Zhang K., Li D. Electromagnetic Theory for Microwaves and Optoelectronics
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Problems 671
Figure 10.25: (a) Problem 10.4, (b) Problem 10.5.
Problems
10.1 A plane wave is normally incident on a ring aperture whose inner and
outer radii are a and b, respectively. Derive the Fraunhofer diffraction
pattern.
10.2 A plane wave is normally incident on a lens whose fo cal length is f.
Derive the field distribution on the focal plane.
10.3 A plane wave is obliquely incident on a square aperture with a side of
length a. The angle between the wave vector and the normal of the
aperture is β. Derive the Fraunhofer diffraction pattern.
10.4 As shown in Fig. 10.25(a), wave sources with identical amplitudes and
phases are distributed uniformly on a spherical surface with a radius of
r
0
. Derive the field distribution on the plane as shown.
10.5 As shown in Fig. 10.25(b) the axially symmetric surface of an antenna
is formulated by z = kr
2
, and the radius of aperture is a. A plane
wave of wavelength λ
0
is incident on it along the axis. Derive the field
distribution at the focal plane.
10.6 Discuss the Fraunhofer diffraction of a plane wave at a circular aperture
on a uniaxial crystal surface by the superposition of eigenmo des.
10.7 In Fig. 10.26(a), a plane wave is obliquely incident on a square aperture
on a uniaxial crystal surface. Derive the Fraunhofer diffraction pattern.
10.8 Discuss the refraction of a normally incident Gaussian beam in a uni-
axial crystal by the superposition of eigenmo des.
10.9 Derive the transformation law of the beam parameter at a dielectric
boundary for an obliquely incident Gaussian beam.
672 10. Scalar Diffraction Theory
Figure 10.26: (a) Problem 10.7, (b) Problem 10.10.
10.10 In Fig. 10.26(b), a Gaussian beam is obliquely incident on a dielectric
slab. Derive the field distribution in and to the right of the slab.
Appendix A
SI Units and
Gaussian Units
A.1 Conversion of Amounts
All factors of 3 (apart from exponents) should, for accurate work, be replaced
by 2.99792456, arising from the numerical value of the velocity of light. [43]
Physical quantity Symbol SI(MKSA) Gaussian
Length l 1 meter (m) 10
2
centimeters (cm)
Mass m 1 kilogram (kg) 10
3
grams (gm or g)
Time t 1 second (sec or s) 1 second (sec or s)
Frequency f 1 hertz (Hz) 1 hertz (Hz)
Force F 1 newton (N) 10
5
dynes
Work, Energy W, U 1 joule (J) 10
7
ergs
Power P 1 watt (W) 10
7
ergs/s
Charge q 1 coulomb (C) 3 × 10
9
statcoulombs
Charge density % 1 C/m
3
3 × 10
3
statcoul/cm
3
Current I 1 ampere (A) 3 × 10
9
statamperes
Current density J 1 A/m
2
3 × 10
5
statamp/cm
2
Potential ϕ 1 volt (V) 10
−2
/3 statvolt
Electric field E 1 V/m 10
−4
/3 statvolt/cm
Electric induction D 1 C/m
2
12π × 10
5
statvolt/cm
Polarization P 1 C/m
2
3 × 10
5
moment/cm
3
Magnetic flux Φ 1 weber (Wb) 10
8
maxwell (Mx)
Magnetic induction B 1 tesla (T) 10
4
gauss (Gs)
Magnetic field H 1 A/m 4π × 10
−3
oersted (Oe)
Magnetization M 1 A/m 10
−3
moment/cm
3
Conductance G 1 siemens (S) 9 × 10
11
cm/s
Conductivity σ 1 S/m 9 × 10
9
1/s
Resistance R 1 ohm (Ω) 10
−11
/9 s/cm
Capacitance C 1 farad (F) 9 × 10
11
cm
Inductance L 1 henry (H) 10
9
cm
674 Appendix A
A.2 Formulas in SI (MKSA) Units and
Gaussian Units
Name of formula SI (MKSA) Gaussian
∇ × E = −
∂ B
∂ t
∇ × E = −
1
c
∂ B
∂ t
Maxwell ∇ × H =
∂ D
∂ t
+ J ∇ ×H =
1
c
∂ D
∂ t
+
4π
c
J
equations
∇ · D = % ∇ · D = 4π%
∇ · B = 0 ∇ · B = 0
Lorentz force F = q(E + v × B) F = q(E +
1
c
v × B)
Constitutional D = ²
0
E + P = ²E D = E + 4πP = ²E
equations B = µ
0
(H + M) = µH B = H + 4πM = µH
J = γE J = γE
Constitutional ² = ²
0
(1 + χ
e
) = ²
0
²
r
² = 1 + 4πχ
e
= ²
r
parameters µ = µ
0
(1 + χ
m
) = µ
0
µ
r
µ = 1 + 4πχ
m
= µ
r
n × (E
2
− E
1
) = 0 n × (E
2
− E
1
) = 0
Boundary n × (H
2
− H
1
) = J
s
n × (H
2
− H
1
) =
4π
c
J
s
equations n · (D
2
− D
1
) = %
s
n · (D
2
− D
1
) = 4π%
s
n · (B
2
− B
1
) = 0 n · (B
2
− B
1
) = 0
Coulomb’s law E =
1
4π²
Z
V
%
r
2
ˆ
rdV
0
E =
1
²
Z
V
%
r
2
ˆ
rdV
0
ϕ =
1
4π²
Z
V
%
r
dV
0
ϕ =
1
²
Z
V
%
r
dV
0
Biot-Savart B =
µ
4π
Z
V
J ×
ˆ
r
r
2
dV
0
B =
µ
c
Z
V
J ×
ˆ
r
r
2
dV
0
law
A =
µ
4π
Z
V
J
r
dV
0
A =
µ
c
Z
V
J
r
dV
0
Poison ∇
2
ϕ = −
%
²
∇
2
ϕ = −4π
%
²
equations
∇
2
A = −µJ ∇
2
A = −
4π
c
µJ
SI Units and Gaussian Units 675
Name of formula SI (MKSA) Gaussian
Wave ∇
2
E − ²µ
∂
2
E
∂ t
2
= 0 ∇
2
E −
²µ
c
2
∂
2
E
∂ t
2
= 0
equations
∇
2
H − ²µ
∂
2
H
∂ t
2
= 0 ∇
2
H −
²µ
c
2
∂
2
H
∂ t
2
= 0
Dynamic B = ∇ × A B = ∇ ×A
potentials
E = −∇ϕ −
∂ A
∂ t
E = −∇ϕ −
1
c
∂ A
∂ t
Lorentz gauge ∇ · A + ²µ
∂ ϕ
∂ t
= 0 ∇· A +
²µ
c
∂ ϕ
∂ t
= 0
D’Alembert ∇
2
ϕ − ²µ
∂
2
ϕ
∂ t
2
= −
%
²
∇
2
ϕ −
²µ
c
2
∂
2
ϕ
∂ t
2
= −4π
%
²
equations
∇
2
A−²µ
∂
2
A
∂ t
2
=−µJ ∇
2
A−
²µ
c
2
∂
2
A
∂ t
2
=−
4π
c
µJ
Retarding ϕ =
1
4π²
Z
V
%(t − r/c)
r
dV
0
ϕ=
1
²
Z
V
%(t − r/c)
r
dV
0
potentials
A=
µ
4π
Z
V
J(t − r/c)
r
dV
0
A=
µ
c
Z
V
J(t − r/c)
r
dV
0
Energy density w =
1
2
(E · D + H · B) w =
1
8π
(E · D + H · B)
Poynting vector P = E × H P =
c
4π
E × H
676 Appendix A
A.3 Prefixes and Symbols for Multiples
Multiple Prefix Symbol
10
−18
atto a
10
−15
femto f
10
−12
pico p
10
−9
nano n
10
−6
micro µ
10
−3
milli m
10
−2
centi c
10
−1
deci d
10 deka da
10
2
hecto h
10
3
kilo k
10
6
mega M
10
9
giga G
10
12
tera T
10
15
peta P
10
18
exa E
Appendix B
Vector Analysis
B.1 Vector Differential Operations
B.1.1 General Orthogonal Coordinates
u
1
, u
2
, u
3
, h
1
, h
2
, h
3
, h
i
=
r
³
∂ x
∂ u
i
´
2
+
³
∂ y
∂ u
i
´
2
+
³
∂ z
∂ u
i
´
2
, i = 1, 2, 3
A =
ˆ
u
1
A
1
+
ˆ
u
2
A
2
+
ˆ
u
3
A
3
∇ϕ =
3
X
i=1
ˆ
u
i
1
h
i
∂ ϕ
∂ u
i
=
ˆ
u
1
1
h
1
∂ ϕ
∂ u
1
+
ˆ
u
2
1
h
2
∂ ϕ
∂ u
2
+
ˆ
u
3
1
h
3
∂ ϕ
∂ u
3
(B.1)
∇ · A =
1
h
1
h
2
h
3
3
X
i=1
∂
∂ u
i
(h
j
h
k
A
i
)
=
1
h
1
h
2
h
3
·
∂
∂ u
1
(h
2
h
3
A
1
) +
∂
∂ u
2
(h
3
h
1
A
2
) +
∂
∂ u
3
(h
1
h
2
A
3
)
¸
(B.2)
∇ × A =
3
X
i=1
ˆ
u
i
1
h
j
h
k
·
∂
∂ u
j
(h
k
A
k
) −
∂
∂ u
k
(h
j
A
j
)
¸
=
ˆ
u
1
1
h
2
h
3
·
∂
∂ u
2
(h
3
A
3
) −
∂
∂ u
3
(h
2
A
2
)
¸
+
ˆ
u
2
1
h
3
h
1
·
∂
∂ u
3
(h
1
A
1
) −
∂
∂ u
1
(h
3
A
3
)
¸
+
ˆ
u
3
1
h
1
h
2
·
∂
∂ u
1
(h
2
A
2
) −
∂
∂ u
2
(h
1
A
1
)
¸
(B.3)
678 Appendix B
∇
2
ϕ =
1
h
1
h
2
h
3
3
X
i=1
∂
∂ u
i
³
h
j
h
k
h
i
∂ ϕ
∂ u
i
´
=
1
h
1
h
2
h
3
·
∂
∂ u
1
³
h
2
h
3
h
1
∂ ϕ
∂ u
1
´
+
∂
∂ u
2
³
h
3
h
1
h
2
∂ ϕ
∂ u
2
´
+
∂
∂ u
3
³
h
1
h
2
h
3
∂ ϕ
∂ u
3
´
¸
(B.4)
∇
2
A = ∇(∇ · A) − ∇ × ∇ × A
=
ˆ
u
1
·
1
h
1
∂ F
0
∂ u
1
−
1
h
2
h
3
³
∂ F
3
∂ u
2
−
∂ F
2
∂ u
3
´
¸
+
ˆ
u
2
·
1
h
2
∂ F
0
∂ u
2
−
1
h
3
h
1
³
∂ F
1
∂ u
3
−
∂ F
3
∂ u
1
´
¸
+
ˆ
u
3
·
1
h
3
∂ F
0
∂ u
3
−
1
h
1
h
2
³
∂ F
2
∂ u
1
−
∂ F
1
∂ u
3
´
¸
(B.5)
where
F
0
= ∇ · A
F
1
= h
1
(∇ × A)
1
=
h
1
h
2
h
3
·
∂
∂ u
2
(h
3
A
3
) −
∂
∂ u
3
(h
2
A
2
)
¸
F
2
= h
2
(∇ × A)
2
=
h
2
h
3
h
1
·
∂
∂ u
3
(h
1
A
1
) −
∂
∂ u
1
(h
3
A
3
)
¸
F
3
= h
3
(∇ × A)
3
=
h
3
h
1
h
2
·
∂
∂ u
1
(h
2
A
2
) −
∂
∂ u
2
(h
1
A
1
)
¸
B.1.2 General Cylindrical Coordinates
u
1
, u
2
, z, h
3
= 1,
∂ h
1
∂ z
= 0,
∂ h
2
∂ z
= 0
∇ϕ =
ˆ
u
1
1
h
1
∂ ϕ
∂ u
1
+
ˆ
u
2
1
h
2
∂ ϕ
∂ u
2
+
ˆ
z
∂ ϕ
∂ z
(B.6)
∇ · A =
1
h
1
h
2
·
∂
∂ u
1
(h
2
A
1
) +
∂
∂ u
2
(h
1
A
2
)
¸
+
∂ A
z
∂ z
(B.7)
∇ × A =
ˆ
u
1
1
h
2
·
∂ A
z
∂ u
2
−
∂
∂ z
(h
2
A
2
)
¸
+
ˆ
u
2
1
h
1
·
∂
∂ z
(h
1
A
1
) −
∂ A
z
∂ u
1
¸
+
ˆ
u
3
1
h
1
h
2
·
∂
∂ u
1
(h
2
A
2
) −
∂
∂ u
2
(h
1
A
1
)
¸
(B.8)
∇
2
ϕ =
1
h
1
h
2
·
∂
∂ u
1
³
h
2
h
1
∂ ϕ
∂ u
1
´
+
∂
∂ u
2
³
h
1
h
2
∂ ϕ
∂ u
2
´
¸
+
∂
2
ϕ
∂ z
2
(B.9)
Vector Analysis 679
∇
2
A = ∇
2
A
T
+
ˆ
z∇
2
A
z
(B.10)
where A
z
is the longitudinal component and A
T
is the transverse 2-
dimensional vector of A
A = A
T
+
ˆ
zA
z
, A
T
=
ˆ
u
1
A
1
+
ˆ
u
2
A
2
∇
2
A
z
=
1
h
1
h
2
·
∂
∂ u
1
³
h
2
h
1
∂ A
z
∂ u
1
´
+
∂
∂ u
2
³
h
1
h
2
∂ A
z
∂ u
2
´
¸
+
∂
2
A
z
∂ z
2
∇
2
A
T
=
ˆ
u
1
µ
1
h
1
∂ F
0
∂ u
1
−
1
h
2
∂ F
z
∂ u
2
+
∂
2
A
1
∂ z
2
¶
+
ˆ
u
2
µ
1
h
2
∂ F
0
∂ u
2
+
1
h
1
∂ F
z
∂ u
1
+
∂
2
A
2
∂ z
2
¶
(B.11)
where
F
0
= ∇ · A
T
=
1
h
1
h
2
·
∂
∂ u
1
(h
2
A
1
) +
∂
∂ u
2
(h
1
A
2
)
¸
F
z
= |∇ × A
T
| =
1
h
1
h
2
·
∂
∂ u
1
(h
2
A
2
) −
∂
∂ u
2
(h
1
A
1
)
¸
B.1.3 Rectangular Coordinates
x, y, z, h
1
= 1, h
2
= 1, h
3
= 1
∇ϕ =
ˆ
x
∂ ϕ
∂ x
+
ˆ
y
∂ ϕ
∂ y
+
ˆ
z
∂ ϕ
∂ z
(B.12)
∇ · A =
∂ A
x
∂ x
+
∂ A
y
∂ y
+
∂ A
z
∂ z
(B.13)
∇×A =
ˆ
x
µ
∂ A
z
∂ y
−
∂ A
y
∂ z
¶
+
ˆ
y
µ
∂ A
x
∂ z
−
∂ A
z
∂ x
¶
+
ˆ
z
µ
∂ A
y
∂ x
−
∂ A
x
∂ y
¶
(B.14)
∇
2
ϕ =
∂
2
ϕ
∂ x
2
+
∂
2
ϕ
∂ y
2
+
∂
2
ϕ
∂ z
2
(B.15)
∇
2
A =
ˆ
x∇
2
A
x
+
ˆ
y∇
2
A
y
+
ˆ
z∇
2
A
z
(B.16)
B.1.4 Circular Cylindrical Coordinates
ρ, φ, z, h
1
= 1, h
2
= r, h
3
= 1
∇ϕ =
ˆ
ρ
∂ ϕ
∂ ρ
+
ˆ
φ
1
ρ
∂ ϕ
∂ φ
+
ˆ
z
∂ ϕ
∂ z
(B.17)
∇ · A =
1
ρ
∂
∂ ρ
(ρA
ρ
) +
1
ρ
∂ A
φ
∂ φ
+
∂ A
z
∂ z
(B.18)
680 Appendix B
∇×A=
ˆ
ρ
·
1
ρ
∂ A
z
∂ φ
−
∂ A
φ
∂ z
¸
+
ˆ
φ
·
∂ A
ρ
∂ z
−
∂ A
z
∂ ρ
¸
+
ˆ
z
1
ρ
·
∂
∂ ρ
(ρA
φ
) −
∂ A
ρ
∂ φ
¸
(B.19)
∇
2
ϕ =
1
ρ
∂
∂ ρ
µ
ρ
∂ ϕ
∂ ρ
¶
+
1
ρ
2
∂
2
ϕ
∂ φ
2
+
∂
2
ϕ
∂ z
2
(B.20)
∇
2
A=
ˆ
ρ
µ
∇
2
A
ρ
−
2
ρ
2
∂ A
φ
∂ φ
−
A
ρ
ρ
2
¶
+
ˆ
φ
µ
∇
2
A
φ
+
2
ρ
2
∂ A
ρ
∂ φ
−
A
φ
ρ
2
¶
+
ˆ
z∇
2
A
z
(B.21)
B.1.5 Spherical Coordinates
r, θ, φ, h
1
= 1, h
2
= r, h
3
= r sin θ
∇ϕ =
ˆ
r
∂ ϕ
∂ r
+
ˆ
θ
1
r
∂ ϕ
∂ θ
+
ˆ
φ
1
r sin θ
∂ ϕ
∂ φ
(B.22)
∇ · A =
1
r
2
∂
∂ r
(r
2
A
r
) +
1
r sin θ
∂
∂ θ
(sin θA
θ
) +
1
r sin θ
∂ A
φ
∂ φ
(B.23)
∇ × A =
ˆ
r
1
r sin θ
·
∂
∂ θ
(sin θA
φ
) −
∂ A
θ
∂ φ
¸
+
ˆ
θ
1
r
·
1
sin θ
∂ A
r
∂ φ
−
∂
∂ r
(rA
φ
)
¸
+
ˆ
φ
1
r
·
∂
∂ r
(rA
θ
) −
∂ A
r
∂ θ
¸
(B.24)
∇
2
ϕ =
1
r
2
∂
∂ r
µ
r
2
∂ ϕ
∂ r
¶
+
1
r
2
sin θ
∂
∂ θ
³
sin θ
∂ ϕ
∂ θ
´
+
1
r
2
sin
2
θ
∂
2
ϕ
∂ φ
2
(B.25)
∇
2
A =
ˆ
r
·
∇
2
A
r
−
2
r
2
³
A
r
+ cot θ A
θ
+ csc θ
∂ A
φ
∂ φ
+
∂ A
θ
∂ θ
´
¸
+
ˆ
θ
·
∇
2
A
θ
−
1
r
2
³
csc
2
θA
θ
−2
∂ A
r
∂ θ
+2 cot θ csc θ
∂ A
φ
∂ φ
´
¸
+
ˆ
φ
·
∇
2
A
φ
−
1
r
2
³
csc
2
θA
φ
−2 csc θ
∂ A
r
∂ φ
−2 cot θ csc θ
∂ A
θ
∂ φ
´
¸
(B.26)
B.2 Vector Formulas
B.2.1 Vector Algebraic Formulas
A · B = B · A (B.27)
A × B = −B × A (B.28)
A · (B × C) = B · (C × A) = C · (A × B) (B.29)
A × (B × C) = (A · C)B − (A · B)C (B.30)
(A × B) · (C × D ) = (A · C)(B · D) − (A · D)(B · C) (B.31)
(A × B) × (C × D ) = (A × B · D)C − (A × B · C)D (B.32)