plane wave reflection and transmission (1)
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Hon Tat Hui Plane Wave Reflection and Transmission
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Plane Wave Reflection and Transmission
1 Normal Incidence at a Conducting Boundary
zy
x
Medium 2 – perfect conductor
( )∞=σ2
Medium 1 – air or dielectric medium
( )01 =σ
z = 0
•
( )11,μ ε
r H
r E
r k̂
•
iE
iH ik̂
( ) ( )0
22
== z z HE
Js ρ s+ + +
Surface current density
Surface charge density
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( ) zii e E z
1-j
0ˆ β xE = ( ) 1-j0
1
ˆ zii
E z e
β
η =H y
Incident fields:
Transmitted fields in medium 2:( ) ( ) 022 == z z HE
Boundary conditions,
( ) ( )
( ) ( )
1 1 20
1 1 20
ˆ 0
ˆ
z
s z
z z
z z
=
=
× − =⎡ ⎤⎣ ⎦
× − =⎡ ⎤⎣ ⎦
n E E
n H H J
Reflected electric field:( ) z
r r e E z1 j
0ˆ β += xE ( ) 1 j0
1
1ˆ z
r r z E e
β
η
+= −H y
( ) ( ) ( )
( ) ( ) ( )
1
1
i r
i r
z z z
z z z
= +
= +
E E E
H H H
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( ) ( ) 000 ||2||1 == E E
Tangential
component of E1
Note that the boundary conditions are prescribed andcannot be theoretically derived or proved.
Note that as only one unknown E r 0 is to be found, weonly need the electric field boundary condition.
( ) ( )
( ) ( )
1|| 1 1 0
2|| 1 2 0
ˆ0
ˆ0
z
z
E z
E z
=
=
= ×
= ×
n E
n E
Let,
Then,
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Total electric field in medium 1:( ) ( ) ( )
z
r
z
i
r i
e E e E
z z z
11 j
0
j-
0
1
ˆˆ β β ++=
+=
xx
EEE
At z = 0, using the boundary condition,
( ) 0 00 00||1 =+⇒= r i E E E
00 ir E E −=∴
( ) ( ) ( )
( )( ) z E j
ee E
z z z
i
z z
i
r i
10
j j-
0
1
sin2ˆ
ˆ 11
β
β β
x
x
EEE
−=
−=
+=+
Total electric field:
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Reflected magnetic field :
( ) ( ) ( )
( ) ( )
zi
z
i
r r
e E
e E
z z
1
1
j0
1
j
0
1
1
ˆ1
ˆˆ1
ˆ1
β
β
η
η
η
+
+
=
−×−=
×−=
y
xz
EzH
Reflected electric field :
( ) zir e E z
1 j
0ˆ β +−= xE
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( ) ( ) ( )
( )
( ) z E
ee E
e E e
E
z z z
i
z zi
z
i
zi
r i
1
1
0
j j-
1
0
j
01
j-
1
0
1
cos2ˆ
ˆ
ˆ1
ˆ
11
11
β η
η
η η
β β
β β
y
y
yy
HHH
=
+=
+=
+=
+
+
Total magnetic field in medium 1:
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Instantaneous fields:
Note that both the total electric and total magnetic fields
in medium 1 are standing waves*.
1. They are ⊥ to each other and 90º out of phase.2. The electric field vanishes at z = -nλ /2, n = 0,1,2,
3. The magnetic field vanishes at z =-(λ /4 + nλ /2).
( ) ( ) ( ) ( )t z E e zt z it j ω β ω sinsin2ˆRe, 1011 xEE ==
( ) ( ){ } ( ) ( )t z E e zt z it j ω β η ω coscos2ˆRe, 1
1
011 yHH ==
* See animation “Formation of a Pure Standing Wave” to show the making of a standing wave from
two traveling waves moving in opposite directions
By setting sin( β 1 z)=0 in E1
By setting cos( β 1 z)=0 in H1
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Total electric and magnetic fields in medium 1
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Example 1A uniform plane wave (Ei, Hi) at a frequency of 100 MHz
travels in air in the + x direction. The electric field is
polarised in the y direction. The wave impinges normally ona perfectly conducting plane at x = 0. The amplitude of the
incident electric field is 6×10-3 V/m and its initial phase is
zero.(a) Write phasor and instantaneous expressions for Ei, Hi.
(b) Write phasor and instantaneous expressions for Er , Hr .
(c) Write phasor and instantaneous expressions for E1, H1 in
air.
(d) Determine the position nearest to the conducting plane
where E1 = 0.
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Solutions(a)
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(b)
( x)
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(d) The electric field vanishes at x = -nλ /2, n = 0,1,2,….Except at the boundary
surface (n = 0), the nearest null will be at n = 1. That is, x = - λ /2 = -1.5 m.
(c)
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2 Normal Incidence at a Dielectric Boundary
r H
r E
r k̂
•
iE
iH ik̂
zy
x
Medium 2 – dielectric medium
( )02 =σ
Medium 1 – air or dielectric medium
( )01 =σ
z = 0
•
( )11,μ ε ( )22 ,μ ε
•
t E
t H t k̂
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Incident, reflected, and transmitted fields:( ) z
ii e E z1-j
0ˆ β xE = ( ) zi
i e E
z 1 j-
1
0ˆ β
η yH =
( ) zr r e E z 1 j
0ˆ β xE = ( ) zr r e E z 1
j
1
0ˆ β
η yH −=
( ) zt t
e E z 2-j
0
ˆ β xE = ( )
zt
t
e E
z 2 j-
2
0ˆ β
η yH =
Medium parameters:
222111 , μ ε ω β μ ε ω β ==
2
22
1
11 ,
ε
μ η
ε
μ η ==
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Boundary conditions:
( ) ( )00 ||2||1 E E = ( ) ( )00 ||2||1 H H =
000 t r i E E E =+2
0
1
0
1
0
η η η
t r i E E E =−The boundary conditions lead to:
Solving for E r 0
and E t 0
,
0
12
120 ir E E
η η
η η
+−
=0
12
20
2it E E
η η
η
+=
( ) ( ) ( ) ( ) ( )00 ,000 0||200||1 t r i E E E E E =+=
( ) ( ) ( )
( ) ( )
2
0||2
1
0
1
0||1
00 ,
000
η η η
t r i E H E E
H =−=
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Define:
τ Γ =+1 Note:
1≤Γ
12
12
0
0 t,coefficienReflectionη η
η η Γ
+−
==
i
r
E
E
12
2
0
0 2 t,coefficienonTransmissiη η
η τ
+==
i
t
E
E
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Transmitted
Reflected
Incident
Using Γ and τ , the field expressions in the media can beexpressed in terms of the incident field amplitude E i0:
( ) zii e E z
1-j
0ˆ β xE = ( )
zii e
E z 1 j-
1
0ˆ β
η yH =
( ) zir e E z 1 j0ˆ β Γ= xE ( ) zir e E z1 j
1
0ˆ β η Γ−=
yH
( ) z
it e E z2-j
0ˆ β
τ xE = ( ) zi
t e
E
z2 j-
2
0
ˆ
β
η
τ
yH =
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Reflected power density, S r
( ) ( ){ } ( )
2
20*
1
2
2 0
1
2
1 1ˆRe Re2 2
1
2
incident power density
ir r
i
E z z
E
η
η
⎧ ⎫⎪ ⎪= × = − Γ⎨ ⎬⎪ ⎪⎩ ⎭
= Γ
= Γ ×
E H z
reflected powerof fractiondensity powerincident
density powerreflected2==Γ
Power Density Relationship
density powerincident2
1
1
2
0 =η
i E
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( ) ( ){ }2
2 0*
2
2
2 01
2 1
2 1
2
1 1ˆRe Re
2 2
1
2
incident power density
i
t t
i
E z z
E
τ η
η τ
η η
η τ
η
⎧ ⎫⎪ ⎪= × = ⎨ ⎬
⎪ ⎪⎩ ⎭
=
= ×
E H z
smitted power tranof fractiondensity powerincident
density powerdtransmitte
2
12 ==η
η τ
Transmitted power density, S t
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Since the media are lossless,
Fraction of reflected power + Fraction of transmitted power = 1
That is,
12
122 =+η
η τ Γ
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( ) ( ){ }
( ) ( ) ( ) ( )
{ }( ) ( ){ } ( ) ( ){ }
( )
*
1 1
* *
* *
2 2
20 0
1 1
2
2 1
2
1Re
2
1Re
21 1
Re Re2 2
2 2
1 incident power density
transmitted power density
incident power density
i r i r
i i r r
i i
z z
z z z z
z z z z
E E η η
η τ
η
= ×
⎡ ⎤⎡ ⎤= + × +⎣ ⎦ ⎣ ⎦
= × + ×
= − Γ
= − Γ ×
=
= ×
E H
E E H H
E H E H
Total average power density in medium 1, S 1
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Total electric field in medium 1:( ) ( ) ( )
( )( ) ( )[ ]( ) ( )[ ]
( )[ ] z je E
z je E
eee E
ee E
e E e E z z z
zi
z
i
z z z
i
z z
i
z
r
z
ir i
1 j-0
1
j-
0
j- j j-
0
j j-
0
j
0
-j
01
sin2ˆ
sin21ˆ
1ˆ
ˆ
ˆˆ
1
1
111
11
11
β Γ τ
β Γ Γ
Γ Γ
Γ
β
β
β β β
β β
β β
+=
++=
−++=
+=
+=+=
x
x
x
x
xxEEE
travelling wave
(amplitude: )0i E τ
standing wave(amplitude: )
02 i E Γ
The total electric field in medium 1 has local maximum and
minimum values but does not go to zero at any location.
|E1|
z
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See animation “Standing Wave (with Different Reflection
Coefficients)” to show the making of a general standing wave from
two traveling waves with different amplitudes (A1 and A2) moving
in opposite directions.
( )
tcoefficienreflectionA2/A1/
2Aˆ1Aˆ
ˆˆwaveresultantTotal
00
j j-
j
0
-j
01
11
11
===Γ
+=
+==
ir
z z
z
r
z
i
E E
ee
e E e E z
β β
β β
xx
xxE
Note that inside the website you can change the ratio of A2/A1 (i.e.,
the reflection coefficient) to investigate the effect on the resultant
standing wave.
Computer Animation of a Standing Wave
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Total magnetic field in medium 1:( ) ( ) ( )
( )( ) ( )[ ]
( )[ ] ze E
eee E
ee
E
z z z
zi
z z zi
z zi
r i
1
j-
1
0
j- j j-
1
0
j j-
1
0
1
cos2ˆ
1ˆ
ˆ
1
111
11
β Γ τ η
Γ Γ η
Γ η
β
β β β
β β
−=
+−+=
−=
+=
y
y
y
HHH
Note: The total fields in medium 2 are the transmitted
waves and they are pure travelling waves.
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Maxima and minima of EM fields:( ) )( ) ( )
( )[ ]θ β β β θ β
β β
−Γ+=
Γ+=
Γ+=
'2 j-' j
0
' j2- j' j
01
j2-j
01
11
11
11
1ˆ
1ˆ'
1ˆ
z z
i
z z
i
z z
i
ee E
eee E z
ee E z
x
xE
xE
( ) ( )
( ) ( )
( )[ ]θ β β
β θ β
β β
η
η
η
−Γ−=
Γ−=
Γ−=
' j2-
1
' j
0
' j2- j
1
' j
01
j2
1
-j
0
1
1
1
1
1
1
1
1ˆ
1ˆ'
1ˆ
z z
i
z z
i
z z
i
ee E
eee E
z
e
e E
z
y
yH
yH
θ
Γ Γ
j
e= z z −='
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( ) ( )θ β −
Γ+='2-j
01 11' z
i e E zE
( ) ( )θ β
η
−Γ−= ' j2-
1
0
111' z
ie
E zH
E1 achieves a maximum at z’ M when such that:
( ) Γ += 101 i M E ' zE
That is, when:
,2,1,0 ,2'2 1 ==− nn z M π θ β
( )1
'2-j 1 =−θ β M ze
magnitude
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real. bothareandlossless,aremediatheIf 21 η η
2 12 1
2 12 1
0, when
0, when
η η η η
η η η η
> >⎧−Γ = ⎨
< ⎧Γ = Γ ⎨
= <⎩
Therefore,
( )1 2 1
1 2 1
2 ' 2 , when2 ' 2 1 , when
M
M
z n z n
β π η η β π η η
= >⎧⎨ = +
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E1 achieves a minimum at z’m when such that:
( ) Γ −= 101 im E ' zE
Using the same analysis as for the maximum position,
( )1 2 11 2 1
2 ' 2 1 , when
2 ' 2 , whenm
m
z n
z n
β π η η
β π η η
⎧ = + >⎨ =
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Table for positions of the maxima and
minima of the EM field in medium 1
|E1
|max
|E1|min
( )( )Γ
Γ
−=
+=
1
1 :that Note
0min1
0max1
i
i
E
E
E
E
z’m z’ M z’
0
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The ratio of |E1|max to |E1|min is called the standing waveratio S :
Γ Γ
−+== 11
min1
max1
EES
1
1
+−=
S
S Γ
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Example 2
A beam of yellow light with a wavelength of 0.6 μm isnormally incident from air ( z > 0) on to a glass ( z < 0) . If
the glass surface is at the plane z = 0 and the relative
permittivity of glass is 2.25, determine:
(a) the locations of the electric field maxima in medium
1 (air),(b) the fraction of the incident power transmitted into
the glass medium.
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Solutions
(a) We determine the medium parameters,
( )
( )
8.012080
1602
2.012080
12080
Ω80
25.2
120~1
Ω 120
~
12
2
12
12
0
0
2
22
0
0
1
1
1
=+
=+
=
−=+−
=+−
=
=−==
−==
π π
π
η η
η τ
π π
π π
η η
η η Γ
π π
ε ε
μ
ε
μ η
π ε
μ
ε
μ
η
r
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Electric-field magnitude is a maximum at (with Γ < 0):
( )
m6.0with
,...2,1,0 24
'
1
11
μ λ
λ λ
=
=+= nn z M
(b) The fraction of the incident power transmitted into
the glass medium is
( ) 96%or96.02.011ely,Alternativ
96.080
1208.0
2/
2
22
2
2
12
1
2
0
2
2
02
2
2
=−=−=
===⎥⎥
⎦
⎤
⎢⎢
⎣
⎡=
Γ
η
η τ
η η τ
avi
av
ii
avi
av
P
P
E E
P
P
1 : Note avavi PP ≠