Download Oblique incidence: Interface between dielectric media

Electromagnetic Fields
Oblique incidence: Interface between dielectric media
Consider a planar interface between two dielectric media.
wave is incident at an angle from medium 1.
A plane
ƒ The interface plane defines the boundary between the media.
ƒ The plane of incidence contains the propagation vector and is
both perpendicular to the interface plane and to the phase planes
of the wave.
Medium 1
Plane of incidence
G
β
Phase
plane
θi
θr
Medium 2
Interface plane
θt
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ε1 = εr1 εo
µ1 = µr1 µo
ε2 = εr2 εo
µ2 = µr2 µo
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Electromagnetic Fields
There are two elementary orientations (polarizations) for the
electromagnetic fields:
ƒ Perpendicular Polarization
The electric field is perpendicular to the plane of incidence and
the magnetic field is parallel to the plane of incidence.
The fields are configured as in the Transverse Electric (TE)
modes.
ƒ Parallel Polarization
The magnetic field is perpendicular to the plane of incidence
and the electric field is parallel to the plane of incidence.
The fields are configured as in the Transverse Magnetic (TM)
modes.
Any plane wave with general field orientation can be obtained by
superposition of two waves with perpendicular and parallel
polarization.
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Electromagnetic Fields
Perpendicular (TE) polarization
Incident
wave
G
Ei
G
Hi
×
Reflected
wave
βz
G
Hr
G
βi
βx
Medium 1 ε1 = εr1 εo
µ1 = µr1 µo
Medium 2 ε2 = εr2 εo
y
Er
×
G
Ht
G
Et
µ2 = µr2 µo
θt ×
βr
×G
θr
θi
G
G
βt
z
Transmitted
wave
x
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Electromagnetic Fields
The electric field phasors for the perpendicular polarization, with
reference to the system of coordinates in the figure, are given by
G
− j βix ⋅ x − j βiz ⋅ z
Ei = E yi e
iˆy
G
− j β rx ⋅ x − j β rz ⋅ z
E r = E yr e
iˆy
G
− j βt x ⋅ x − j βt z ⋅ z
E t = E yt e
iˆy
The propagation vector components in medium 1 are expressed as
G
β i = βix2 + βiz2 = β 1= ω µ1ε1
βix = β 1cos θi
βiz = β 1sin θi
G
2
2
β r = β rx
+ β rz
= β1
β rx = − β 1cos θ r
β rz = β 1sin θ r
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Electromagnetic Fields
The propagation vector components in medium 2 are expressed as
G
β t = βtx2 + βtz2 = β 2= ω µ2ε 2
βtx = β 2 cos θt
βtz = β 2 sin θt
The magnetic field components can be obtained as
G
Hi =
G
Hr =
G
Ht =
G
G
βi × E i
ω µ1
=
G
G
βr × Er
ω µ1
ω µ2
η1
=−
G
G
βt × E t
E yi
=
− j βix x − j βiz z
ˆ
ˆ
( − sin θi ix + cosθi iz ) e
E yr
η1
E yt
η2
− j β rx x − j β rz z
ˆ
ˆ
( sin θr ix + cosθr iz ) e
− j βt x x − j βt z z
ˆ
ˆ
( − sin θt ix + cosθt iz ) e
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Electromagnetic Fields
Assuming that the amplitude of the incident electric field is given,
to completely specify the problem we need to find the amplitude of
reflected and transmitted electric field.
The boundary condition at the interface (x = 0) states that the
tangential electric field must be continuous. Because of the
perpendicular polarization, the tangential field is also the total field
x = 0)
E yi e
− j βi z z
+ E yr e
− j β rz z
= E yt e
− j βt z z
The relation above must be valid for any choice of “z” and we must
have (phase conservation law)
βiz = β rz = βtz
The first equality indicates that the reflected angle is the same as
the incident angle.
βiz = β rz ⇒ β1 sin θi = β1 sin θ r
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⇒ θ i = θr
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Electromagnetic Fields
The second equality provides the transmitted angle
βiz = βtz ⇒
⇒ θ t = sin
−1
β1 sin θi = β 2 sin θt
 µ1 ε1

sin θ i 

 µ 2 ε2



Snell's Law
Since we have also
e
− j βi z z
=e
− j β rz z
=e
− j βt z z
the boundary condition for the electric field becomes
E yi + E yr = E yt
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Electromagnetic Fields
The tangential magnetic field must also be continuous at the
interface. This applies in our case to the z−components
= Hzt
Ey i
Ey r
Ey t
cos θi −
cos θi =
cos θt
Hzi
η1
+
Hzr
η1
⇒ Ey i − Ey r
η2
η1 cos θt
Ey t
=
η2 cos θi
Solution of the system of boundary equations gives
Γ⊥ ( E ) =
E yr
E yi
η2 cos θi − η1 cos θt
=
η2 cos θi + η1 cos θt
E yt
2η2 cos θi
τ ⊥ (E) =
=
E yi η2 cos θi + η1 cos θt
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Reflection coefficient
Transmission coefficient
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Electromagnetic Fields
For the magnetic field, we can define the reflection coefficient as
H zr
Hr
Γ⊥ ( H ) =
=−
H zi
Hi
In terms of electric field, the magnetic field components are
H zr =
H zi =
− E yr
η1
E yi
η1
cos θi = − H r cos θi
cos θi = H i cos θi
The reflection coefficient for the magnetic field is then
Γ⊥ ( H ) =
−Ey r
Ey i
© Amanogawa, 2006 – Digital Maestro Series
η1 cos θt − η2 cos θi
= −Γ ⊥ ( E ) =
η2 cos θi + η1 cos θt
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Electromagnetic Fields
The transmission coefficient is defined as
Ht
τ ⊥ (H ) =
Hi
The magnetic field components are
Ht =
Hi =
Ey t
η2
Ey i
η1
The transmission coefficient for the magnetic field is then
H t η1
2η1 cos θi
τ ⊥ (H ) =
= τ ⊥ (E) =
H i η2
η2 cos θi + η1 cos θt
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Electromagnetic Fields
Parallel (TM) polarization
Incident
wave
G
Ei
G
Hi
×
βz
Reflected
wave
G
βx
G
Hr
βi
Medium 1 ε1 = εr1 εo
µ1 = µr1 µo
y
Medium 2 ε2 = εr2 εo
×
θr
θi
µ2 = µr2 µo
×
βr
G
Er
z
θt × G
G
Et
G
HG t
Transmitted
wave
βt
x
© Amanogawa, 2006 – Digital Maestro Series
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Electromagnetic Fields
The magnetic field phasors for the parallel polarization are given by
G
− j βix ⋅ x − j βiz ⋅ z
iˆy
Hi = H yi e
G
− j β rx ⋅ x − j β rz ⋅ z
iˆy
H r = H yr e
G
− j βt x ⋅ x − j βt z ⋅ z
iˆy
H t = H yt e
and the electric field components can be obtained as
G
Ei = −
G
Er
G
Et
G
G
βi × Hi
− j βix x − j βiz z
ˆ
ˆ
= η1H yi ( sin θi ix − cos θi iz ) e
ωε1
G G
− j β rx x − j β rz z
βr × Hr
=−
= η1H yr ( sin θ r iˆx + cos θ r iˆz ) e
ωε1
G G
− j βt x x − j βt z z
βt × H t
ˆ
ˆ
=−
= η2 H yt ( sin θt ix − cos θt iz ) e
ωε 2
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Electromagnetic Fields
Also for parallel polarization one can verify that the same
relationships between angles apply, as found earlier for the
perpendicular polarization, including Snell’s law
θi = θ r
θ t = sin
−1
 µ1 ε1

sin θ i 

 µ 2 ε2



We have again two boundary conditions at the interface. One
condition is for continuity of the tangential magnetic field
H yi + H yr = H yt
© Amanogawa, 2006 – Digital Maestro Series
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Electromagnetic Fields
A second condition is for continuity of the tangential electric field
E zi
+
E zr
=
E zt
−η1 cos θi H yi + η1 cos θi H yr = −η2 cos θt H yt
η2 cos θt
⇒ H yi − H yr =
H yt
η1 cos θi
From the equations provided by the boundary conditions we obtain
the reflection and transmission coefficients for the magnetic field of
a wave with parallel polarization as
H yr
η1 cos θi − η2 cos θt
Γ& ( H ) =
=
H yi η1 cos θi + η2 cos θt
H yt
2η1 cos θi
τ& (H ) =
=
H yi η1 cos θi + η2 cos θt
© Amanogawa, 2006 – Digital Maestro Series
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Electromagnetic Fields
The reflection coefficient for the electric field is defined as
E zr
Er
Γ& ( E ) =
=−
E zi
Ei
The tangential components of the electric field can be expressed in
terms of magnetic field as
E zr = Er cos θi = η1 cos θi H yr
E zi = − Ei cos θi = −η1 cos θi H yi
The reflection coefficient for the electric field is
E zr
H yr
η2 cos θt − η1 cos θi
Γ& ( E ) =
=−
= −Γ& ( H ) =
η1 cos θi + η2 cos θt
E zi
H yi
© Amanogawa, 2006 – Digital Maestro Series
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Electromagnetic Fields
The transmission coefficient for the electric field is defined as
τ& (E) =
Et
Ei
The electric field components are given by
E t = η2 H yt
Ei = η1H yi
The transmission coefficient for the electric field becomes
Et η2 H yt η2
τ& (E) = =
= τ& (H )
Ei η1H yi η1
2η1 cos θi
2η2 cos θi
η2
=
=
η1 η1 cos θi + η2 cos θt η1 cosθi + η2 cosθt
© Amanogawa, 2006 – Digital Maestro Series
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Electromagnetic Fields
Considerable simplifications are possible for the common case of
nonmagnetic dielectric media with
µ1 = µ2 = µo
First of all, Snell’s law becomes
θ t = sin
−1
 µ1 ε1


−1  ε1
sin θ i  = sin 
sin θ i 

 µ 2 ε2


 ε2


or, equivalently
sin θi
ε 2 n2
=
=
sin θt
ε1 n1
(n = index of refraction )
Snell’s law provides then a useful recipe to express the reflection
and transmission coefficients only with angles, thus eliminating the
explicit dependence on medium impedance.
© Amanogawa, 2006 – Digital Maestro Series
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Electromagnetic Fields
After some trigonometric manipulations, we obtain the following
table of simplified coefficients for electric and magnetic field
sin (θi − θt )
Γ ⊥ ( E ) = −Γ ⊥ ( H ) = −
sin (θi + θt )
tan (θi − θt )
Γ& ( E ) = −Γ& ( H ) = −
tan (θi + θt )
ε1 2sin θt cos θi
τ ⊥ (E) = τ ⊥ (H )
=
ε 2 sin (θi + θt )
ε1
2sin θt cos θi
τ& (E) = τ& (H )
=
ε 2 sin (θi + θt ) cos (θi − θt )
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Electromagnetic Fields
Power flow
The time-average power flow normal to the interface must be
continuous. We can express this as
1 Ei2
1 Er2
1 Et2
cos θi −
cos θi =
cos θt
2 η1
2 η1
2 η2
incident power − reflected power = transmitted power
We define the reflection and transmission coefficients for the timeaverage power as
R=
reflected power
T=
transmitted power
incident power
=
incident power
© Amanogawa, 2006 – Digital Maestro Series
Er2
Ei2
Et2 η1 cos θt
= 2
Ei η2 cos θi
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Electromagnetic Fields
The following conversion formulas relate power and electric field
coefficients
R = Γ2 ( E )
η1 cos θt
T = τ (E)
η2 cos θi
2
Note that reflection and transmission coefficients for the timeaverage power are always real positive quantities. The following
power conservation condition is always verified
R +T =1
Since the power flow normal to the interface is considered, the
results obtained above apply equally to perpendicular and parallel
polarization.
© Amanogawa, 2006 – Digital Maestro Series
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Electromagnetic Fields
Non−magnetic perfect dielectric media
Case ε 2 > ε1
G
βi
Medium 1 ε1 = εr1 εo
G
θi
θi
µ1 = µo
Medium 2 ε2 = εr2 εo
y
µ2 = µo
×
G
θt
βt
βr
z
x
From Snell’s law
ε1 sin θi = ε 2 sin θt
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ε 2 > ε1 ⇒ θt < θi
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Electromagnetic Fields
Since θ
t
< θ i there is always a transmitted (refracted) beam.
The transmission coefficients are always positive
ε1 2sin θt cos θi
τ ⊥ (E) = τ ⊥ (H )
=
>0
ε 2 sin (θi + θt )
ε1
2sin θt cos θi
τ& (E) = τ& (H )
=
>0
ε 2 sin (θi + θt ) cos (θi − θt )
⇒ transmitted and incident wave are in phase at the boundary.
τ ⊥ (E) =
© Amanogawa, 2006 – Digital Maestro Series
E yt
E yi
E yi
E yt
×
×
H zi
H zt
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Electromagnetic Fields
Perpendicular polarization
sin (θi − θt )
Γ⊥ ( E ) = −
sin (θi + θt )
sin (θi − θt )
Γ⊥ ( H ) =
sin (θi + θt )
ƒ The reflection coefficient for the electric field is always negative
E yi and E yr have always phase difference of 180°
ƒ The reflection coefficient for the magnetic field is always positive
H zi and H zr are always in phase
Γ⊥ ( E ) =
E yr
E yi
Γ⊥ ( H ) =
© Amanogawa, 2006 – Digital Maestro Series
H zr
H zi
E yi
×
H zi
E yr H zr
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Electromagnetic Fields
Parallel polarization
tan (θi − θt )
Γ& ( E ) = −
tan (θi + θt )
tan (θi − θt )
Γ& ( H ) =
tan (θi + θt )
When θi + θt < 90° ⇒ tan (θi + θt ) > 0
E zi and E zr have phase difference of 180°
H yi and H yr are in phase
Ezi H yi
×
H yr Ezr
×
When θi + θt > 90° ⇒ tan (θi + θt ) < 0
E zi and E zr are in phase
H yi and H yr have phase difference of 180°
Ezi H yi Ezr
×
© Amanogawa, 2006 – Digital Maestro Series
H yr
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Electromagnetic Fields
When θi + θt = 90° ⇒ tan (θi + θt ) → ∞
the reflection coefficients vanish (TOTAL TRANSMISSION)
tan (θi − θt )
tan (θi − θt )
Γ& ( E ) = −
→ 0 ; Γ& ( H ) =
→0
tan (θi + θt )
tan (θi + θt )
For θi + θt = 90° ⇒ θi = θ B (Brewster angle)
and ⇒ cosθ B = sin θt
From Snell’s law
sin θ B sin θ B
ε2
−1 ε 2
=
= tan θ B =
⇒ θ B = tan
ε1
ε1
sin θt cos θ B
© Amanogawa, 2006 – Digital Maestro Series
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Electromagnetic Fields
Case ε 2 < ε1
G
βi
Medium 1 ε1 = εr1 εo
G
θi
θi
µ1 = µo
Medium 2 ε2 = εr2 εo
y
µ2 = µo
βr
G
×
βt
z
θt
x
From Snell’s law
ε1 sin θi = ε 2 sin θt
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ε 2 < ε1 ⇒ θt > θi
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Electromagnetic Fields
For angles of incidence such that
sin θt < 1
we have for perpendicular polarization
sin (θi − θt ) sin (θt − θi )
Γ⊥ ( E ) = −
=
>0
sin (θi + θt ) sin (θi + θt )
E yi and E yr are always in phase
sin (θi − θt )
sin (θt − θi )
Γ⊥ ( H ) =
=−
<0
sin (θi + θt )
sin (θi + θt )
H zi and H zr have always phase difference of 180°
E yi
×
© Amanogawa, 2006 – Digital Maestro Series
H zi
H zr E yr
×
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Electromagnetic Fields
For parallel polarization
tan (θ i − θ t ) tan (θ t − θ i )
Γ& ( E ) = −
=
tan (θ i + θ t ) tan (θ i + θ t )
tan (θ i − θ t )
tan (θ t − θ i )
Γ& ( H ) =
=−
tan (θ i + θ t )
tan (θ i + θ t )
When θi + θt < 90° ⇒ tan (θi + θt ) > 0
H
E
yi
zi
E zi and E zr are in phase
×
Ezr
H yr
H yi and H yr have phase difference of 180°
When θi + θt > 90° ⇒ tan (θi + θt ) < 0
E zi and E zr have phase difference of 180°
E zi H yi H yr Ezr
H yi and H yr are in phase
×
×
© Amanogawa, 2006 – Digital Maestro Series
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Electromagnetic Fields
When θi + θt = 90° ⇒ tan (θi + θt ) → ∞
Also in this case the reflection coefficients vanish and we have
TOTAL TRANSMISSION
tan (θt − θi )
tan (θt − θi )
Γ& ( E ) =
→ 0 ; Γ& ( H ) = −
→0
tan (θi + θt )
tan (θi + θt )
Total transmission occurs again, for parallel polarization only, at
the Brewster angle
θi = θ B = tan
© Amanogawa, 2006 – Digital Maestro Series
−1
ε2
ε1
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Electromagnetic Fields
When
ε2
ε2
sin θi =
sin θt =
⇒ sin θt = 1 ⇒ θt = 90°
ε1
ε1
we have a limit condition for TOTAL REFLECTION, valid for both
polarizations. This particular angle of incidence is called
critical angle
θi = θc = sin
−1
ε2
ε1
θc
θ t = 90°
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Electromagnetic Fields
For angles of incidence beyond the critical angle
ε1
>1
θi > θc ⇒ sin θt = sin θi
ε2
⇒ cos θt = imaginary
± , choose "− "
⇓
cos θt = − 1 − sin 2 θt = − j
ε1
ε2
2
sin θi −
ε2
ε1
The negative sign is selected, in order to obtain the proper wave
vector in medium 2, as shown later.
© Amanogawa, 2006 – Digital Maestro Series
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Electromagnetic Fields
The reflection and transmission coefficients become complex
Γ ⊥ ( E ) = −Γ ⊥ ( H ) =
cos θi + j sin 2 θi − ε 2 ε1
cos θi − j sin 2 θi − ε 2 ε1
ε2
− cos θi − j sin 2 θi − ε 2 ε1
ε1
Γ& ( E ) = −Γ& ( H ) =
ε2
cos θi − j sin 2 θi − ε 2 ε1
ε1
ε1
2cos θi
τ ⊥ (E) = τ ⊥ (H )
=
ε 2 cos θ − j sin 2 θ − ε ε
i
i
2 1
ε1
2 ε 2 / ε1 cos θi
τ& (E) = τ& (H )
=
ε 2 ε 2 cos θ − j sin 2 θ − ε ε
i
i
2 1
ε1
© Amanogawa, 2006 – Digital Maestro Series
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Electromagnetic Fields
If we consider the coefficients for time-average power flow, we
have, for both polarizations
R = Γ ( E ) ⋅ Γ* ( E ) = 1
T = 1− R = 0
This means that incident and reflected waves carry the same timeaverage power, and no power is transmitted to medium 2. But this
does not mean that the field disappears in medium 2. The
instantaneous power that enters medium 2 is eventually reflected
back to medium 1.
The electric field phasor of the transmitted wave has the form
E t = Et e
− j βt x ⋅ x − j βt z ⋅ z
e
© Amanogawa, 2006 – Digital Maestro Series
= Et e− j β 2 cosθt ⋅ x e− j β 2 sin θt ⋅ z
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Electromagnetic Fields
The wave vector components are

ε1
ε2 
2
sin θi − 
β 2 cos θt = ω µoε 2  − j
ε2
ε1 

ε2
ε2
2
= − jω µoε1 sin θi −
= − j β1 sin θi −
= − jα t
ε1
ε1
2
ε1
sin θi = β1 sin θi = βi z
β 2 sin θt = ω µoε 2
ε2
The field in medium 2 corresponds to a surface wave, moving along
the z−direction and exponentially decaying (evanescent) along the
x−direction
E t = Et e
− j ( − jαt ⋅ x ) − j βi z ⋅ z
© Amanogawa, 2006 – Digital Maestro Series
e
= Et e
−α t ⋅ x − j βi z ⋅ z
e
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Electromagnetic Fields
The surface wave moves parallel to the surface, with a phase
velocity equal to the apparent phase velocity along z of the incident
wave in medium 1
v p2 =
v p1
sin θi
= v pz > v p1
For the surface wave, planes of constant amplitude are parallel and
planes of constant phase are normal to the interface. These planes
do not coincide, therefore the surface wave is a nontransverse
wave.
θi > θc
Constant amplitude planes
Et
Constant phase planes
x
© Amanogawa, 2006 – Digital Maestro Series
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Electromagnetic Fields
If you consider a beam incident on the interface, it is found that the
power is totally reflected but after penetrating for some distance
into medium 2. The reflected beam emerges displaced by a
distance D (called Goos-Hänchen shift, discovered in 1947)
D
θi > θc
From experiments, the displacement is found to be
D ≈ 0.52 ε 2
2π
ε2
β1 sin θi −
ε1
© Amanogawa, 2006 – Digital Maestro Series
2
= 0.52 ε 2
2π
αt
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Electromagnetic Fields
Examples:
Medium 1
θB = ?
µ = µo
air
ε = εo
Medium 2
water
−1 ε 2
θ B = tan
ε1
ε =
{
µ = µo
80ε o microwaves
1.8ε o
optical
θ B = tan −1 80 ≅ 83.6° microwaves
θ B = tan −1 1.8 ≅ 53.3° optical
© Amanogawa, 2006 – Digital Maestro Series
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Electromagnetic Fields
At the Brewster angle
θi + θt = θ B + θt = 90°
6.38°
⇒ θt ≅ 
36.7°
microwaves
optical
Verification with Snell’s law
ε1 sin θ B = ε 2 sin θt
θt = sin
−1  sin θ B 
© Amanogawa, 2006 – Digital Maestro Series


6.38°
≅
ε 2  36.7°
microwaves
optical
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Electromagnetic Fields
Medium 1
water
θB = ?
ε =
{
µ = µo
80ε o microwaves
1.8ε o
optical
Medium 2
θ B = tan
−1
ε2
ε1
air
µ = µo
ε = εo
θ B = tan
−1
1
≅ 6.38° microwaves
80
θ B = tan
−1
1
≅ 36.7° optical
1.8
© Amanogawa, 2006 – Digital Maestro Series
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Electromagnetic Fields
At the Brewster angle
θi + θt = θ B + θt = 90°
83.6°
⇒ θt ≅ 
53.3°
microwaves
optical
Verification with Snell’s law
ε1 sin θ B = ε 2 sin θt
−1  sin θ B ε 2  83.6°
θt = sin 
≅
ε o  53.3°

© Amanogawa, 2006 – Digital Maestro Series
microwaves
optical
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Electromagnetic Fields
Medium 1
θc = ?
µ = µo
air
ε = εo
Medium 2
θc = sin
−1
ε2
ε1
water
ε =
{
µ = µo
80ε o microwaves
1.8ε o
optical
The total reflection angle does not exist since
ε 2 > ε1
© Amanogawa, 2006 – Digital Maestro Series
205
Electromagnetic Fields
Medium 1
µ = µo
water
θc = ?
θc = sin
−1
ε2
ε1
ε =
{
80ε o microwaves
1.8ε o
optical
Medium 2
air
µ = µo
ε = εo
θc = sin
−1
1
≅ 6.4193° microwaves
80
θc = sin
−1
1
≅ 48.19°
1.8
© Amanogawa, 2006 – Digital Maestro Series
optical
206
Electromagnetic Fields
Medium 1
θ i = 60°
µ = µo
ε = 4ε o
Medium 2
air
µ = µo
ε = εo
Consider a perpendicularly polarized wave.
ƒ Find the Brewster angle and the critical angle:
θ B = tan
θc = sin
−1 1
4
−1 1
© Amanogawa, 2006 – Digital Maestro Series
4
= tan
= sin
−1  1 
  ≈ 26.565°
2
−1  1 
  = 30°
 2
207
Electromagnetic Fields
ƒ Find the components of the incident propagation vector and of
the x-component of the transmitted propagation vector in terms
of
βo = ω µoε o
βix
βiz
2βo
= β1 cos θi = ω µo 4ε o cos 60° =
= βo
2
3
= β1 sin θi = ω µo 4ε o sin 60° = 2 β o
= 3β o
2
βtx = βt2 − βtz2
βtx = ±
N
βt = β o
βtz = βiz = 3β o
β o2 − 3β o2 = − j 2β o = − jα t
choose
"−"
© Amanogawa, 2006 – Digital Maestro Series
208
Electromagnetic Fields
ƒ In the second medium, find the distance at which the field
strength is 1/e of that at the interface
1
1
d=
=
αt
2β o
ƒ What is the value of the magnitude of the reflection coefficient at
the interface?
The reflection coefficient is a complex quantity when the incident
angle exceeds the critical angle. Because of total reflection we
know that it must be
Γ⊥ ( E ) = 1
since the time-average power reflection coefficient is
R = Γ⊥ ( E ) 2 = 1
© Amanogawa, 2006 – Digital Maestro Series
209