Chapter 3
Plane Waves
This chapter describes the behavior of plane waves at dielectric interfaces, as well as the reflection of plane
waves at multiple layers.
3.1 Plane waves at dielectric interfaces
An optical fiber is shown in fig. 3.1. It consists of a fiber core with the refraction index n1 and a core
cladding with the refraction index n2 < n1 . In such an optical fiber the wave is guided by the constant total
reflection. For typical dimensions of the optical fiber ( e.g. fiber diameter 125 µm) there is a reflection e.g.
every millimeter as shown in fig. 3.1. Hence, there are about 106 reflexions per kilometer. With assumed
additional losses of less than 0.1 dB/km , the reflectivity per reflection has to be greater than 99, 999998% ,
so that ’total’ reflection is really required. What should the waveguide properties be in order to ensure this
amount of total internal reflection?
T o ta lre fle x io n
n
2
n
L ic h ts tra h l
1
A u s b re itu n g s ric h tu n g (z )
Figure 3.1: Principle of a waveguide with n2 < n1
To investigate the total reflection, the reflection of a plane wave at a dielectric boundary, in accordance with
fig. 3.2, should be considered in detail. The vectors ⃗k1 , ⃗k1′ and ⃗k2 , shown in fig. 3.2, describe the wave
⃗ is assumed perpendicular to the
vector of the incident, reflected and transmitted wave. In this example, E
1
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plane of incidence. Therefore the wave vectors are equal to:
( )2 ( )2
⃗k1 = ⃗k1′ = k 2 n2
0 1
( )2
⃗k2 = k 2 n2
0 2
(3.1)
For example: The vector ⃗k1 has an x- and a z-component. Thus:
 
kx1

⃗k1 =  0 

kz
2
2 + k 2 = k 2 n2 . Furthermore it is:
with ⃗k1 = kx1
z
0 1
(3.3)
(3.2)
kz
kz
cos(θ1 ) = =
k0 n1
⃗k1 k
H
1 '
k
q
j
H
2
E
E
H
k
(3.4)
j
2
2
1
q
1
1
E
n
n
1
2
z
y
x
Figure 3.2: A plane wave incident on a boundary surface
The components of the incident wave yield
E y = E y1 exp (−j kx1 x − j kz z)
sin (θ1 )
ZF 1
cos(θ1 )
H x = −E y
ZF 1
H z = Ey
(3.5)
(3.6)
(3.7)
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The characteristic wave impedances on both sides of the interface are
√
√
1 µ0
1 µ0
ZF 1 =
, ZF 2 =
n1 ε 0
n2 ε0
3
(3.8)
At the interface ( x = 0 ) the tangential field components must be continuous. ⇒ The z-dependent components of the incident, reflected and transmitted waves have to be equal.
⇒ Hence, the z-components of ⃗k1 , ⃗k1′ and ⃗k2 are equal (kz ).
For the transmitted wave there is ⃗kz = cos(θ2 ) = k0knz 2 and therefore yields
|k2 |
This is Snell’s law:
kz = k0 n1 cos(θ1 ) = k0 n2 cos(θ2 )
(3.9)
n1
cos(θ2 )
sin(φ2 )
=
=
n2
cos(θ1 )
sin(φ1 )
(3.10)
If cos(θ1 ) > nn21 or sin(φ1 ) > nn21 , there won’t be a real solution in eq. (3.9) for a real angle θ2 . The wave no
longer penetrates into medium 2. Total reflection occurs. The critical angle for total reflection is:
sin(φ1g ) = cos(θ1g ) =
n2
n1
(3.11)
As long as it is θ1 < θ1g and hence φ1 > φ1g , total reflection takes place.
3.1.1
Determination of the reflection coefficient for plane waves at dielectric interfaces
E
′
The reflection coefficient rE is defined as rE = Ey1 . The index E of the reflection coefficient rE
y1
specifies in this case that the E-field is polarized perpendicular to the plane of incidence. The demand for
the continuity of the tangential field components at the interface yields the following boundary conditions
E y1 + E y1′ = E y2
(3.12)
H z1 + H z1′ = H z2
(3.13)
The indexing 1, 1 ’and 2 refer to the incident, the reflected and the transmitted waves. In both media, wave
impedances are now introduced:
√
√
E y1
E y1′
µ0
k0 µ0
1
Z1E =
=
(3.14)
=−
=
H z1
H z1′
n1 sin(θ1 ) ε0
kx1 ε0
√
√
E y2
µ0
k0 µ0
1
Z2E =
=
(3.15)
=
H z2
n2 sin(θ2 ) ε0
kx2 ε0
Now the components of the H-field in eq. (3.13) can be expressed by the components of the E-field. For
that eq. (3.14) and eq. (3.15) have to be inserted into eq. (3.13):
E y1
Z2E
Z2E
− E y1′
= E y2
Z1E
Z1E
(3.16)
Inserting this expression into eq. (3.12) and transforming it, the reflection factor rE results in
rE =
n1 cos(φ1 ) − n2 cos(φ2 )
Z2E − Z1E
n1 sin(θ1 ) − n2 sin(θ2 )
=
=
Z2E + Z1E
n1 sin(θ1 ) + n2 sin(θ2 )
n1 cos(φ1 ) + n2 cos(φ2 )
(3.17)
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Making use of Snell’s Law (eq. (3.10)) with n1 = n2 sin(φ2 )/ sin(φ1 ) , eq. (3.17) can be transformed to:
rE =
sin(φ2 ) cos(φ1 ) − cos(φ2 ) sin(φ1 )
sin(φ2 − φ1 )
=
sin(φ2 ) cos(φ1 ) + cos(φ2 ) sin(φ1 )
sin(φ2 + φ1 )
(3.18)
In Analogy for a wave with the H-field perpendicular to the incident plane the reflection factor rH can be
determined. Firstly, this result leads to the following field components of the incident wave:
H y = H y1 exp (−j kx1 x − j kz z)
(3.19)
E z = −H y · sin (θ1 ) · ZF 1
(3.20)
E x = H y · cos(θ1 ) · ZF 1
(3.21)
with the characteristic wave impedances
Z1H
Z2H
With rH =
E z1′
E z1
H
√
√
E z1
E z1′
sin(θ1 ) µ0
kx1
µ0
=−
=
=
= 2
H y1
H y1′
n1
ε0
n1 · k0 ε0
√
√
sin(θ2 ) µ0
kx2
E
µ0
= 2
= − z2 =
H y2
n2
ε0
n2 · k0 ε0
(3.22)
(3.23)
′
= − Hy1 it results:
rH =
y1
n1 sin(θ2 ) − n2 sin(θ1 )
n1 cos(φ2 ) − n2 cos(φ1 )
Z2H − Z1H
=
=
Z2H + Z1H
n1 sin(θ2 ) + n2 sin(θ1 )
n1 cos(φ2 ) + n2 cos(φ1 )
(3.24)
With Snell’s Law (3.10) eq. (3.24) is redefined to:
rH =
cos(φ2 + φ1 ) · sin(φ2 − φ1 )
cos(φ2 − φ1 ) · sin(φ2 + φ1 )
(3.25)
For the given refraction numbers n1 and n2 the reflection factors can now be determined as a function of the
incident angle. Fig. 3.3 shows two examples.
In fig. 3.3 the so called Brewster-angle φ1B is represented, wherein rH = 0 results for the reflexion factor
φ1 = φ1B . Corresponding to eq. (3.25), φ1B results when φ1 + φ2 = π/2 . Hence, in this case, the
reflected and transmitted beams are perpendicular to each other. Thus:
tg(φ1B ) =
n2
n1
(3.26)
The Brewster angle is utilized e.g. for anechoic laser resonators (Fig. 3.4).
For n1 = 1 (air) it is simply tg(φ1B ) = n2 .
⃗
Example: For n2 = 1.5 the Brewster angle results in φ1B = 56.3◦ . Since only the polarization with H
perpendicular to the incident plane penetrates into the gas discharge chamber losslessly, the emitted laser
beam is linearly polarized.
√
For φ1 > φ1g the wave number kx2 = k0 · n22 − n21 sin2 (φ1 ) , as well as Z2H becomes imaginary.
⇒ |rE | = |rH | = 1
(3.27)
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Figure 3.3: Magnitude of the amplitude reflection coefficients at an interface of glass/air, shown as a function
of the incident angle φ. (a) the incident wave proceeds in air, (b) in glass.
S p ie g e l
G a s e n tla d u n g s ra u m
j
1 B
Figure 3.4: Schematic of the laser resonator
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Assuming a total reflection, the field in region 2 (refraction index n2 ) decreases exponentially with the
penetration depth
1
λ
dx =
= √
(3.28)
j · kx2
2π n2 sin2 (φ ) − n2
For
√
1
1
2
n21 − n22 = 0.2 and φ1 = 90◦ , the result is for example dx = 0.8 · λ .
3.2 Reflection on multi layers
The reflection of a plane wave at multiple layers in the x-direction is examined, where each layer has a
different refractive index. m layers each with a refraction number ni ( i = 1...m ) are assumed (fig. 3.5).
z
j
x
1
n
1
n
2
n
3
n
n
d
i
m
i
Figure 3.5: m-Layers with the refraction numbers ni and the thicknesses di
If the incident angle φ1 is given, the x-component of the wave number in the i-th layer kxi , based on the
constant wave component in z-direction kz = k0 n1 sin(φ1 ) , will be determined as:
√
kxi = k0 n2i − n21 sin2 (φ1 )
(3.29)
and the characteristic wave impedances ZiE and ZiH , as shown in eq. (3.14), eq. (3.15), eq. (3.22) and eq.
(3.23), are
√
k0 µ0
ZiE =
(3.30)
kxi ε0
√
kxi
µ0
ZiH =
.
(3.31)
2
ni k0 ε0
Since the wave resistance of the layers can be calculated directly, it is possible to develop an equivalent
electronic circuit (fig. 3.6).
In order to calculate the reflection coefficient at the first layer, the terminating resistor has to be transformed
to the input (e.g. with the Smith chart, see lecture Hochfrequenztechnik I for further details). Hence the
problem is solved in general. A particularly simple and special case is λ4 -layers.
Accordingly it applies for λ4 -layers: kxi · di = π2 . In this case it is possible to simplify to following
expression:
2
ZiE
(3.32)
Zai =
Za(i+1)
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Z
1 E
Z
k
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Z
2 E
x 2
Z
iE
k
a
x i
d
k
(m -1 )E
Z
7
m E
x (m -1 )
i
B e re c h n u n g d e s R e fle x io n s fa k to rs
in d ie s e r E b e n e
⃗ perpendicular to the incident plane
Figure 3.6: equivalent electronic circuit for E
d i= p / ( 2 k x i)
Z
a i
Figure 3.7: equivalent resistance for λ/4-layers
to transform step-by-step the resistance Za(i+1) a layer forward to Zai (fig. 3.7). With multiple λ/4 -layers
this transformation can be applied recursively, leading to Za . Then the reflection factor can be calculated by
rE =
Za − Z1E
.
Za + Z1E
(3.33)
As an example, a dielectric layer can be determined, so that the transition from medium 1 to medium 3 is
completely non-reflective, i.e. the wave couples without reflection into the dielectric with n3 (fig. 3.8). For
simplicity, a perpendicular wave incidence (φ1 = 0) is assumed. In this case, the wave impedances are
given as:
√
1 µ0
ZiE = ZiH = ZF i =
(3.34)
ni ε 0
ZF 2 and ZF 3 are merged to the equivalent impedance Za .
√
ZF2 2
n3 µ0
Za =
(3.35)
= 2
ZF 3
n2 ε 0
Hence it results Za = ZF 1 for rE = rH = 0. And furthermore
√
√
ZF 2 = ZF 1 · ZF 3 ⇒ n2 = n1 · n3
(3.36)
Also the thickness of the layer can be determined, since following relation applies:
d2 · kx2 = d2 · k0 · n2 = d2 · n2
2π ! π
=
λ
2
(3.37)
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s e n k re c h t e in fa lle n d e
e b e n e W e lle
n
n
1
2
n
3
d 2= l /(4 n 2)
Z
Z a= Z
F 2
2
Z
F 1
/Z
F 2
Z
F 3
F 3
B e re c h n u n g d e s R e fle x io n s fa k to rs
in d ie s e r E b e n e
Figure 3.8: e.g.: λ/4-layer for anti-reflection
Hence the thickness of the layer can be evaluated as
d2 =
λ
4 · n2
(3.38)