EECS 117
Lecture 24: Oblique Incidence on
Dielectrics
University of California, Berkeley
Oblique Incidence (TM waves)
Ei
Er
Hi
Hr
ki
kr
θi
θr
n1,ε1,µ1
n2,ε2,µ2
x
z
θt
kt
Ht
Et
TE Waves (cont.)
A transverse magnetic (TM) wave, also so called the Ppolarized wave, has the electric field oscillating in the plane of
incidence and the magnetic field out of the plane. The plane of
incidence is defined to be the one made by the incident wave
vector (propagation direction vector) and the line normal to the
surface of discontinuity.
In general, a field (either electric or magnetic) of a wave can be
described by
r
r
r
jk • r ˆ
F ( x, z ) = F e
f
r
r
ˆ is a unit vector along the field F ,
k
where
is
the
wave
vector,
f
r
r
and r is a general space vector, i.e., r = xxˆ + yyˆ + zzˆ
Incident, Reflected and Transmitted Waves
For the incident electric field,
r r
r
r
jk i • r
jn1 kkˆi • r
Ei ( x, z ) = Ei e
eˆi = Ei e
eˆi
= Ei e
jn1 k sin θ i x
e
jn1 k cos θ i z
(cosθ i xˆ − sin θ i zˆ )
= Ei x xˆ + Ei z zˆ
k is the wave number of the field in vacuum, and k1 =
For the reflected and transmitted fields,
ω
v
=
ω
c / n1
r
Er ( x, z ) = Er e jn1k sin θ r x e − jn1k cosθ r z (− cos θ r xˆ − sin θ r zˆ )
r
Et ( x, z ) = Et e jn2 k sin θt x e jn2 k cosθt z (cos θ t xˆ − sin θ t zˆ )
= n1k
Boundary Conditions
Applying the boundary condition for tangential electric field at
z = 0, we have
Eix + Erx = Et x
Ei cos θ i e jn1k sin θi x − Er cos θ r e jn1k sin θ r x = Et cos θ t e jn2 k sin θt x
Just like in the case of oblique incidence on a perfect
conductor, this condition needs to be satisfied independent of x.
This is possible by equating the exponents and coefficients
separately:
(1)
n1k sin θ i = n1k sin θ r = n2 k sin θ t
Ei cos θ i − Er cos θ r = Et cos θ t
(2)
Law of Reflection and Refraction
The first equality in (1) gives the law of reflection, i.e.,θ i = θ r
and the second equality in (2) gives the Snell’s law, namely,
n1 sin θ i = n1 sin θ r = n2 sin θ t
For most of the materials, which are non-magnetic, µ1= µ2= µ0.
Then,
sin θ t n1
ε 1µ1
ε1
ε 1µ 2 η 2
=
=
≈
≈
=
sin θ i n2
ε 2 µ2
ε2
ε 2 µ1 η1
Total Reflection
If n1 > n2 or η1 < η 2 , we have total reflection for incident
angle larger than the critical angle θc , where
 n2 sin π / 2 
ε2
−1
−1 η1
 ≈ sin
θ c = sin 
≈ sin
n1
ε1
η2


−1
For θi > θc
ε1
sin θ t =
sin θ i > 1
ε2
ε1 2
cos θ t = 1 − sin θ t = ± j
sin θ i − 1
ε2
2
Total Reflection (cont.)
The transmitted electric and magnetic fields are then equal to:
r
jn 2 k sin θ t x jn 2 k cos θ t z
(cosθt xˆ − sin θt zˆ )
Et ( x, z ) = Et e
e
= Et e jβ tx x e jα tz z (cosθ t xˆ − sin θ t zˆ )
r
H t ( x, z ) = H t e jn 2 k sin θ t x e jn 2 k cos θ t z yˆ = H t e jβ tx x e jα tz z yˆ
where
ε1
ε1 2
β tx = n2 k
sin θ i , α tz = n2 k
sin θ i − 1
ε2
ε2
The wave propagates in x direction only!! It does penetrate
into medium 2, with its amplitude decaying exponentially inside
the medium. This surface wave is called an evanescent wave.
Reflection and Transmission Coefficients
The equation (2) obtained from applying the boundary condition
to the tangential electric field can be used to find the reflection
and transmission coefficients.
But we need one more piece of information, which we can get
from applying the tangential magnetic field boundary condition:
Hi e
jn1k sin θ i x
+ Hr e
jn1k sin θ r x
= Ht e
jn2 k sin θ t x
Hi + Hr = Ht
We know that H and E are related to each other through
impedance of the medium. Thus, we have
H i = − Ei / η1 , H r = − Er / η1 , H t = − Et / η 2
Coefficients (cont.)
We need the negative signs
r in
r the equations because with our
definition of direction of Er , Hrmust be negative according to
the right hand rule so that E × H is pointing the same direction as
does.
Now we have two simultaneous equations:
Ei cos θ i − Er cos θ r = Et cos θ t
Ei / η1 + Er / η1 = Et / η 2
Solving these two equations, we can expressions for the
reflection and transmission coefficients.
ΓTM
Er η1 cosθ i − η 2 cosθ t
=
=
Ei η1 cosθ i + η 2 cosθ t
Coefficients (cont.)
τ TM
Et η 2
2η 2 cos θ i
=
= (1 + ΓTM ) =
Ei η 2
η1 cos θ i + η 2 cos θ t
We can also define the coefficients in terms of the tangential
components of the electric fields. They have expressions:
ΓZ TM =
τZ
TM
=
E rx
Ei x
Et x
Ei x
− Er cosθ r η 2 cosθ t − η1 cosθ i
=
=
η1 cosθi + η 2 cosθt
Ei cosθ i
Et cosθ t
2η 2 cosθ t
=
=
Ei cosθ i η1 cosθ i + η 2 cosθ t
Brewster Angle (TM Waves)
There is a situation that we don’t have reflection off the boundary,
when the numerator of ΓTM is equal to 0, or η1 cos θ i = η 2 cos θ t
The incident angle that this happens is called Brewster angle. It can
be shown that
sin θ BTM =
2
1
1 + (η 2 /η1 )
2
1 − µ 2ε1 / µ1ε 2
1
=
2 ≈
1 + ε1 / ε 2
1 − (ε1 / ε 2 )
or
θB
TM
≈ tan −1 ε1 / ε 2 ≈ tan −1 n1 / n2
Oblique Incidence (TE Waves)
For a TE wave, which has the electric field oscillating out of the
plane of incidence and magnetic field parallel to the plane, we can
obtain the same set of quantities as in TM waves quickly by swapping
E and H: E → H , H → E with some cautions.
In the TM case, we applied the boundary condition to the tangential
electric field. Here we need to apply the condition to the tangential
magnetic field. Because both conditions require the field to be
continuous across the boundary, we have
H i cosθi e jn1k sin θ i x − H r cosθ r e jn1k sin θ r x = H t cosθt e jn 2 k sin θ t x
which also yield the laws of reflection and refraction. If the incident
angle is larger than the critical angle (of course n1 > n2), the
transmitted wave is evanescent.
Boundary Conditions for TE Waves
Analogous to the TM case, we obtain this from the boundary
condition for tangential magnetic fields:
H i cosθ i − H r cosθ r = H t cosθ t
Just like what we had in TM case, we also have this equality
due to the boundary condition for tangential electric fields:
Ei e
jn1 k sin θ i x
+ Er e
jn1 k sin θ r x
= Et e
jn 2 k sin θ t x
Ei + Er = Et
The electric and magnetic fields are related:
H i = − Ei / η1 , H r = − Er / η1 , H t = − Et / η 2
Boundary Conditions (cont.)
So we have these two simultaneous equations in the TE case:
(E
i
cosθ i − Er cosθ r ) /η1 = Et /η 2 cosθ t
Ei + Er = Et
These two equations yield the expressions of reflection and
transmission coefficients:
τ TE
Er η 2 cosθ i − η1 cosθ t
ΓTE =
=
Ei η 2 cosθ i + η1 cosθ t
Et
2η 2 cosθ i
=
= 1 + ΓTE =
Ei
η 2 cosθ i + η1 cosθ t
Since the electric fields are tangential to the boundary,
ΓZ TE = ΓTE , τ Z TE = τ TE
Brewster Angle (TE Waves)
We can also define Brewster angle just like before. In this case,
we have
sin θ BTM
2
1 − µ1ε 2 / µ 2ε1
=
2
1 − (µ1 / µ 2 )
For most of the materials which have µ1= µ2= µ0, this angle
usually does not exist.
Impedance for Waves at Oblique Incidence
Since in many cases the tangential components of the fields are
continuous across the boundary, it is useful to define a
characteristic wave impedance referred to the z direction (as in
our setup) in terms of the components in planes transverse to that
direction. as the ratio of electric to magnetic field components in
planes parallel to the boundary. In other words, in the TM case,
ηz
TM
Ex +
Ex −
=
=−
= η cosθ
H y+
H y−
ηz = −
TE
Ey +
H x+
=
Ey −
H x−
= η secθ
+ and – refer, respectively, to incident and reflected wave. The
sign of the ratio is chosen for each wave to yield a positive
impedance.
Reflection and Transmission Revisit
If we substitute these new definitions into the reflection and
transmission coefficient expressions, we can see that these
expressions are just what we learned in the transmission line.
ΓZ TM
τZ
η 2 cosθt − η1 cosθi η 2, Z − η1, Z
=
=
η1 cosθi + η 2 cosθt η1, Z + η 2, Z
2η 2, Z
2η 2 cosθ t
=
=
η1 cosθ i + η 2 cosθ t η1, Z + η 2, Z
TM
TM
TM
TM
TM
TM
TM
ΓZ TE
TM
η 2 cosθ i − η1 cosθ t η 21, Z − η1, Z
= ΓTE =
=
η 2 cosθi + η1 cosθt η1, Z + η 2, Z
TE
TE
τ Z = τ TE
TE
2η 2, Z TE
2η 2 cosθ i
=
=
η 2 cosθ i + η1 cosθ t η 2, Z TE + η1, Z TE
TE
TE
Reflection and Transmission Revisit (cont.)
In general,
η 2, Z − η1, Z
ΓZ =
η1, Z + η 2, Z
τZ
2η 2, Z
=
η1, Z + η 2, Z
The impedance defined by the fields’ tangential components is
analogous to the impedance in transmission lines.