends and absorption by the fiber
material (not perfect dielectric).
sin 40 o = .64 = .67 (sin 73o )
Wave incidence at oblique angles
2
Dr. S. Cruz-Pol
Optical fibers have cylindrical fiber core with
index of refraction nf, surrounded by another
cylinder of lower, nc < nf , called a cladding.
Exercise: optical fiber
o  An optical fiber (in air)
is made of fiber core
with index of refraction
of 1.52 and a cladding
with an index of
refraction of 1.49.
o  Find the acceptance
sin ! a !
angle:
[Figure from Ulaby, 1999]
o  For total reflection:
Acceptance angle
n
sin θ 3 ≥ sin θ c = 2
n1
θ 2 + θ 3 = 90
o
sin θ a ≤
(n 2f − nc2 )
Er
Et
Ers = Ero (cos θ r xˆ + sin θ r zˆ )e − jβ1 ( x sinθ r − z cosθ r )
H rs = −
Ero
η1
kr
e − jβ1 ( x sinθ r − z cosθ r ) yˆ
Ei
Ets = Eto (cos θ t xˆ − sin θ t zˆ )e − jβ 2 ( x sinθt + z cosθt )
H ts =
Eto
η2
Which components are tangent to the
interface between two surfaces?
o  y and x
At z = 0 (interface):
x
Medium 1 : ε1 , µ1
e − jβ1 ( x sinθi + z cosθi ) yˆ
θi
ki
θt
y
z
z=0
Medium 2 :
ε2, µ2
Reflection and Transmission
Coefficients: Parallel Incidence
o  Reflection
E
η cosθ t − η1 cosθ i
Γ|| = ro = 2
Eio η 2 cosθ t + η1 cosθ i
E
2η 2 cosθ i
τ || = to =
Eio η 2 cosθ t + η1 cosθ i
τ || = (1 + Γ|| )
cosθ i
cosθ t
Wave incidence at oblique angles
xˆ : Eio (cos θ i )e − jβ1 ( x sinθi + z cosθi ) + Ero (cos θ r )e − jβ1 ( x sinθ r − z cosθ r ) = Eto (cos θ t )e − jβ 2 ( x sinθt + z cosθt )
kt
θr
e − jβ 2 ( x sinθt + z cosθt ) yˆ
where
(1.52 "1.49 2 )
1
Equating for continuity, the tangent
fields
o  It s defined as E is || to incidence plane
η1
=
Answer: 17.5 degrees
n0
Eis = Eio (cos θ i xˆ − sin θ i zˆ )e − jβ1 ( x sinθi + z cosθi )
Eio
n0
2
Parallel polarization
H is =
(n 2f " nc2 )
z
yˆ :
Eio
η1
e − jβ1 ( x sinθi + z cosθi ) −
Ero
η1
e − jβ1 ( x sinθi − z cosθi ) =
Eto
η2
e − jβ 2 ( x sinθt + z cosθt )
xˆ : Eio cosθ i + Ero cosθ r = Eto cosθ t
yˆ :
Eio
η1
−
Ero
η1
=
Eto
η2
Reflection and Transmission
Coefficients: Perpendicular Incidence
Γ⊥ =
Ero η 2 cos θ i − η1 cos θ t
=
Eio η 2 cos θ i + η1 cos θ t
τ⊥ =
Eto
2η 2 cos θ i
=
Eio η 2 cos θ i + η1 cos θ t
1 + Γ⊥ = τ ⊥
3
Dr. S. Cruz-Pol
Java Animation
Exercise
o  http://www.amanogawa.com/archive/
Oblique/Oblique-2.html
A plane wave in air with
Ei = yˆ10e − j (3 x + 4 z ) [V / m]
Is incident upon planar surface of nonmagnetic
dielectric material with εr=4, on z>0, Find
o  The polarization of the incident wave
o  The angle of incidence
o  The time-domain expressions for the reflected
electric and magnetic fields.
o  The average power density carried by the wave
in the dielectric medium.
Brewster angle, θB
Exercise
o  Is defined as the incidence angle at which the
reflection coefficient is 0 (all transmission).
A plane wave in free space with
Ei = (10 yˆ + 5zˆ) cos(ωt + 2 y − 4 z )[V / m]
Is incident upon planar surface of nonmagnetic lossless
dielectric material with εr=4, on z>0, Find
o  The polarization of the incident wave
o  The angle of incidence and transmission
o  The reflection and transmission coefficients.
o  The E field in reflected and in dielectric
Answers :
o  Brewster angle
26.6 o , 12.9 o , − 0.295, 0.65,
(−2.946 yˆ + 1.47 zˆ ) cos(ωt + 2 y + 4 z ),
Γ|| =
η 2 cosθ t − η1 cosθ B
=0
η 2 cosθ t + η1 cosθ B
η 2 cosθ t − η1 cosθ B = 0
sin θ B|| =
1 − (ε 1µ 2 / ε 2 µ1 )
1 − (ε 1 / ε 2 ) 2
* θB is
known as
the
polarizing
angle
o  The Brewster angle does not exist for
perpendicular polarization for nonmagnetic
materials.
(6.5 yˆ + 3.2 zˆ ) cos(ωt + 2 y − 8.7 z ),63.4 o
Reflection vs. Incidence angle.
Exercise (Brewster angle)
Reflection vs.
incidence angle
for different
types of soil and
parallel or
perpendicular
polarization.
A wave in air is incident upon the flat boundary of a
nonmagnetic soil medium with εr=4, at θi=50o.
o  Find reflection and transmission coefficients for both
incident polarizations, and the Brewster angle.
o  Answers:
Γ|| = −0.16, τ || = 0.58
Γ⊥ = −0.48, τ ⊥ = 0.52
EMAG
Wave incidence at oblique angles
θ B|| = 63.4o
4
Dr. S. Cruz-Pol
Summary
Property
Reflection
coefficient
Transmission
coefficient
Relation
Normal
Incidence
η −η
Γ= 2 1
η 2 + η1
τ=
2η 2 i
η 2 + η1t
τ = 1+ Γ
Power
Reflectivity
R= Γ
Power
Transmissivity
T = 1− R
Snell s Law:
2
n
sin θ t = 1 sin θ i
n2
Perpendicular
Parallel
Γ⊥ =
η cosθ t − η1 cosθ i
η 2 cosθ i − η1 cosθ t
Γ|| = 2
η 2 cosθ t + η1 cosθ i
η 2 cosθ i + η1 cosθ t
τ⊥ =
2η 2 cosθ i
2η 2 cosθ i
τ =
η 2 cosθ i + η1 cosθ t || η 2 cosθ t + η1 cosθ i
τ ⊥ =1 + Γ⊥
R⊥ = Γ⊥
2
T⊥ = 1 − R⊥
τ || = (1 + Γ|| )
R|| = Γ||
cosθ i
cosθ t
2
T|| = 1 − R||
where n2 = µ r 2ε r 2
Wave incidence at oblique angles
5