Download interference, diffraction and scattering of electromagnetic waves

Chapter 13
INTERFERENCE, DIFFRACTION
AND SCATTERING OF
ELECTROMAGNETIC WAVES
13.1
Introduction
Waves having the same frequency (thus the same wavelength) can interfere with each other. Interference is the fundamental nature of waves, and it is not an exaggeration to state that what can
interfere with themselves is de…ned to be a wave. Today, it is well known that light is a wave, but it
was not so obvious before Young devised a simple, but very convincing, experiment to demonstrate
the wave nature of light. Interference is a phenomenon caused by superposition of more than one
waves all having the same frequency. If two waves have the same phase (0; 2 ; 4 ; ) the total
amplitude is doubled, while if the phase di¤erence is (or 3 ; 5 ; ), they cancel each other out.
The phase di¤erence is, therefore, the key agent to control the total wave amplitude. Even for a
large number of waves, the total amplitude can easily be calculated by phase-vectorial summation
of each wave. Interference pattern produced by multiple antennas, for example, can be analyzed
by phase vector summation of radiation electric …elds due to each antenna. Often, it becomes
necessary to superpose (or integrate) in…nitely many waves. Such cases occur when we wish to
analyze di¤raction of waves by either an obstacle or an opening on an opaque screen. Strictly
speaking, di¤raction of electromagnetic waves (in contrast to sound waves which are longitudinal)
should be analyzed as vector boundary value problems. An alternative (somewhat arti…cial but
equally rigorous) method will be employed in our study to facilitate di¤raction calculations.
13.2
Antenna Array and Radiation from an Aperture
A single dipole antenna creates a radiation pattern symmetric about its axis. This is, of course,
a desirable feature for commercial broadcasting. The symmetric radiation pattern can be greatly
modi…ed if there are more than one antenna, all driven by a common generator. A sharper direc1
d
θ
d sinθ
Figure 13-1: Antenna array (N = 4) with separation distance d:
2φ
φ/2
R
R
ET
φ
E0
Figure 13-2: E0 = 2R sin ( =2) :ET = 2R sin (4 =2) : Then ET = E0
sin (2 )
:
sin ( =2)
tivity can be achieved by suitably arranging a larger number of antennas. Consider an antenna
array consisting of N antennas each emitting electromagnetic …elds of equal amplitude E0 . The
case N = 4 is shown in Fig. 13-1. If the separation between two neighboring antennas is d, the
path di¤erence between the distances from each antenna to the observer is
path di¤erence = d sin
(13.1)
where is the angle measured from the direction perpendicular to the array, as shown in Fig. ??.
Then, the phase di¤erence between the electric …elds due to neighboring antennas is
= kd sin =
2
d sin
(13.2)
and the total electric …eld is
ET = E0 + E0 ej + E0 e2j + E0 e3j
2
(13.3)
Adding the N (= 4) waves vectorially as illustrated in Fig. 13-2, the magnitude of ET can be readily
found as
N
sin
2
(13.4)
ET = E0
sin
2
This can be derived by eliminating R between
1
ET = R sin
2
1
; E0 = R sin
2
N
2
2
The angular dependence of the Poynting ‡ux becomes
N
2
sin2
f( )=
(13.5)
2
sin
2
For example, when there are four (N = 4) antennas with a separation interval of half wavelength
(d = =2), the function f ( ) becomes (note kd = in this case)
f( )=
sin2 (2 sin )
sin
sin2
2
(13.6)
which is shown in Fig. 13-3. As the number of antenna increases, the directivity becomes sharper.
The concept of directivity is equally applicable when the array is used as a receiver (or detector).
In radio astronomy, an antenna array of extremely sharp directivity is required in order to pinpoint
precisely the location of a radio source. In general, the angular resolution is inversely proportional
to the total length of the antenna array, or in the case optical devices, to the diameter of the device,
such as lenses and mirrors.
sin2 (2 sin )
sin
sin2
2
Let us increase N to a very large number maintaining the array length a = (N 1)d ' N d
constant. Then, in Eq. (13.3) becomes small, and the function in Eq. (13.4) may be approximated
by
N
sin
sin
2
'N
sin
2
where
=
a
3
sin
2
-16 -14 -12 -10 -8 -6 -4 -2
-2
2
4
6
8
10 12 14 16
Figure 13-3: Directivity of 4-antenna array when d = =2: The main lobe in hte directions of
and : The secondary lobe is at = 48:6 :
=0
The radiation pattern becomes proportional to
2
sin
f( )=
The function f ( ) has its peak at = 0 (f (0) = 1) and its …rst zero occurs at =
. If the
width a (or the aperture size of optical devices) is much larger than the wavelength , the function
f ( ) is sharply peaked within the angle
<
c
=
a
Application of the preceding analysis is found in optical di¤raction. Consider a slit opening in an
opaque screen, as shown in Fig. ??. In analyzing the radiation …eld behind the opaque screen, we
may assume a large number of coherent radiation sources at the opening. The power going through
the opening is evidently
P
= c"0 Ei2 a (W/m)
(13.7)
l
per unit length along the slit, where Ei is the electric …eld amplitude of the plane wave incident on
the screen. Behind the screen, the wave radiated from the slit is expected to be cylindrical,
Erad = p
Ed
ej(!t
2 k
k ) sin
(13.8)
where Ed is the amplitude of the di¤racted wave corrected for the geometrical spreading of power.
p
(If the slit is long enough, the di¤racted wave is cylindrical. Hence, the factor 1= 2 k .) Therefore,
4
the power per unit length is
P
l
where the integral
Z
=2
c"0 Ed2 sin
=2 2 k
Z 1
c"0 Ed2
sin
'
2 k a 1
a
= c"0 Ed2
(ka)2
=
Z
1
2
d
2
d
(13.9)
2
sin
d =
1
has been used. Equating Eq. (13.7) and (13.9), we …nd
Ed = kaEi
(13.10)
In general, radiation or di¤raction electric …eld due to an opening in an opaque ‡at screen is
given by (Smythe, 1948)
Z
j
E=
k (n Es ) dS
(13.11)
2 r
where Es is the electric …eld at the opening, k = ker , and n is the unit vector normal to the
opening, as shown in Fig. ??. Note that only the electric …eld component tangential to the
opening surface contributes to the di¤racted …eld because of the term n E in Eq. (13.11). The
…eld on the opening can be approximated by the incident electric …eld only if the opening size is
much larger than the wavelength. This condition is well satis…ed in optical wavelength regime, but
not in microwave wavelength regime. When the wavelength is comparable with or larger than the
opening size, a rigorous analysis incorporating the boundary conditions for electric and magnetic
…elds becomes necessary.
sin x 2
x
Di¤raction inevitably blurs images formed by optical instruments. This is unavoidable as long
as light has wave nature. The smallest angular resolution is of order
'
(13.12)
a
where a is the aperture size. For a long slit, a corresponds to the slit width. For a circular opening,
replace a by the diameter d. The resolution of circular apertures, such as optical lenses and mirrors,
is given by
= 1:22
d
(13.13)
where the numerical factor 1.22 is due to the di¤erence between the cylindrical waves in the case
5
a
θ
Figure 13-4: For a slit with opening a; a large number of sources N with separation distance
d = a=N may be assumed. Then we take the limit N ! 1:
y
1.0
0.8
0.6
0.4
0.2
-10
-8
-6
-4
-2
0
2
4
6
Figure 13-5: Intensity of di¤racted light, (sin = )2 ; as a function of
occurs at
= ; or sin '
= =a (
1) :
6
8
=
10
x
a
sin : The …rst zero
of a long slit and spherical waves in the case of a circular aperture. It is clear that to improve
resolution, either the aperture size a is to be increased, or the wavelength is to be made shorter.
In astronomy, the merit of optical telescopes of larger aperture is mainly in the higher resolving
power, that is, the capability to distinguish stars at further distances. Of course, a larger telescope
can collect more light, and the image is brighter, but this is not of primary importance because
images can be intensi…ed electronically. In optical microscopes, di¤raction limits the maximum
magni…cation up to about 600. The only way to improve resolution is to use shorter wavelength,
as in electron microscopes, and x-ray lithography which is under rapid development for microelectronics applications (e.g., VLSI). Quantum mechanics has revealed that an object having a
momentum mv is associated with a quantum mechanical wavelength given by
=
h
mv
where h = 6:64 10 34 J sec is Planck constant. For example, an electron accelerated to an energy
of 100 keV (= 1:6 10 14 J) has wave nature with a wavelength 4 10 10 cm = 0.04 Å. This
is orders of magnitude shorter than the wavelengths in the visible range,
' 4000 7000 Å.
Therefore, an electron microscope can "see" much better than optical microscopes. (Of course,
electron microscopes cannot be used for living cells because the energetic electron beam will kill
them.)
13.3
Long Slit: Exact Analysis (Advanced)
Figure 13-6: Di¤raction by a long slit of width a in a conducting screen. In short wavelength limit
ka
1; the electric …eld in the slit opening may be approximated by that of the incident wave.
Smythe’s formula in Eq. (13.55) is applicable.
7
Let a plane wave be incident on a ‡at opaque screen having a long slit with an opening width
a: The problem is essentially two dimensional and the di¤racted wavevector k may be assumed to
lie in the x y plane. Since the slit extends from z 0 = 1 to 1; we are not allowed to assume
r
r0 and the di¤racted electric …eld has to be calculated from
Z
E(r) = 2 (n E) r0 Gdy 0
(13.14)
where G is the two-dimensional Green’s function satisfying
(r2 + k 2 )G2 =
Solution for G2 is
r0 ) =
2 (r
i (1) h p
G2 = H0 k x2 + (y
4
(1)
x0 ) (y
(x
y 0 )2
y0)
i
(13.15)
(13.16)
where H0 (x) is the Hankel function of the …rst kind. Its asymptotic form is
(1)
H0 (x)
Using n0 =
'
r
2 ikx
e
kx
i4
;
x
1
(13.17)
n (unit normal directed into the di¤racted region) and noting
r 0 G2 =
we rewrite Eq. (13.14) as
E(r) =
rG2
Z
2r
G(n0 E)dy 0
(13.18)
Substituting
(1)
H0
h p
k x2 + (y
y 0 )2
where
=
and
r
2 ik
e
k
r
2 ik
=
e
k
'
p
i4
ik y 0
i4
iky 0 sin
x2 + y 2
(13.19)
(13.20)
is the angle between k and x-axis, we obtain
E(r) '
with
i
a
k
2
0
(n
r
2 ik
E0 )
e
k
i4
sin
(13.21)
de…ned by
=
ka
sin
2
=
a
sin
(13.22)
The function sin = is the familiar Fraunhofer di¤raction form factor to describe the angular ( )
8
dependence of the …eld intensity. The radiation (di¤racted) power per unit length of the slit is
Z
P
=
l
=2
=2
c"0 jEj2 d
(13.23)
In short wavelength limit ka
1; the integration limits may be approximated by from
and we obtain an expected result
P
= c"0 jE0 j2 a;
l
ka
1
1 to 1;
(13.24)
In the opposite limit ka
1; the electric …eld in the slit opening cannot be replaced by the
incident …eld because the current induced in the conducting plate greatly disturbs the incident
…eld. The problem is dual of scattering of electromagnetic waves by a thin, long conducting plate.
13.4
Yagi Antenna
TV antennas have more than one element. One element is connected to the cable but others are
not. Standing waves are still excited in those unconnected elements through mutual impedances
and a¤ect the radiation directivity of the antenna. The unconnected elements are called either
directors or re‡ectors depending on the e¤ects on the antenna directivity.
Consider a half-wavelength dipole antenna. Its radiation intensity is symmetric about the
antenna without any preferred directivity. Consider a second antenna placed parallel to the …rst
one at a distance d: The …rst antenna is driven by an rf generator but the second antenna is passive.
Circuit-wise, we have
V = Z11 I1 + Z12 I2
and
0 = Z22 I2 + Z12 I1
where Z11 = Z22 are the self impedances (approximately equal to 73
impedance which is given by
Z12
Z0
=j
4
Here
R1 =
s
Z
=4
e
ikR1
R1
=4
+
e
z
+ d2 ; R2 =
4
ikR2
cos (kz) dz
R2
2
) and Z12 is the mutual
s
(13.25)
2
z+
4
+ d2
(13.26)
The mutual impedance Z12 above can be …gured out from the radiation impedance of =2 antenna
of radius a derived in Chapter 11,
Zrad
Z0
=j
4
Z
=4
=4
e
ikR1
R1
9
+
e
ikR2
R2
cos (kz) dz
(13.27)
passive
active
λ/2
d
f ( φ)
φ
Yagi
antenna
(top view)
Figure 13-7: Two =2 dipole antennas, one driven and other passive, separated by a distance d:
Current is excited in the passive antenna through mutual impednace between the antennas.
where
R1;2 =
s
2
z
4
+ a2
(13.28)
The angular dependence of radiation pattern becomes proportional to
f ( ) = I1 + I2 e
jkd cos
2
= I12 1
Z12
e
Z22
2
jkd cos
(13.29)
where is the azimuthal angle about the antenna elements. By adding more passive elements,
radiation pattern can be made sharp (higher gain). The Yagi antenna was invented in the early
1930’s.If d = 0:2 ; Z12 ' 50 j20 . (Con…rm this by using Eq. (13.25).) Assuming Z22 = 73 ;
we have
50 j20 j0:4 cos 2
e
f( )= 1
73
Polar plot of radiation directivity is shown in Fig. 13.4.
10
0.8
0.6
0.4
0.2
-0.50-0.25
-0.2
0.25 0.50 0.75 1.00 1.25 1.50
-0.4
-0.6
-0.8
13.5
Rigorous Di¤raction Formula (Advanced)
In the preceding Section, a few examples of interference and di¤raction have been qualitatively
studied. Di¤raction (and scattering) analysis of electromagnetic waves is essentially a boundary
value problem. If electromagnetic …elds (E, B) are speci…ed on a closed surface, the …elds o¤ the
surface are uniquely determined, as in the case of the boundary value problems for scalar potential
. (See Chapter 4.) Because both electric and magnetic …elds are vectors, and an electric …eld
on a boundary is likely to be a source for both electric and magnetic …elds o¤ the surface, we
expect analysis will be rather involved. Main complication arising from the vectorial nature of the
electromagnetic …elds is that the …elds on a given closed surface are, in general, not unidirectional.
At an observing point, the electric …eld has a magnitude and direction uniquely determined by the
boundary …elds. In other words, we are talking about a vectorial transformation from the nonunidirectional boundary …elds to a unique …eld at a given observing point. Vectorial transformation
involves a matrix (or tensor) and a rigorous analysis for vector boundary value problems should
indeed be done along this line as shown by Stratton and Chu (1940). Before that, (not very
rigorous) approximation based on scalar functions had been used. In some cases (e.g., for cases
of unidirectional boundary …elds), such approximation indeed turns out to be satisfactory, but
usefulness of scalar …eld approximation is rather limited because of the very fact that E and B are
vectors.
Earlier than the work by Stratton and Chu, Schelkuno¤ (1936) introduced the concept of
magnetic charge to facilitate vector di¤raction analysis. The magnetic charge had been speculated
by Dirac (1932) in relativistic quantum electrodynamics (Dirac’s magnetic monopole). Although
the absence of magnetic charges appears to be well established by various experimental (null) results
(Maxwell was right!), the arti…cial introduction of magnetic charges and their current allows one
to make a rigorous analysis of the electromagnetic boundary value problems without resorting to
the tensorial vector transformation.
In Chapter 8, we have seen that a surface current Js (A/m) causes a discontinuity in the
tangential component of the magnetic …eld H,
n
Ht = Js
11
(13.30)
This indicates that a tangential magnetic …eld on a speci…ed surface can be replaced by an electric
surface current given in Eq. (13.30), and the vector potential due to the electric surface current
can be found from
Am =
0
4
Z
e
jkjr r0 j J
jr
s
r0 j
dS 0 =
Z
0
4
jkjr r0 j
e
jr
r0 j
n
Ht dS 0
(13.31)
Substituting this into the Maxwell’s equation IV,
r
B =r
r
Am =
j
1
r
4 "0 !
r
Z
1 @E
c2 @t
(13.32)
we …nd the electric …eld
E (r) =
jkjr r0 j
e
jr
r0 j
n
Ht dS 0
(13.33)
This takes care of the part of di¤racted …elds due to the tangential component of the magnetic …eld
on a speci…ed surface. Of course, the magnetic …eld appearing in Eq. (13.31) should be the retarded
…eld. Unfortunately, we do not have anything to achieve a similar replacement for the tangential
electric …eld, Et , because the tangential components of the electric …eld are always continuous.
(Recall the boundary condition for the electric …eld.) This di¤erence between the magnetic and
electric …eld is due to the asymmetry in the Maxwell’s equations,
r
r
H = J e + "0
E=
0
@E
@t
@H
@t
(13.34)
(13.35)
Notice the lack of a term in Eq. (13.35) corresponding to the conduction current term in Eq.
(13.34). It is obviously related to (or required from) another Maxwell’s equation
r B=0
which is a postulate put forward by Maxwell based on the experimental lack of magnetic charges.
A question arises as to the possibility of introducing …ctitious magnetic charges without a¤ecting
the well developed electrodynamic formulation.
Let us assume the presence of magnetic charge density m de…ned through
r B=
m
(Wb/m3 )
(13.36)
We also assume that the magnetic charge is conserved
@ m
+ r Jm = 0
@t
where Jm =
mv
(13.37)
is the magnetic charge current density. These are similar to the familiar electric
12
counterparts,
r E=
1
"0
(13.38)
e
@ e
+ r Je = 0
@t
The magnetic scalar potential
(13.39)
and electric vector potential are introduced as follows:
=
Z
1
4
0
Z
"0
F=
4
jkjr r0 j
e
jr
jkjr r0 j J
e
m
r0 j
m
dV 0
r0 j
jr
dV 0
(13.40)
(13.41)
which are dual of the familiar pair,
1
=
4 "0
A=
In terms of four potentials,
0
4
Z
Z
jkjr r0 j
e
jr
jkjr r0 j J
e
jr
e
r0 j
r0 j
e
dV 0
dV 0
(13.42)
(13.43)
; ; A and F, the electric and magnetic …elds are now given by
E=
r
@A
@t
1
r
"0
F
(13.44)
H=
r
@F
1
+ r
@t
0
A
(13.45)
Using the Maxwell’s equations together with the following supplementary conditions (Lorenz gauge)
r A+
1 @
1 @
= 0; r F + 2
=0
2
c @t
c @t
(13.46)
we can readily derive four inhomogeneous wave equations for the four potentials,
r2
r2
r2
r2
1 @2
c2 @t2
1 @2
c2 @t2
1 @2
c2 @t2
1 @2
c2 @t2
=
A=
=
e
"0
(13.47)
0 Je
(13.48)
m
(13.49)
0
F=
"0 J m
(13.50)
We thus have seen that introducing the magnetic charge m and its ‡ow Jm does not a¤ect the
physical potentials A and . Of course, the new potentials, and F, are created by the …ctitious
13
magnetic charge and its ‡ow, and they are not physically meaningful potentials.
Usefulness of the …ctitious potentials, and F, in vector boundary value problems may be seen
as follows. Taking the curl of Eq. (13.44), recalling the Lorenz gauge in Eq. (13.46), and noting F
satis…es the inhomogeneous wave equation, Eq. (13.50), we …nd
r
@B
@t
E=
(13.51)
Jm
The part of E due to the time varying magnetic …eld is the physical electric …eld, and that due to
the magnetic current Jm is the …ctitious …eld. In vector boundary value problems, the tangential
component of the electric …eld may be replaced by a surface magnetic current, Jms , de…ned by
Et =
n
(13.52)
Jm
Then, the electric vector potential F due to the magnetic current is given by
"0
F=
4
Z
e
jkjr r0 j
r0 j
jr
ndS 0
Et
(13.53)
which is dual of the magnetic vector potential due to the electric surface current in Eq. (13.43).
Adding the contribution from the magnetic current to the di¤racted electric …eld, we thus obtain
the following formula for the electric …eld,
E (r) =
=
1
c2
r (r A)
r F
!
"0
Z
0
1
e jkjr r j
j
r r
n
4 "0 !
jr r0 j
j
1
Ht dS +
r
4
0
Z
e
jkjr r0 j
jr
r0 j
n
Et dS 0
(13.54)
This result completely agrees with a more orthodox method without resorting to the …ctitious …elds
developed by Stratton and Chu (1939).
At r
r0 , Eq. (13.54) may be approximated by
ej(!t kr)
E (r) = j
4 r
Z
Z
k
k
(k
(n
Ht ))
k
(n
0
Et ) ejk r dS 0
p
where Z =
0 = 0 = 377 , and k = (!=c)n with n being the unit radial vector. Also, for a ‡at
boundary (such as an aperture on an opaque screen), Eq. (13.54) can be further simpli…ed as
1
E (r) =
r
2
Z
e
aperture
jkjr r0 j
jr
r0 j
n
Et dS
(13.55)
This formula is due to Smythe whose derivation was also based on an equivalent magnetic current.
Example 1: Di¤raction by a circular aperture
14
Figure 13-8: Geometry of circular aperture of radius a in an opaque screen.
Radiation from a circular aperture on a large conducting plate is a classical problem and of
practical importance. In short wavelength limit ka
1 (a the hole radius), the …eld in the
aperture may be approximated by that of the incident wave E0 : The di¤racted electric …eld
is given by
Z
0
i
0
ikr
E(r) =
k (n E0 )e
e ik r dS;
(13.56)
2 r
where in the geometry shown,
k r0 = kr0 sin sin
0
:
The integral can be performed as follows:
Z 2
0
0
r0 dr0
e ikr sin sin d
0
0
Z a
= 2
J0 (kr0 sin )r0 dr0
Z
a
0
0
=
2 a
J1 (ka sin ):
k sin
(13.57)
The di¤racted electric …eld thus reduces to
E(r) =
ia2
k
r
(n0 E0 )eikr
J1 (ka sin )
:
ka sin
(13.58)
p
Since the amplitude of the Bessel function J1 (x) decreases with x in a manner J1 (x) _ 1= x;
the electric …eld is appreciable only at small : In order of magnitude, di¤racted Poynting
‡ux has an angular spread about the z axis,
sin '
15
'
1
ka
1:
The …rst intensity minimum occurs at the …rst root of J1 (x) = 0; ka sin ' 3:83; or
'
3:83
= 1:22 ;
ka
D
D = 2a (diameter of the hole).
(13.59)
This is the familiar di¤raction limited resolving power of circular apertures such as mirrors
in telescopes and parabolic microwave antennas. The radiation power associated with the
di¤racted …eld is
Z
2
jEj2 d
P = c"0 r
Z =2 2
J1 (ka sin )
2
2
= c"0 jE0 j 2 a
sin d :
(13.60)
sin2
0
Since ka
1; the upper limit of the integral may be extended to 1 and sin
approximated by : Noting
Z
1
J12 (ka )
0
may be
1
d = ; (independent of ka);
2
we …nd
P = c"0 jE0 j2 a2 :
This is an expected result in the short wavelength limit (ka
the incident power going through the circular aperture.
(13.61)
1) and simply corresponds to
In long wavelength limit ka
1; the above analysis breaks down completely, for the electromagnetic …elds in the aperture are entirely di¤erent from those associated with the incident wave.
For normal incidence on a conducting plate, the electric …eld vanishes at the plate but the
magnetic …eld is doubled because of complete re‡ection. This does not mean there exists a
tangential magnetic …eld of H =2H0 in the aperture because the aperture should not a¤ect
the incident magnetic …eld and the tangential component of the magnetic …eld in the aperture
is H = H0 :What we can do is to …nd an e¤ective magnetic dipole moment of the hole using
2H0 as the unperturbed …eld as we did in analyzing the leakage of magnetic …eld through a
hole in a superconducting plate. The dipole moment of the hole is
8a3
3
m=
2H0 =
16a3
H0 ;
3
(13.62)
which radiates at a power
P
=
=
where the factor
1
2
1
1 2! 4 m2
2 4 "0 3c5
64
(ka)4 a2 Z0 jH0 j2 ;
27
(13.63)
is due to radiation into half space (behind the screen). The transmission
16
cross-section is
t
'
64
P
4 2
4 6
2 = 27 (ka) a _ k a :
Z0 jH0 j
(13.64)
The dependence _ k 4 (or ! 4 ) is the common feature of Rayleigh scattering of electromagnetic waves by small objects (in this case an aperture). _ a6 indicates extremely sensitive
dependence of the cross section on the hole radius. To order k 6 ; the cross section is
'
64
22
(ka)4 a2 1 + (ka)2 +
27
25
; ka
1:
(13.65)
Example 2: Radiation from an open end of a rectangular waveguide (TE10 mode) We
consider radiation from an open end of a rectangular waveguide having a cross section a b in
which a TE10 mode is excited (Fig. 13-9). The reader may prematurely jump on the Smythe’s
formula, Eq. (13.55), because the opening, on which the boundary …elds are speci…ed, is ‡at.
Unfortunately, the Smythe’s formula is not applicable here, because for the formula to be
applicable, the …elds must be known everywhere on an in…nitely large ‡at surface. In this
problem, the …elds on the plane containing the opening are not known except for the opening
itself.
Figure 13-9: Radiation from an open end of a rectangular waveguide. The electric …eld in the
opening is Ey (x) = E0 sin( x=a) and the tangential component of the magnetic …eld is Hx (x) =
E(x)=Z10 where Z10 is the impedance of the TE10 mode.
The TE10 mode is described by the following …eld components tangential to the opening (see
Chapter 12),
x j!t
Ey = E0 sin
e
a
E0
x j!t
Hx =
sin
e
Z10
a
17
where Z10 is the impedance for the TE10 mode de…ned in Chapter 12,
Z10 =
r
0
"0
q
(fc =f )2
1
We assume an observing point at (r; ; ) in the spherical coordinates shown in Fig. 13-9
where r
a; b. The vector k is thus described by (k; ; ) and scalar product k r0 becomes
k r0 = k(x0 sin cos + y 0 sin sin )
(13.66)
where x0 and y 0 are the coordinates on the opening. (They are the dummy variables for the
integrations to be carried out.) The integral required for the di¤racted …eld is
I( ; ) =
Z
0
=
a
0
dx
Z
b
dy 0 sin
0
j
k sin sin ( =a)2
x0
a
exp jk x0 sin cos + y 0 sin sin
a
k 2 sin2
cos2
1 + ejka sin
cos
and the electric …eld radiated from the opening is given by
80
1
0
<
jk
cos
E (r; ; ) =
E0 I ( ; ) @ q
+ 1A sin e + @ q
:
4 r
2
1 (fc =f )
1
The angular dependence on
example, the function
and
1
2
(fc =f )
sin
(13.67)
1
+ cos A cos e
(13.68)
is similar to the case of long slit, Eq. (13.12). For
1
kb sin sin
1
ejkb sin
can be cast into
iej
where
ejkb sin
1
kb
2
sin sin
sin
sin B
B
1
B = kb sin sin
2
The Fraunhofer factor B dictates the angular distribution of the radiation power.
13.6
Scattering by Small Objects
When a plane electromagnetic wave is incident on a material object, electrons contained in the
object are forced to oscillate by the electric …eld of the wave, and they re-radiate electromagnetic
waves. ( When the object is a ‡at conductor, the re-radiation is called re‡ection!) Any objects can
re-radiate when exposed to electromagnetic waves, as long as they contain electrons. (Ion response
is weak because of its large mass.) Re- radiation of electromagnetic waves is called scattering when
18
9
=
;
the object size is su¢ ciently smaller than the incident wavelength. In the opposite limit (when the
size is comparable or larger than the wave length), the phenomenon is called di¤raction. In spite of
di¤erent terminologies, the physical mechanism is the same, and all caused by interference among
the waves re-radiated from various points in the object. However, di¤raction is in general more
di¢ cult to analyze.
Thomson Scattering by Free Electron
Let us consider motion of a free electron placed in an electromagnetic …eld having an electric
…eld E0 ej!t and magnetic …eld B0 ej!t . The equation of motion for the electron is
m
dv
=
dt
e E0 ej!t + v
B0 ej!t
(13.69)
The magnetic force is of second order, because both v and B are oscillating with time, and can be
ignored in the lowest order approximation. Then the acceleration on the electron is given by
e
E0 ej!t
m
a=
(13.70)
Recalling the Larmor’s formula for the radiation power,
P =
1 2 e2 a2
4 "0 3 c3
(13.71)
we …nd the re-radiation power in terms of the incident electric …eld
P =
1 2 e4 E02
4 "0 3 m2 c3
(13.72)
The ratio of this power to the incident Poynting ‡ux
Si = c"0 E02
(13.73)
has the dimensions of area,
P
=
Si
where
re =
=
8 2
r
3 e
1 e2
= 2:8
4 "0 mc2
10
(13.74)
15
m2
(13.75)
is de…ned as the classical radius of the electron. Of course, it is impossible to assign a physical
dimension for an electron. The "radius" de…ned above is a mere consequence of scattering of
electromagnetic waves by a single electron. However, the radius found above is not totally absurd
because it also follows from the requirement that the potential energy associated with an electron
be …nite and not to exceed the rest energy of the electron. The scattering cross-section in Eq.
(13.75) was derived by Thomson who also discovered the electron.
Scattering by a Conducting Sphere
The procedure developed above for a single electron can be applied to the case of a small
19
conducting sphere having a radius a much smaller than the incident wavelength, a
, except in
this case, a magnetic dipole moment is also induced in addition to the electric dipole. When the
size of scattering object is smaller than the incident wavelength, the object simply sees oscillating
electric and magnetic …elds. The phase di¤erence due to the …nite size is ignorable. The induced
electric and magnetic dipoles are also oscillating with time and they re-radiate electromagnetic
waves. The electric dipole moment can be approximated by the static formula derived in Chapter
2,
p = 4 "0 a3 E0 ej!t
(13.76)
The only di¤erence between the static case and the present case is that the electric …eld is oscillating
with a frequency high enough to make the skin depth smaller than the sphere radius,
a
.
(Otherwise, the dipole moment should be modi…ed.)
The magnetic dipole moment corresponds to that of a superconducting sphere placed in a static
magnetic …eld which has been worked out in Chapter 7,
2 a3 H0 ej!t
m=
(13.77)
where H0 is the incident magnetic …eld. Note that for a plane incident wave, E0 ? H0 , and they
p
is the impedance of free space . Once each
are related by E0 = ZH0 , where Z =
0 = 0 = 377
dipole moment is known, the re-radiated power from each dipole can be calculated independently
as if they are incoherent
1 2! 4 p2
Pe =
(13.78)
4 "0 3c3
Pm =
1 2! 4 m2
4 "0 3c5
(13.79)
and the total re-radiation (or scattered) power is given by
P
= 4 "0
=
2! 4 a6 2 ! 4 a6 2
E +
H
3c3 0 6c5 "20 0
10 6 4
a k c"0 E02
3
(13.80)
where H0 = E0 =Z and ! = ck have been substituted. The reradiation power from the electric
dipole is four times larger than the magnetic dipole part. This is because of the factor 2 di¤erence
in the original dipole moments in Eqs.(13.76) and (13.77). The scattering cross-section is given by
=
8 +2 6 4
a k ; ka
3
1
(13.81)
Note that the scattering cross-section is proportional to k 4 a6 . Higher frequency components are
scattered much more e¤ectively than lower frequency components. Also, is extremely sensitive
to the radius, a6 !
As the wavelength decreases (or the frequency increases), the cross section in Eq. (13.81)
20
sigma 0.0025
0.0020
0.0015
0.0010
0.0005
0.0000
0.00
0.05
0.10
0.15
0.20
ka
Figure 13-10: Normalized scattering cross-section =2 a2 vs. ka (< 0:2): In this long wavelength
regime, dipole radiation dominates ' 103 k 4 a6 (Rayleigh’s law).
becomes invalid. The case in which the wavelength is comparable with the sphere radius is most
di¢ cult to analyze and will not be attempted here. (See Phys 812 Notes.) The largest cross-section
occurs when ka ' 1 where max ' 1:2 2 a2 , as shown in Fig. 13-11.Scattering cross-section of
an ideally conducting sphere as a function of x = ka: In the region ka
1; follows the Rayleigh’s
4
6
law, _ k a : In the region ka & 1; is insensitive to ka (Mie scattering). Scattering cross-section
of an ideally conducting sphere as a function of x = ka: In the region ka
1; follows the
4
6
Rayleigh’s law, _ k a : In the region ka & 1; is insensitive to ka (Mie scattering).
In the limit of su¢ ciently short wavelength, ka
1, analysis becomes straightforward again,
because in this limit, the geometrical optics applies. Consider a perfectly conducting sphere of a
radius a placed in a high frequency plane wave, as shown in Fig. 13-12. One half of the sphere
surface is illuminated, while the other half is dark (shadow). Scattering by the illuminated surface
is nothing but re‡ection by a spherical surface, and we can write down the cross-section as
ill
= a2
because the plane wave "sees" the maximum obstacle cross-section a2 . However, the total crosssection is twice of ill , The reason is as follows. Before the sphere is introduced, the …eld associated
with the plane wave is everywhere. After the sphere is introduced, the illuminated surface evidently
re‡ects the wave, and the shadow surface becomes dark. The total …eld on the shadow surface is
obviously zero. We interpret this zero …eld as cancellation between the incident …eld and the
boundary …eld for scattered (di¤racted) …eld required for calculating the scattered …eld in terms of
the boundary …eld as established in the preceding Section. Therefore, re-radiation from the shadow
surface does occur even though there are no …elds there!
Scattering from the illuminated surface occurs uniformly and the Poynting ‡ux is independent
21
sigma
1.0
0.8
0.6
0.4
0.2
0.0
0
1
2
3
4
5
Figure 13-11: Normalized scattering cross-section
ka & 1; is insentitive to k (Mie scattering).
6
7
8
9
10
ka
=2 a2 vs. ka up to ka = 10: In the region
of the angular location, as expected from geometric optics. However, scattering from the shadow
surface is essentially di¤raction by a circular aperture and is sharply peaked in the forward direction.
In fact, the shape of the shadow surface is immaterial, as long as its projection has an area a2 . For
example, the total scattering cross-section of a conducting hemisphere is still given by 2 a2 as long
as the plane wave falls parallel to the axis of the hemisphere. Di¤raction by an object is equivalent
to that due to a complementary void on an opaque screen having the same shape as the projection
area of the object. This interesting theorem is due to Babinet, and called Babinet’s principle. To
see why quantitatively, let us recall the Smyhte’s formula applicable to an aperture on an opaque
screen, For a circular aperture, the integral reduces to that along the radius from r0 = 0 to a, the
aperture radius. However,
Z
Z
Z
a
1
dr =
0
0
1
(13.82)
a
The …rst term in the RHS merely reproduces the incident, unperturbed …eld. The second term
is the di¤racted …eld due to an circular obstacle having the same radius a as the aperture. Eq.
(13.82) clearly states that the di¤raction by an aperture is equivalent to that due to an obstacle
having the same projected area, except that the …eld polarity is reversed. (The scattered power is
una¤ected, however.) Of course, observing the sharply peaked forward scattering (di¤raction) by
an obstacle is more di¢ cult than the case of aperture on an opaque screen, because of the strong
background incident wave.
Scattering by a small dielectric sphere can be analyzed similarly. If the sphere radius is much
smaller than the wavelength of the incident wave, the sphere essentially sees an oscillating electric
22
shadow
surface
illuminated
surface
a
incident wave
Poynting flux
by illuminated surface
by shadow surface
Figure 13-12: In short wavelength limit ka
1; scattering by the illuminated surface of the
conducting sphere is isotropic and independent of the angle as expected from re‡ection at a spherical
mirrror. The shadow surface causes sharply peaked forward scattering which is equivalent to
di¤raction by a circular aperture (Babinet’s principle).
…eld, and an oscillating electric dipole moment
p = 4 "0
where E0 is the incident electric …eld and
power can be readily found as
P = 4 "0
" "0 3
a E0 ej!t
" + 2"0
(13.83)
is the permittivity of the sphere. Then the radiation
2! 4
3c3
" "0
" + 2"0
2
a6 E02
(13.84)
and the scattering cross-section is
=
8
3
" "0
" + 2"0
2
k 4 a6
(13.85)
Note that this contains the same factor k 4 a6 as in the case of the conducting sphere. In the limit
! 1, Eq. (13.85) reduces to the electric dipole part of the cross-section of the conducting
sphere given in Eq. (13.81). This is an expected result, because a highly permittive body is
mathematically identical to a conducting body as we saw in Chapter 3. Physically, however, they
23
are quite di¤erent. For example, no magnetic dipole moments can be induced in a dielectric sphere
no matter how large is the permittivity. This is obvious because a dielectric sphere is still an
insulator even if the permittivity is large, and no conduction current can ‡ow, in contrast to a
conducting sphere.
Air molecules act as electric dipoles when exposed to solar radiation. They scatter solar radiation in random directions as the …eld polalization of solar radiation is random. Since the scattered
power is proportional to k 4 = (!=c)4 , we can see that blue light is scattered much more e¤ectively
than red light. This explains why the sky is blue. What we observe as the color of the sky is the
result of predominant scattering of blue light. This explanation is based on scattering by a single
dielectric sphere. But the density of air molecules is su¢ ciently large so that for one air molecule,
we can always …nd another molecule which precisely cancel the scattered …eld. For light scattering
by a gas, density nonuniformity due to thermal ‡uctuations is essential. This was pointed out by
Einstein.
24