Lecture 19
Non-Normal Incidence of Waves at Interfaces
In this lecture you will learn:
• What happens when waves strike an interface between two different media
coming at an angle
• Reflection and transmission of waves at interfaces
• Application of E-field and H-field boundary conditions
• Total internal reflection
• Brewster’s angle
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
Waves at Interfaces – TE and TM Waves
εi
r
ki
µo
Transverse Electric
(TE) wave
Transverse Magnetic
(TM) wave
x
εt
µo
Ei
Hi
r
ki
Ei
Hi
z=0
z
r
Plane of Incidence: The plane containing the incident wavevector k i and a vector that
is normal to the interface is called the plane of incidence (in the figure above the x-z
plane is the plane of incidence)
TE Wave: If the E-field of the wave is perpendicular to the plane of incidence then the
wave is called a TE-wave
TM Wave: If the H-field of the wave is perpendicular to the plane of incidence then the
wave is called a TM-wave
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
1
TE Wave - Wavevectors
r
kr
Transverse Electric
(TE) wave
x
Hr
r
kt
Et
Er
θr
θt
Ht
θi
εi
r
ki
µo
εt
Ei
µo
Hi
z
z=0
r
ki = kix xˆ + kiz zˆ = k i [sin(θ i ) xˆ + cos(θ i ) zˆ ]
r
k r = k rx xˆ + k rz zˆ = k r [sin(θ r ) xˆ − cos(θ r ) zˆ ]
r
kt = ktx xˆ + ktz zˆ = kt [sin(θt ) xˆ + cos(θt ) zˆ ]
ki = ω
n
µo ε i = ω i
c
kr = ki = ω
kt = ω
µo ε i = ω
nt
c
µo ε t = ω
ni
c
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
TE Wave – First Boundary Condition
r
kr
Transverse Electric
(TE) wave
Hr
x
r
kt
Et
Er
θr
θt
Ht
θi
εi
µo
r
ki
εt
Ei
µo
Hi
z=0
r r
r r
r r
E (r )
= yˆ Ei e − j k i . r + yˆ Er e − j k r . r
z <0
r r
r r
E (r )
= yˆ Et e − j kt . r
z>0
Use boundary conditions:
z
r
k i = k i [sin(θ i ) xˆ + cos(θ i ) zˆ ]
r
k r = k r [sin(θ r ) xˆ − cos(θ r ) zˆ ]
r
kt = kt [sin(θt ) xˆ + cos(θt ) zˆ ]
(1) At z = 0 the E-field parallel to the interface must be continuous across the
interface for all x
This gives:
Ei e − j k i sin(θ i ) x + E r e − j k r sin(θ r )x = Et e − j kt sin(θt )x
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
2
TE Wave – Phase Matching Condition
x
Hr
r
kr
Er
Transverse Electric
(TE) wave
r
kt
Et
θr
θt
Ht
θi
εi
µo
r
ki
εt
Ei
µo
Hi
z
z=0
Ei e − j k i sin(θ i ) x + E r e − j k r sin(θ r )x = Et e − j kt sin(θt )x
The only way the above boundary condition can be satisfied for all x is if all the xdependent phase factors are the same (this is called “phase matching”)
ki sin(θ i ) = kr sin(θ r ) = kt sin(θt )
kix = krx = ktx
The first equality gives ( using ki = kr ) :
sin(θ i ) = sin(θ r )
⇒
θi = θr
angle of incidence equals the
angle of reflection
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
TE Wave – Snell’s Law
Hr
r
kr
x
Er
Transverse Electric
(TE) wave
r
kt
Et
θr
θt
Ht
θi
εi
µo
r
ki
εt
Ei
µo
Hi
z=0
ki sin(θ i ) = kr sin(θ r ) = kt sin(θt )
z
k ix = k rx = ktx
The second equality gives:
k i sin(θ i ) = kt sin(θt )
n
n
⇒ ω i sin(θ i ) = ω t sin(θt )
c
c
⇒ ni sin(θ i ) = nt sin(θt )
E i e − j k i sin(θ i )x + E r e − j k r sin(θ r )x = Et e − j kt sin(θ t )x
⇒ E i + E r = Et
Snell’s Law
(1)
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
3
TE Wave – Second Boundary Condition
r
kr
Transverse Electric
(TE) wave
x
Hr
r
kt
Et
Er
θr
θt
Ht
θi
εi
r
ki
µo
εt
Ei
µo
Hi
z
z=0
(2) At z = 0 the H-field component parallel to the interface must be continuous for all x
r r
H (r )
r r
H (r )
r
r
r
r
E
E
= kˆi × yˆ i e − j k i . r + kˆr × yˆ r e − j k r . r
z <0
ηi
ηi
(
z >0
)
(
)η
E
= kˆt × yˆ t e
)
t
− xˆ cos(θ i )
⇒
(
r r
− j kt . r
Ei
ηi
r
k i = k i [sin(θ i ) xˆ + cos(θ i ) zˆ ]
r
k r = k r [sin(θ r ) xˆ − cos(θ r ) zˆ ]
r
kt = kt [sin(θt ) xˆ + cos(θt ) zˆ ]
E
E
e − j k i sin(θ i )x + xˆ cos(θ r ) r e − j k r sin(θ r )x = − xˆ cos(θt ) t e − j kt sin(θt )x
ηi
ηt
⎡E E ⎤
E
cos(θ i ) ⎢ i − r ⎥ = cos(θt ) t
ηt
⎣ ηi ηi ⎦
(2)
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
TE Wave – Reflection and Transmission Coefficients
r
kr
Transverse Electric
(TE) wave
Hr
x
r
kt
Et
Er
θr
θt
Ht
θi
εi
µo
r
ki
εt
Ei
µo
Hi
z=0
The solution is:
z
k
n cos(θt )
η cos(θt )
2 iz
2 i
2 t
Et
ktz
nt cos(θ i )
ηi cos(θ i )
T =
=
=
=
Ei ηt cos(θt ) + 1 k iz + 1 ni cos(θt ) + 1
Transmission
ktz
nt cos(θ i )
ηi cos(θ i )
coefficient
Reflection
coefficient
ηt cos(θt )
−1
E r ηi cos(θ i )
Γ=
=
=
Ei ηt cos(θt ) + 1
ηi cos(θ i )
k iz
ni cos(θt )
−1
−1
ktz
n cos(θ i )
= t
k iz
ni cos(θt )
+1
+1
ktz
nt cos(θ i )
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
4
TM Wave - Wavevectors
x
r
kr
Transverse Magnetic E
r
(TM) wave
r
kt
Et
Hr
θr
Ht
θt
θi
εi
r
ki
Ei
µo
εt
Hi
µo
z
z=0
r
k i = k ix xˆ + k iz zˆ = k i [sin(θ i ) xˆ + cos(θ i ) zˆ ]
r
k r = k rx xˆ + k rz zˆ = k r [sin(θ r ) xˆ − cos(θ r ) zˆ ]
r
kt = ktx xˆ + ktz zˆ = kt [sin(θt ) xˆ + cos(θt ) zˆ ]
ki = ω
n
µo ε i = ω i
c
kr = ki = ω
kt = ω
µo ε i = ω
µo ε t = ω
nt
c
ni
c
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
TM Wave – First Boundary Condition
x
r
kr
Transverse Magnetic E
r
(TM) wave
θr
r
kt
Et
Hr
Ht
θt
θi
εi
µo
r
ki
Ei
εt
Hi
z=0
r r
r r
r r
H (r )
= yˆ Hi e − j k i . r + yˆ Hr e − j k r . r
z<0
r r
r r
H (r )
= yˆ Ht e − j kt . r
z >0
Use boundary conditions:
µo
z
r
k i = k i [sin(θ i ) xˆ + cos(θ i ) zˆ ]
r
k r = k r [sin(θ r ) xˆ − cos(θ r ) zˆ ]
r
kt = kt [sin(θt ) xˆ + cos(θt ) zˆ ]
(1) At z = 0 the H-field parallel to the interface must be continuous across the
interface for all x
This gives:
Hi e − j k i sin(θ i ) x + Hr e − j k r sin(θ r ) x = Ht e − j kt sin(θt )x
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
5
TM Wave – Phase Matching Condition
x
r
kr
Transverse Magnetic E
r
(TM) wave
r
kt
Et
Hr
θr
Ht
θt
θi
εi
µo
r
ki
Ei
εt
Hi
µo
z
z=0
Hi e − j k i sin(θ i ) x + Hr e − j k r sin(θ r ) x = Ht e − j kt sin(θt )x
The only way the above boundary condition can be satisfied for all x is if all the xdependent phase factors are the same (this is called “phase matching”)
k i sin(θ i ) = k r sin(θ r ) = kt sin(θt )
k ix = k rx = ktx
The first equality gives ( using ki = kr ) :
sin(θ i ) = sin(θ r )
θi = θr
⇒
angle of incidence equals the
angle of reflection
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
TM Wave – Snell’s Law
x
r
kr
Transverse Magnetic E
r
(TM) wave
θr
r
kt
Et
Hr
θt
Ht
θi
εi
µo
r
ki
Ei
εt
Hi
z=0
k i sin(θ i ) = k r sin(θ r ) = kt sin(θt )
µo
z
k ix = k rx = ktx
The second equality gives:
k i sin(θ i ) = kt sin(θt )
n
n
⇒ ω i sin(θ i ) = ω t sin(θt )
c
c
⇒ ni sin(θ i ) = nt sin(θt )
Hi e − j k i sin(θ i )x + Hr e − j k r sin(θ r )x = Ht e − j kt sin(θt )x
⇒ Hi + Hr = Ht
Snell’s Law
(1)
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
6
TM Wave – Second Boundary Condition
x
r
kr
Transverse Magnetic E
r
(TM) wave
r
kt
Et
Hr
θr
Ht
θt
θi
εi
r
ki
Ei
µo
εt
Hi
µo
z
z=0
(2) At z = 0 the E-field component parallel to the interface must be continuous for all x
r r
E (r )
(
)
(
)
r
r
(
r
)
r
= − kˆi × yˆ ηi Hi e − j k i . r − kˆr × yˆ ηi Hr e − j k r . r
z <0
r
k i = k i [sin(θ i ) xˆ + cos(θ i ) zˆ ]
r r
r r
r
E (r )
= − kˆt × yˆ ηt Ht e − j kt . r
k r = k r [sin(θ r ) xˆ − cos(θ r ) zˆ ]
z >0
r
kt = kt [sin(θt ) xˆ + cos(θt ) zˆ ]
xˆ cos(θ i )ηi Hi e − j k i sin(θ i ) x − xˆ cos(θ r )ηi H r e − j k r sin(θ r ) x = xˆ cos(θt )ηt Ht e − j kt sin(θ t )x
⇒
cos(θ i )(ηi H i − ηi Hr ) = cos(θt )ηt Ht
(2)
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
TM Wave – Reflection and Transmission Coefficients
x
r
kr
Transverse Magnetic E
r
(TM) wave
r
kt
Et
Hr
θr
θt
Ht
θi
εi
µo
r
ki
Ei
εt
Hi
The solution is:
T
Transmission
coefficient
Reflection
coefficient
TM
z=0
µo
z
η cos(θt )
ε k
n cos(θt )
2 t iz
2 i
2 t
Ht
ηt cos(θ i )
ε i ktz
ni cos(θ i )
=
=
=
=
Hi ηi cos(θt ) + 1 ε t k iz + 1 nt cos(θt ) + 1
ni cos(θ i )
ηt cos(θ i )
ε i ktz
ηi cos(θt )
−1
H
η cos(θ i )
=
ΓTM = r = t
Hi ηi cos(θt ) + 1
ηt cos(θ i )
ε t kiz
nt cos(θt )
−1
−1
ε i ktz
n cos(θ i )
= i
ε t kiz
nt cos(θt )
+1
+1
ε i ktz
ni cos(θ i )
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
7
Snell’s Law
ni sin(θ i ) = nt sin(θt )
ni
nt
θi
θt
θi
• If ni < nt then θt < θi and the transmitted wave bends towards the normal
ni
nt
θi
θt
θi
• If ni > nt then θt > θi and the transmitted wave bends away from the normal
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
Total Internal Reflection – Critical Angle
If ni > nt then θt > θi
ni
nt
If θi is increased, then θt
will eventually become 90o
θi
θt
θi
The value of θi for which θt
is 90o is called the critical
angle θc
ni sin(θ i ) = nt sin(θt )
⇒
⎛π ⎞
ni sin(θc ) = nt sin⎜ ⎟
⎝2⎠
⇒
n
sin(θc ) = t
ni
What if θi is increased beyond θc ?
When θi is increased beyond θc the wave is not transmitted but is completely
(100%) reflected at the interface back into the medium of incidence
This phenomenon is called total internal reflection – it happens for both TE and TM
waves
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
8
Total Internal Reflection – Phase Matching
ni > nt and θi > θc
ni
We need to consider in more
detail what happens when the
angle of incidence is greater
than the critical angle
nt
θt
θi
θi
k ix = k rx = ktx
The phase matching condition gives us:
⇒
ktx = k ix = k i sin (θ i )
⇒
2
= k i2 sin2 (θ i )
ktx
=
ω2
c2
ni2 sin2 (θ i )
We also know the dispersion relation for medium “t ”:
kt =
⇒
ω
c
kt2 =
⇒
nt
2
2
ktx
+ ktz
=
ω2
c2
ω2
c2
nt2
nt2
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
Total Internal Reflection – Evanescent Wave
ni > nt and θi > θc
ni
θi
If θi is larger than θc the
wave in medium “t” is
evanescent in the z-direction
⇒
ω2
c
2
c
2
nt2 − ktx
=
ktz =
ω
c
ω2
c2
[nt2 − ni2 sin2 (θi )]
nt2 − ni2 sin2 (θ i )
-ve when θi > θc
For θi > θc one can write:
ω
nt
θi
2
The previous two equations imply: ktz
=
ktz = − j
Et
The z-component of the
wavevector has become
completely imaginary
''
ni2 sin2 (θ i ) − nt2 = − j ktz
The E-field (assuming TE wave) in medium “t ” is then:
r r
E (r )
z >0
= yˆ Et e − j ktx
x
''
e − ktz z
The field is evanescent in the
z-direction in medium “t ”
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
9
Total Internal Reflection – Evanescent Wave
ni > nt and θi > θc
ni
nt
Et
θi
θi
The E-field (assuming TE wave) in medium “t ” is:
r r
E (r )
z >0
⇒
= yˆ Et e − j ktx
r r
E (r , t )
x
''
e − ktz z = yˆ Et e j φ e − j ktx
x
''
e − ktz z
''
z >0
= yˆ Et e − ktz z cos(ω t − ktx x + φ )
The wave is propagating along the interface (in the x-direction) but decaying
(without spatial oscillations) in the z-direction
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
Total Internal Reflection – Reflection Coefficient
ni > nt and θi > θc
If θi is larger than θc the
wave is completely reflected
back into the medium of
incidence
ni
Et
θi
nt
θi
For θi > θc one can write:
ktz = − j
ω
c
''
ni2 sin2 (θ i ) − nt2 = − j ktz
The z-component of the
wavevector has become
completely imaginary
The reflection coefficient for the E-field (assuming TE wave) is:
k iz
−1
''
E r ktz
k − ktz k iz + jktz
Γ=
=
= iz
=
= e jϕ
''
k iz
Ei
+ 1 k iz + ktz k iz − jktz
ktz
⇒
The phase ϕ of the reflection
coefficient Γ in total internal
reflection is called the GoosHanschen phase-shift
Γ =1
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
10
TE and TM Waves: Reflection Coefficients
TE Waves
TM Waves
ε t k iz
−1
H
ε i ktz
TM
r
Γ
=
=
Hi ε t k iz + 1
ε i ktz
k iz
−1
Er
ktz
Γ=
=
k iz
Ei
+1
ktz
Question: Can one ever get the reflection coefficient to go to zero (very desirable to
get rid of unwanted reflections in optics)?
n
ki = ω i
c
ki
n
kt = ω t
c
kt
k ix
k iz
k i ≠ kt
ktx
ktz
( if
ni ≠ nt
But k ix = ktx
⇒ k iz ≠ ktz
⇒
)
(Phase matching)
Γ is never zero for TE waves
But one can have Γ=0 for TM waves if:
ε t kiz = ε i ktz
⇒ nt cos(θ i ) = ni cos(θt )
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
TM Waves: Brewster’s Angle
For θi = θB
One can have Γ=0 for TM
waves if:
ni
nt
θi
nt cos(θ i ) = ni cos(θt )
θt
θi
Snell’s law gives:
ni sin(θ i ) = nt sin(θt )
The above two equations will have a solution if and only if:
sin(θ i ) = cos(θt )
and
This happens when: θ i + θt =
cos(θ i ) = sin(θt )
π
2
The angle of incidence for which this happens is called the Brewster’s angle θB :
n
tan(θ i ) = t
ni
⇒
⎡ nt ⎤
⎥
⎣ ni ⎦
θ B = tan−1 ⎢
ECE 303 – Fall 2005 – Farhan Rana – Cornell University
11