Jaesang Lee
Dept. of Electrical and Computer Engineering Seoul National University
(email: [email protected])
Electromagnetics
<Chap. 8> Plane Electromagnetic waves Section 8.9 ~ 8.10
(1st of week 5)
Textbook: Field and Wave Electromagnetics, 2E, Addison-Wesley
Chap. 8 | Contents for 1
stclass of week 5
Review of the last class
Sec 9. Normal incidence at multiple dielectric interfaces
• Wave impedance
• Impedance transformation
Chap. 8 | Review (Terminologies)
Definition
• “Polarization” direction of EM wave = Direction of E-field
• Plane of incidence
- Plane that is perpendicular to the surface
- Plane that contains a vector of propagation direction
• Transverse Electric (TE) wave - E-field ⊥ Plane of incidence
• Transverse Magnetic (TM) wave - H-field ⊥ Plane of incidence
x
y z
Er
Hr
anr
Ei
ani
Hi
Plane of incidence
(xz plane)
Transverse Electric (TE) wave (E ⊥ plane of incidence)
x
y z
Er
Hr
anr
Ei
ani
Hi
Plane of incidence Transverse Magnetic (TM) wave
(H ⊥ plane of incidence) TEM wave propagation
(In transmission lines)
Chapter. 9
TE & TM wave propagation (In waveguides)
Chapter. 10
(Image source: http://www.uta.edu/optics/research/ellipsometry/ellipsometry.htm)
Chap. 8 | Review (TE wave)
x
y z
Er
Hr
anr
Ei
ani
Perfect conductor
Medium 1 (σ1 = 0)
z = 0
Medium 2 (σ2 = ∞) Hi
θr
θi
Transverse Electric (TE) wave
• Boundary Condition
→ [total E-field in Medium 1 at z = 0] = [total E-field in Medium 2 at z = 0]
E
1( ) x,0 = E
i( ) x,0 + E
r( ) x,0
⎡⎣ ⎤⎦ = ⎡⎣ E
2( ) x,0 = 0 ⎤⎦
→ a
y( E
i0e
− jβ1xsinθi+ E
r0e
− jβ1xsinθr) = 0
• Snell’s law of reflection
∴ E
r0= − E
i0, θ
r= θ
i Snell’s law of reflection Phase is shifted by 180º• Microscopic interpretation of “metallic reflection”
- Boundary condition for H-field at z = 0
a
n2× ( H
1− H
2) = J
sJ
s= a
y2 E η
i01
cos θ
ie
− jβ1xsinθi- induced Js results in the reflected wave in medium 1
- Microscopically, incident wave absorbed by free electrons - Oscillating electrons (Js) re-radiate EM wave
- Reflected wave cancels the incident wave in the wall
Chap. 8 | Review (Reflection and Transmission)
Two different dielectric media
x
y z
Er
Hr
anr
Ei
ani
Medium 1 (ε1, μ1)
z = 0
Medium 2 (ε2, μ2) Hi
Et
ant
Ht
E
r0= η
2− η
1η
2+ η
1E
i0→ Γ =
E
r0E
i0= η
2− η
1η
2+ η
1• Er0 and Et0 in terms of Ei0
E
t 0= 2 η
2η
2+ η
1E
i0→ τ = E
t0E
i0= 2 η
2η
2+ η
1Reflection Coefficient
Transmission Coefficient
• Relationship between Γ and τ
∴ 1 + Γ = τ
• Complex Γ and τ (i.e. complex η1 and η2)
→ phase shift introduced upon transmission and reflection
• If medium 2 is a perfect conductor (i.e. η2 =0)
→ Γ = –1 → Er0 = –Ei0
→ τ = 0 → Et0 = 0
P
av1= 1
2 Re ( E
1× H
1*) = a
zE 2 η
i021
1 − Γ
2( )
⎡
⎣ ⎢ ⎤
⎦ ⎥ = P
av2= 1
2 Re ( E
2× H
2*) = a
zE 2 η
i022
τ
2⎡
⎣ ⎢ ⎤
⎦ ⎥
• EM energy
(If both media are lossless)
P
av1= P
av2→ ∴ 1 − Γ
2= η
1η
2τ
2Chap. 8 | Examples of Multiple Dielectric layers
Practical use of “several layers of dielectric media”
“Anti-reflective (AR)” coating on the lens minimizes the ambient reflection
Radome (Radar + dome)
permits EM waves through enclosure with little reflection
Determining proper dielectric materials (μ, ε) and their thicknesses
→
Design & engineering problems
(Image source: All about vision) (Image source: Wikipedia)
Chap. 8 | Total reflection in a stack of three dielectric media (1/2)
Total reflection in a stack of three dielectric media
x
z Er
Hr
anr
Ei
ani
Medium 1 (ε1, μ1)
z = 0 Hi
E2+
an2+
H2+
z = d E2-
H2-
an2-
Et
ant
Ht
Medium 2 (ε2, μ2)
Medium 3 (ε3, μ3)
E
r0= η
2− η
1η
2+ η
1E
i0 For a “two layer” situation (Last class or see Sec. 8-8)E
r0≠ η
2− η
1η
2+ η
1E
i0 For a “three layer” situation(∵multiple reflections in medium 2 at z = 0 and d)
Our interest!
Er
infinite multiple reflections!
• How do we determine Er0?
• One option to obtain Er0
→ By using boundary conditions!
Chap. 8 | Total reflection in a stack of three dielectric media (2/2)
E
1= E
i+ E
r= a
x( E
i0e
− jβ1z+ E
r0e
jβ1z)
H
1= H
i+ H
r= a
y1
η
1E
i0e
−jβ1z
− E
r0e
jβ1z( )
⎧
⎨ ⎪
⎩ ⎪
• Total electric and magnetic fields in each medium
E
2= E
2++ E
2−= a
x( E
2+e
− jβ2z+ E
2−e
jβ2z)
H
2= H
2++ H
2−= a
y1
η
1E
2+e
−jβ2z
− E
2−e
jβ2z( )
⎧
⎨ ⎪
⎩ ⎪
∵ H = a
n× E η
Medium 1
Medium 2
E
3= a
xE
3e
− jβ3zH
3= a
yE
3η
3e
− jβ3z⎧
⎨ ⎪
⎩⎪
Medium 3
x
z Er
Hr
anr
Ei
ani
Medium 1
z = 0 Hi
E2+
an2+
H2+
z = d E2-
H2-
an2-
Et
ant
Ht
Medium 2 Medium 3
4 unknowns
(E
r0, E
2+, E
2-, E
3) 4 B.C. equations
At z = 0 : E
1( ) 0 = E
2( ) 0 H
1( ) 0 = H
2( ) 0
⎧ ⎨
⎪
⎩⎪
At z = d : E
2( ) d = E
3( ) d H
2( ) d = H
3( ) d
⎧ ⎨
⎪
⎩⎪
Chap. 8 | Wave impedance (1/2)
Wave impedance of the total field
Z z ( ) ! Total E
x( ) z
Total H
y( ) z ( ) Ω
: Ratio of the total electric field intensity to the total magnetic field intensity
Why do we care about the impedance?
• In Chapter 9 <Theory and Applications of Transmission Lines>
“Transmission line lumped element equivalent circuit”
Transmission line can be modeled as circuit elements Transmission Line:
A passage with a certain geometry and components (μ, ε, σ)
where EM wave travels through
Parallel plate Two-wire Coaxial
Wave impedance vs. intrinsic (characteristic) impedance
• Intrinsic (characteristic) impedance (η or Z0)
- Used for a single wave propagating in an “unbounded” medium - i.e. there is no reflected wave
- equivalent to infinitely long transmission line
• Wave impedance (Z)
- Used for waves propagating across many different media where reflection occurs
Chap. 8 | Wave impedance (2/2)
x
Er
Hr
anr
Ei
ani
Medium 1
z = 0
Medium 2
Hi
Et
ant
Ht
z
E
1( ) z = a
xE
1x( ) z = a
x( E
i0e
− jβ1z+ E
r0e
jβ1z) = a
xE
i0( e
− jβ1z+ Γ e
jβ1z)
H
1( ) z = a
yH
1y( ) z = a
yE
i0η
1e
−jβ1z
− E
r0η
1e
jβ1z
⎛
⎝⎜
⎞
⎠⎟ = a
yE
i0η
1e
−jβ1z
− Γ e
jβ1z( )
⎧
⎨ ⎪
⎩ ⎪
∵ E
r0E
i0= Γ
Z
1( ) z = E
1x( ) z
H
1y( ) z = η
1e
− jβ1z
+ Γ e
jβ1ze
− jβ1z− Γ e
jβ1zExample: Two dielectrics situation
• Wave impedance • Wave impedance at z = -l
Z
1( ) − l = η
1e
jβ1l+ Γ e
− jβ1le
jβ1l− Γ e
− jβ1l= η
1η
2cos β
1l + j η
1sin β
1l η
1cos β
1l + j η
2sin β
1l
∵ Γ = η
2− η
1η
2+ η
1• If η2 = η1
Z
1( ) − l = η
1∵ No reflected wave involved
→ Wave impedance of total field = intrinsic impedance
η2
η1
Z
1( ) − l
Wave impedance at z = -l,
Wave impedance in medium 2
• Total field in medium 2
- Result of multiple reflections at z = 0 and z = d
- Grouped into a wave traveling in +z and other in -z directions
Chap. 8 | Impedance transformation (1/2)
x
z Er
Hr
anr
Ei
ani
Medium 1
Hi
z = d
Medium 2 Medium 3
Z
1( ) − l = η
1η
2cos β
1l + j η
1sin β
1l
η
1cos β
1l + j η
2sin β
1l
(Wave impedance in medium 1 for two dielectrics)Z
2( ) 0 = η
2η
3cos β
2d + j η
2sin β
2d η
2cos β
2d + j η
3sin β
2d
η
2→ η
3, η
1→ η
2β
1→ β
2, l → d
• Equivalent situation for the wave traveling in medium 1:
- It encounters a discontinuity at z = 0
- Discontinuity → Unbounded medium with an intrinsic impedance Z2(0)
Z
2( ) 0
Wave impedance at z = 0,
Medium 1 “Unbounded” medium characterized by Z2(0)
η2 η3
η1
Z2(0) Ei
ani
Hi
Er
Hr
anr
• Wave impedance in medium 2 at z = 0
Chap. 8 | Impedance transformation (2/2)
Impedance transformation
Γ
0= Z
2( ) 0 − η
1Z
2( ) 0 + η
1Γ = η
3− η
1η
3+ η
1Without medium 2
Z
2( ) 0 = η
2η
3cos β
2d + j η
2sin β
2d η
2cos β
2d + j η
3sin β
2d
where
• Transforming η3 into Z2(0)
- With suitable choices of d and η2
• In many applications, Γ0 and Ero are the only quantities of interest
E
r0= Γ
0E
i0 whereΓ
0= Z
2( ) 0 − η
1Z
2( ) 0 + η
1“Effective” reflection coefficient at z = 0
Γ
0= E
r0E
i0= Z
2( ) 0 − η
1Z
2( ) 0 + η
1Γ = η
a− η
bη
a+ η
bc.f.) for two dielectrics a, b
Medium 1 “Unbounded” medium characterized by Z2(0)
Z2(0) Ei
ani
Hi
Er
Hr
anr
x
z Er
Hr
anr
Ei
ani
Medium 1
Hi
z = d
Medium 2 Medium 3
Z
2( ) 0
Wave impedance at z = 0,
η2 η3
η1
With medium 2
Chap. 8 | Impedance transformation Example (1/2)
Q. 8-12: To obtain zero “effective” reflection coefficient at z = 0, What should be d and η
2?
Γ
0= E
r0E
i0= Z
2( ) 0 − η
1Z
2( ) 0 + η
1= 0 Z
2( ) 0 = η
2η
3cos β
2d + j η
2sin β
2d
η
2cos β
2d + j η
3sin β
2d = η
1η
2( η
3cos β
2d + j η
2sin β
2d ) = η
1( η
2cos β
2d + j η
3sin β
2d ) η
3cos β
2d = η
1cos β
2d ! (1)
η
22sin β
2d = η
1η
3sin β
2d ! (2)
Equating real and imaginary parts separately,
For Equation (a) to be satisfied,
η
3cos β
2d = η
1cos β
2d η
1= η
3! case A
cos β
2d = 0 ! case B
Chap. 8 | Impedance transformation Example (2/2)
Q. 8-12: To obtain zero “effective” reflection coefficient at z = 0, What should be d and η
2?
1. Case A (η1 = η3)
Equation (2) becomes
η
22sin β
2d = η
12sin β
2d
η
1= η
2= η
3 : Trivial case (Not interesting)sin β
2d = 0 → d = n π
β
2= n
λ
22 , n = 0,1,2...
Dielectric 2 is called half-wave dielectric window Narrow-band device??
η
3cos β
2d = η
1cos β
2d η
1= η
3! case A
cos β
2d = 0 ! case B
η
2( η
3cos β
2d + j η
2sin β
2d ) = η
1( η
2cos β
2d + j η
3sin β
2d ) η
3cos β
2d = η
1cos β
2d ! (1)
η
22sin β
2d = η
1η
3sin β
2d ! (2)
2. Case B (η1 ≠ η3)
cos β
2d = 0 → d = ( 2 n + 1 ) π
β
2= ( 2 n + 1 ) λ
24 , n = 0,1,2...
And
sin β
2d ≠ 0.
∴ from Equation (2), we getη
22= η
1η
3or η
2= η
1η
3Dielectric 2 is called quarter-wave impedance transformer
Jaesang Lee
Dept. of Electrical and Computer Engineering Seoul National University
(email: [email protected])
Electromagnetics
<Chap. 8> Plane Electromagnetic waves Section 8.9 ~ 8.10
(2nd of week 5)
Textbook: Field and Wave Electromagnetics, 2E, Addison-Wesley
Chap. 8 | Contents for 2nd week of week 5
Sec 10. Oblique incidence at a plane dielectric interfaces
• Snell’s law of reflection & refraction
• Total internal reflection
• Brewster’s angle: TM vs. TE
Chap. 8 | Snell’s law
x
O z
Medium 1 (ε1, μ1)
z = 0
Medium 2 (ε2, μ2) θr
θi
B θt
O’
A’
A
AO, A ′ O ′ , B O ′ :
Intersections of the wave fronts (equi-phase surface) of incident, reflected, transmitted waves• Snell’s Law of Reflection
A O ′ = O A ′ :
Length of the trajectory of incident wave = reflected wave (∵Same phase velocity)→ ⎡⎣ A O ′ = O O ′ cos 90 (
!− θ
r) ⎤⎦ = ⎡⎣ O A ′ = O O ′ cos 90 (
!− θ
i) ⎤⎦
• Snell’s Law of Refraction
t
OB! "!!= t
! "A!!!O′:
Time for transmitted wave to travel from O to B= Time for incident wave to travel from A to O’
∴ θ
i= θ
r→ t
OB! "!!= OB
u
p2⎡
⎣ ⎢
⎢
⎤
⎦ ⎥
⎥ = t
OB! "!!= A O ′
u
p1⎡
⎣ ⎢
⎢
⎤
⎦ ⎥
⎥ → OB
A O ′ = O O ′ sin θ
tO O ′ sin θ
i=
u
p2u
p1∴ sin θ
tsin θ
i=
u
p2u
p1= β
1β
2=
n
1n
2∵ u
p= dz
dt = ω β
∵ n " c u
pRefractive index
: Ratio of speed of light to that in the medium
Chap. 8 | Snell’s Law of Refraction
• Nonmagnetic media (μ1 = μ2 = μ0)
sin θ
tsin θ
i=
u
p2u
p1= β
1β
2=
n
1n
2Formula for Snell’s law of refraction
sin θ
tsin θ
i=
u
p2u
p1= µ
1ε
1µ
2ε
2=
ε
1ε
2=
ε
r1ε
r2=
n
1n
2= η
2η
1η = µ ε
• Medium 1 is free space (εr1 = 1, n1 = 1)
sin θ
tsin θ
i=
1
ε
r2=
1
n
2= η
2120 π
• Relationship between angles and refractive indices
Medium 1 (n1)
Medium 2 (n2)
θi
θt Propagation to “denser” medium (n2 > n1 → θt < θi)
→ Ray bent toward the normal
Medium 1 (n1)
Medium 2 (n2)
θi
θt
Propagation to “less dense” medium (n2 < n1 → θt > θi)
→ Ray bent away from the normal
n = c
u
p= µε
µ
eε
0= µ
rε
rChap. 8 | Total reflection
Total reflection
Medium 1 (n1)
Medium 2 (n2)
θi
θt
θi = θc
• Wave is incident on a less dense medium 2 from medium 1 (ε1 > ε2 → n1 > n2)
- θi < θc: both reflection + refraction
- θi = θc: Reflection + refraction along the surface - θi > θc: Total reflection
θc: critical angle when θt = 90º θt
sin θ
tsin θ
i=
ε
1ε
2sin θ
c= ε
2ε
1θ
i= θ
c, θ
t= π
What will happen to refracted wave when θ
i> θ
c? 2
x
z = 0 z θr
θi
θt
Surface wave
ant
ani
anr
sin θ
t= ε
1ε
2sin θ
i= sin θ
isin θ
c> 1 cos θ
t= 1 − sin
2θ
t= − j ε
1ε
2sin
2θ
i− 1
• Refracted wave has a direction vector ant such that
a
nt= a
xsin θ
t+ a
zcos θ
te
− jβ2ant⋅R= e
− jβ2(
xsinθt+zcosθt)
= e
− j β2 sinθi ε1 ε2
⎛
⎝⎜
⎞
⎠⎟ x
e
− β2 ε1
ε2 sin2θi−1
⎛
⎝⎜
⎞
⎠⎟z
= e
− jβRxe
−αRzEvanescent wave
• Mathematical interpretation
• Evanescent wave
- Propagating along the surface (i.e. along x direction) → Surface wave - Decaying exponentially away from the surface (i.e. in z direction)
Chap. 8 | Total reflection
Example 8-14: Optical fiber
• A dielectric rod or fiber of a transparent material that can be used to guide light or electromagnetic wave under the conditions of total internal reflection
Question
: What is minimum dielectric constant of guiding medium (ε1) so that incident wave at any angles can be confined within the rod?
θi
θt θ1
ε2 = ε0
ε1 = εrε0 • For total internal reflection,
sin θ
c< sin θ
1= sin π
2 − θ
t⎛ ⎝⎜ ⎞
⎠⎟ = cos θ
t! (1)
• On the other hand, from Snell’ law of refraction,
sin θ
tsin θ
i=
ε
2ε
1=
1
ε
r→ sin θ
t= sin θ
i1
ε
r! (3)
1
sin θ
c=
ε
1ε
2= ε
r! (2)
and
• By plugging (3) and (2) into (1),
sin θ
c< cos θ
t→ 1
ε
r< 1 −
sin
2θ
iε
r→ ε
r> 1 + sin
2θ
i≤ 2 ∴ ε
r≥ 2 → n ≥ 1.4
Glass or Quartz
Chap. 8 | Oblique incidence of TE wave at dielectric boundary (1/4)
sin θ
tsin θ
i=
u
p2u
p1= β
1β
2=
n
1n
2Snell’s law of refraction
Polarization dependence
sin θ
c= ε
2ε
1Critical angle
Transverse Electric wave
x
y z
Er
Hr
anr
Ei
ani
Medium 1 z = 0
Medium 2 Hi
θr
θi
Snell’s law of refraction, critical angle → Polarization-independent Reflection and transmission coefficients → Polarization-dependent
E
i( ) x, z = a
yE
i0e
− jβ1(
xsinθi+zcosθi)
H
i( ) x, z = E η
i01
− a
xcos θ
i+ a
zsin θ
i( ) e
− jβ1(
xsinθi+zcosθi)
⎧
⎨ ⎪
⎩ ⎪
• Incident wave
• Reflected wave
• Transmitted wave
E
r( ) x, z = a
yE
r0e
− jβ1(
xsinθr−zcosθr)
H
r( ) x, z = E η
r01
a
xcos θ
r+ a
zsin θ
r( ) e
− jβ1(
xsinθr−zcosθr)
⎧
⎨ ⎪
⎩ ⎪
θt
ant
E
t( ) x, z = a
yE
t0e
− jβ2(
xsinθt+zcosθt)
H
t( ) x, z = E η
t02
− a
xcos θ
t+ a
zsin θ
t( ) e
− jβ2(
xsinθt+zcosθt)
⎧
⎨ ⎪
⎩ ⎪
E
r0, E
t0, θ
r, θ
t4 Unknown variables
θr
Chap. 8 | Oblique incidence of TE wave at dielectric boundary (2/4)
Boundary condition
• At z = 0, Tangential components of E and H should be continuous
E
i( ) x, z = a
yE
i0e
− jβ1(
xsinθi+zcosθi) E
r( ) x, z = a
yE
r0e
− jβ1(
xsinθr−zcosθr) E
t( ) x, z = a
yE
t 0e
− jβ2(
xsinθt +zcosθt)
⎧
⎨ ⎪⎪
⎩ ⎪
⎪
H
i( ) x, z = E η
i01
− a
xcos θ
i+ a
zsin θ
i( ) e
− jβ1(xsinθi+zcosθi)H
r( ) x, z = E η
r01
a
xcos θ
r+ a
zsin θ
r( ) e
− jβ1(xsinθr−zcosθr)H
t( ) x, z = E η
t02
− a
xcos θ
t+ a
zsin θ
t( ) e
− jβ2(xsinθt−zcosθt)⎧
⎨
⎪ ⎪
⎪
⎩
⎪ ⎪
⎪
E
iy( ) x,0 + E
ry( ) x,0 = E
ty( ) x,0
→ E
i0e
− jβ1xsinθi+ E
r0e
− jβ1xsinθr= E
t0e
− jβ2xsinθtThis should satisfy for all x →
all three exponential factors should be equal
β
1x sin θ
i= β
1x sin θ
r= β
2x sin θ
t! (1) E
i0+ E
r0= E
t0! (2)
⎧ ⎨
⎩
→ Phase matching
H
ix( ) x,0 + H
rx( ) x,0 = H
tx( ) x,0 1
η
1( − E
i0cos θ
ie
− jβ1xsinθi+ E
r0cos θ
re
− jβ1xsinθr) = E η
r02
cos θ
te
− jβ2xsinθtβ
1x sin θ
i= β
1x sin θ
r= β
2x sin θ
t! (1) 1
η
1( E
i0− E
r0) cos θ
i= E η
t 02
cos θ
t! (3)
⎧
⎨ ⎪
Try solving equation for az components!
⎩⎪
Chap. 8 | Oblique incidence of TE wave at dielectric boundary (3/4)
Boundary condition
β
1x sin θ
i= β
1x sin θ
r= β
2x sin θ
t! (1)
sin θ
i= sin θ
r→ θ
i= θ
rsin θ
tsin θ
i=
β
1β
2=
u
p2u
p1= n
1n
2E
i0+ E
r0= E
t0! (2) 1
η
1( E
i0− E
r0) cos θ
i= E η
t02
cos θ
t! (3)
⎧
⎨ ⎪
⎩⎪
: Snell’s law of reflection
: Snell’s law of refraction
Γ
TE= E
r0E
i0= η
2cos θ
i− η
1cos θ
tη
2cos θ
i+ η
1cos θ
t=
η2
cosθt
( ) − (
η1 cosθi)
η2
cosθt
( ) + (
η1 cosθi)
τ
TE= E
t 0E
i0= 2 η
2cos θ
iη
2cos θ
i+ η
1cos θ
t=
2 (
η2 cosθt)
η2
cosθt
( ) + (
η1 cosθi)
c.f.)
Γ = η
2− η
1η
2+ η
1c.f.)
τ = 2 η
2η
2+ η
1∴ η
1→ η
1cos θ
i, η
2→ η
2cos θ
t• If θi = θt = 0, expressions ΓTE and τTE reduce to those for normal incidence
• Relationship between ΓTE and τTE
1 + Γ
TE= τ
TE• If medium 2 is a perfect conductor (η2 = 0), - ΓTE = -1 (Er0 = -Ei0), τTE = 0 (Et0 = 0)
Chap. 8 | Oblique incidence of TE wave at dielectric boundary (4/4)
Brewster angle
Γ
TE= E
r0E
i0= η
2cos θ
i− η
1cos θ
tη
2cos θ
i+ η
1cos θ
t=
η2
cosθt
( ) − (
η1 cosθi)
η2
cosθt
( ) + (
η1 cosθi) = 0 → cos η θ
1 TE= cos η
2θ
t! (1)
• Incident angle (θTE) that makes reflection coefficient zero
According to Snell’s law of refraction,
cos θ
t= 1 − sin
2θ
t= 1 − n
1n
2sin
2θ
TE! (2) ∵ sin θ
tsin θ
TE=
n
1n
2cos θ
TE= η
1η
2cos θ
t= η
1η
21 −
n
1n
2sin
2θ
TE→ 1 − sin
2θ
TE= η
12η
221 −
n
1n
2sin
2θ
TE⎛
⎝⎜
⎞
⎠⎟
∴ sin
2θ
TE= 1 − η
12η
221 − η
12n
1η
22n
2= 1 − µ
1ε
2µ
2ε
11 − µ
1µ
2( )
2By plugging (2) into (1), we get
θTE: Brewster’s angle of no reflection for TE wave
✴For nonmagnetic media (μ1 = μ2 = μ0)
• Right side of the equation diverges to infinity
• Brewster’s angle for TE wave DOES NOT exist for such case
Chap. 8 | Oblique incidence of TM wave at dielectric boundary (1/3)
Transverse Magnetic wave
x
y z
Hr
Er
anr
Hi
ani
Medium 1 z = 0
Medium 2 Ei
θr
θi
E
i( ) x, z = E
i0( a
xc
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