#### Transcript Wave incidence

```Wave Incidence
at Oblique angles
Sandra Cruz-Pol
ECE Dept. UPRM
Quiz VIERNES
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Wave Incidence
I.
Normal incidence
Wave arrives at 90o from the surface

II.
Standing waves
Oblique incidence (lossless)
Wave arrives at an angle

Snell’s Law and Critical angle

Two types: Parallel or Perpendicular

Brewster angle
Oblique incidence
First, we need to define:
 Expression for uniform plane wave traveling in
any direction
x
Medium 2 : e2, m2
Medium 1 : e1 , m1
 The Normal , an
 Plane of incidence
 Angle of incidence
qr
qt
qi
y
z=0
z
Oblique incidence
 Uniform plane wave in general form

 

j ( k r t )
E (r , t )  Eo cos( k  r  t )  Re[ Eo e
]

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
Where:
k 2  k x2  k y2  k z2   2 me
 For lossless unbounded media, k = b
Oblique Incidence
x
Medium 2 : e2, m2
Medium 1 : e1 , m1
kr
kt
qr
qi
ki
qt
y
z=0
z
Look at the direction of travel
Medium 1 : e1 , m1
kr
ki
kiz
Medium 2 : e2, m2
kt
qr
qi
kix
x
qt
y
z=0
z
Expression for Waves
Ei  Eio cos( kix x  kiz z  t )
Er  Ero cos( k rx x  k rz z  t )
Et  Eto cos( ktx x  ktz z  t )
qi
ki
where
kiz
kix
k i  k ix2  k iz2  b1   m1e 1
k ix  b1 sin q i
k iz  b1 cos q i
Tangential
E must be Continuous



Ei ( z  0)  Er ( z  0)  Et ( z  0)
wi = wr = wt = w
kix = krx = ktx = kx
kiy = kry = kty = ky
From this we know that
frequency is a property of
the wave. So is color.
So 700nm is not always red!!
Give example of bathing suit.
kix  k rx
b1 sin q i  b1 sin q r
Snell Law
 Equating, we get
or
kix = ktx
b1 sinqi = b2 sinq t
Also written as,
n1 sinqi = n2 sinq 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.
c
n1 = = er1
u1
Critical angle, qc
…All is reflected
n2
sin q c  sin q t [q t  90 o ]
n1
n2
er2


n1
e r1
 When qt =90o, the refracted
wave flows along the
(for m1  m 2 )
surface and no energy is
transmitted into medium 2. Example
 The value of the angle of
er 2
sin 90 o
incidence corresponding to sin q c =
er1
this is called critical angle,
qc.
4
o
sin 42 =
1) = .67
(
 If qi > qc, the incident wave
9
is totally reflected.
sin 50 o = .77 = .67 (sin?? o )
sin 40 o = .64 = .67 (sin 73o )
Fiber optics
 Light can be guided with total
reflections through thin dielectric
rods made of glass or transparent
plastic, known as optical fibers.
 The only power lost is due to
reflections at the input and output
ends and absorption by the fiber
material (not perfect dielectric).
Optical fibers have cylindrical fiber core with
index of refraction nf, surrounded by another
cylinder of lower, nc < nf , called a cladding.
 For total reflection:
Use Snell and critical angle to derive:
n2
sin q 3  sin q c 
n1
q 2  q 3  90o
[Figure from Ulaby, 1999]
Acceptance angle
sin q a 
(n 2f  nc2 )
n0
Exercise: optical fiber
An optical fiber (in air) is
index of refraction of 1.52
index of refraction of 1.49.
 Find the acceptance
angle:
2
sin q a £
(n f - nc2 )
n0
(1.52 2 -1.49 2 )
=
1
Parallel polarization
 It’s defined as E is || to incidence plane
Eis  Eio (cos q i xˆ  sin q i zˆ )e  jb1  x sinqi  z cosqi 
H is 
Eio
1
e  jb1  x sinqi  z cosqi  yˆ
Er
Et
Ers  Ero (cos q r xˆ  sin q r zˆ )e  jb1  x sinq r  z cosq r 
H rs  
Ero
1
H ts 
Eto
2
kr
e  jb1  x sinq r  z cosq r  yˆ
Ets  Eto (cos q t xˆ  sin q t zˆ )e
e  jb 2  x sinqt  z cosqt  yˆ
x
Medium 1 : e1 , m1
 jb 2  x sinq t  z cosq t 
Ei
qr
qi
ki
qt
y
kt
z
z=0
Medium 2 :
e2, m2
Equating for continuity, the tangent
fields
Which components are tangent to the
interface between two surfaces?
 y and x
At z = 0 (interface):
xˆ : Eio (cos q i )e  jb1  x sinqi  z cosqi   Ero (cos q r )e  jb1  x sinq r  z cosq r   Eto (cos q t )e  jb 2  x sinqt  z cosqt 
yˆ :
Eio
1
e  jb1  x sinqi  z cosqi  
Ero
1
e  jb1  x sinqi  z cosqi  
Eto
2
e  jb 2  x sinqt  z cosqt 
xˆ : Eio cos q i  Ero cos q r  Eto cos q t
yˆ :
Eio
1

Ero
1

Eto
2
Reflection and Transmission
Coefficients: Parallel Incidence
 Reflection
Ero  2 cos q t  1 cos q i
|| 

Eio  2 cos q t  1 cos q i
Eto
2 2 cos q i
 || 

Eio  2 cos q t  1 cos q i
where
 ||  (1  || )
cos q i
cos q t
Reflection and Transmission
Coefficients: Perpendicular Incidence
Ero  2 cos q i  1 cos q t
 

Eio  2 cos q i  1 cos q t
Eto
2 2 cos q i
 

Eio  2 cos q i  1 cos q t
1     
Java Animation
 http://www.amanogawa.com/archive
/Oblique/Oblique-2.html
Exercise
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 er=4, on z>0, Find
 The incidence polarization of this wave
 The angle of incidence
 The time-domain expressions for the reflected
electric and magnetic fields.
 The average power density carried by the wave
in the dielectric medium.
Brewster angle, qB
 Is defined as the incidence angle at
which the reflection coefficient is 0
(all transmission).
G || =
h2 cosq t - h1 cosq B
=0
h2 cosq t + h1 cosq B
h2 cosq t - h1 cosq B = 0
sin q B|| =
1- (e1m 2 / e 2 m1 )
1- (e1 / e 2 )2
* qB is
known as
the
polarizing
angle
 The Brewster angle does not exist for
perpendicular polarization for
nonmagnetic materials.
Exercise
A plane wave in free space with
Ei  (10 yˆ  5zˆ) cos(t  2 y  4 z )[V / m]
Is incident upon planar surface of nonmagnetic lossless
dielectric material with er=4, on z>0, Find
 The polarization of the incident wave
 The angle of incidence and transmission
 The reflection and transmission coefficients.
 The E field that is reflected and in dielectric
 Brewster angle
26.6 o, 12.9 o, - 0.295, 0.65,
(-2.946 ŷ +1.47ẑ)cos(w t + 2y + 4z),
(6.5 ŷ + 3.2 ẑ)cos(w t + 2y - 8.7z), 63.4o
Reflection vs. Incidence angle.
Reflection vs.
incidence angle
for different
types of soil and
parallel or
perpendicular
polarization.
Exercise (Brewster angle)
A wave in air is incident upon the flat boundary of a
nonmagnetic soil medium with er=4, at qi=50o.
 Find reflection and transmission coefficients for both
incident polarizations, and the Brewster angle.
||  0.16,  ||  0.58
  0.48,    0.52
EMAG
q B||  63.4o
Summary: wave incidence
Property
Reflection
coefficient
Transmission
coefficient
Relation
Normal
Incidence
 
 2 1
 2  1

2 2 i
 2  1t
R 
Power
Transmissivity
T  1 R
sin q t 
 2 cos q t  1 cos q i
 2 cos q t  1 cos q i
 2 cos q i  1 cos q t
 2 cos q i  1 cos q t
 
2 2 cos q i
2 2 cos q i
 || 
 2 cos q i  1 cos q t
 2 cos q t  1 cos q i
  1  
R  
2
n1
sin q i
n2
Parallel
 
 1  
Power
Reflectivity
Snell’s Law:
Perpendicular
2
T  1  R
where
n2  m r 2e r 2
|| 
 ||  1  || 
R||  ||
cos q i
cos q t
2
T||  1  R||
```