1
Scattering, Interference and Diffraction
Scattering and diffraction are closely related topics of wave optics. When the wavelength of
radiation is large compared to the dimensions of the target, scattering can be described by
radiation from induced electric dipoles. Diffraction occurs when propagating waves encounter
any objects. Its effects are generally most pronounced for waves where the wavelength is
roughly similar to the dimensions of the diffracting objects.
3.1. Scattering at Long Wavelengths
Light scattering is caused by the interaction of electromagnetic waves with material systems.
When the size of the scattering object is small compared to the wavelength of the radiation
(π βͺ π), we can describe the scattering as following (Fig 9.1): (i) the incident electromagnetic
waves induce an oscillating electric dipole in the scattering object and (ii) the electric dipole
emits radiation.
We write the incident fields of a plane monochromatic wave of frequency π ( β πβπππ‘ ) as
ππ = π0πΈ0ππππ§0β π±
ππ =1π π§0 Γ ππ
(3.1)
where π = π/π, π§0 is a unit vector in the direction of incidence, and π0 is the incident
polarization vector. These fields induce an electric dipole moment π© in the small scatterer and
the dipole radiates energy in all directions. In the far zone, the scattered fields are expressed as
dipole radiation (Eq. 2.28),
ππ = ππΈπ ππππ§β π± =1
4πν0π2 ππππ
π(π§ Γ π©) Γ π§
ππ =1π π§ Γ ππ
(3.2)
where π§ is a unit vector in the direction of observation, π is the polarization vector, and π is the
distance away from scatterer. We define the differential scattering cross section as the ratio of
Fig 3.1 Scattering at long wavelengths:
radiation by induced electric dipole
2
the power radiated in the direction of observation per unit solid angle to the incident energy
flux:
ππ
πΞ©(π§, π; π§0, π0) =
π2 12 πν0|πβ β ππ |
12 πν0|ππ
β β ππ| (3.3)
Substituting Eq. 3.2 into Eq. 3.3, we obtain
ππ
πΞ©(π§, π; π§0, π0) =
π4
(4πν0πΈ0)2|πβ β π©|2 (3.4)
The differential cross section varies with wave number as π4 (or in wavelength as πβ4), which is
known as Rayleighβs law.
Scattering by a dielectric sphere
We consider a small dielectric sphere of radius π with a uniform isotropic dielectric constant
νπ(π). The electric dipole moment induced by the incident field is
π© = 4πν0 (νπ β 1
νπ + 2) π3ππ
(3.5)
Then the differential cross section is
ππ
πΞ©(π§, π; π§0, π0) = π4π6 |
νπ β 1
νπ + 2|
2
|πβ β ππ|2 (3.6)
The scattered radiation is linearly polarized in the plane defined by π and π§ (Fig 3.2).
Typically the incident radiation is unpolarized, i.e., an equal incoherent mixture of π0 = ππ₯
(polarization parallel to the scattering plane) and π0 = ππ¦ (polarization perpendicular to the
scattering plane):
(i) for π0 = ππ₯, πβ₯ = cos π ππ₯ β sin π ππ§
(ii) for π0 = ππ¦, πβ₯ = ππ¦.
Therefore, we can write
ππ
πΞ©=
ππβ₯
πΞ©+
ππβ₯
πΞ© (3.7)
where
Fig 3.2 Polarization and propagation
vectors for the incident and scattered
radiation
3
ππβ₯
πΞ©=
π4π6
2 |νπ β 1νπ + 2|
2
cos2 π
ππβ₯
πΞ©=
π4π6
2 |νπ β 1νπ + 2|
2
(3.8)
Thus the total differential cross section is
ππ
πΞ©= π4π6 |
νπ β 1
νπ + 2|
2 1
2(1 + cos2 π) (3.9)
and the total cross section is
π = β« (ππ
πΞ©) πΞ© =
8π
3π4π6 |
νπ β 1
νπ + 2|
2
(3.10)
We can see that the scattered radiation is partially polarized. Its polarization Ξ (π) is defined as
Ξ (π) =
ππβ₯
πΞ©β
ππβ₯
πΞ©ππβ₯
πΞ©+
ππβ₯
πΞ©
=sin2 π
1 + cos2 π (3.11)
Ξ (π) β 1, i.e., the scattered radiation is linearly polarized, around π = π/2. This explains how
the blue sky, accounted for by Rayleigh scattering, exhibits partial polarization.
3.2. Coherence and Interference
Coherence and incoherence
Suppose that we have two dipole oscillators operation at exactly the same frequency π (Fig
3.3). If the phase between the oscillators remains absolutely fixed [Fig. 3.3(a)], the amplitude of
electric field at any point in space will remain constant in time. The intensity pattern averaged
over a cycle of the oscillation will not vary with time. These two oscillators are then completely
coherent with respect to one another.
Fig 3.3
4
Suppose, on the other hand, that the two oscillators are going very fitfully in stops and starts so
that every so often the phase between them would jump. If we look at the combined electric
field from the two oscillators at any given point in space, it would vary in amplitude each time
such a jump in phase occurs. If the jumps take place in intervals which are short compared with
the time constant of our detectors, we would average over them and see no net constructive or
destructive interference, just the sum of the independent intensities [Fig. 3.3(b)].
Interference from an array of discrete dipoles
We examine the interference pattern at a great distance π from a simple linear array of π
parallel dipoles (π©π = πππ₯ , π = 0,1, β¦ π β 1) spaced a distance π (βͺ π) apart (see Fig. 3.4).
The electric dipole field (Eq. 3.2) can be written as π = πΈπ₯ππ₯, where
πΈπ₯(π, π, π‘) =ππ2
4πν0
ππ(ππβππ‘)
πβ πβππππ sin π
πβ1
π=0
=ππ2
4πν0
ππ(ππβππ‘)
π
1 β πβπππΌ
1 β πβππΌ (3.12)
with πΌ β‘ ππ sin π. Then we find the intensity
πΌ(π, π) =1
2πν0|πΈπ₯|2 =
ππ2π4
32π2ν0π2
sin2 ππΌ2
sin2 πΌ2
(3.13)
The intensity pattern for sets of π = 2, 4, and 10 is shown in Fig. 3.5. Radiation of a given
frequency can be made more and more directional by increasing the number of phased
antennas involved in the transmission.
Fig 3.5 Intensity and normalized intensity of radiation versus πΆ = ππ π¬π’π§ π½ for sets of π΅ = π, π, and 10
synchronous dipole antennas.
Fig 3.4 Array of oscillating dipoles
5
Passage of EM radiation through matter
We reexamine EM wave propagation in matter making use of coherent electric dipole radiation.
An incoming electric field induces a dipole moment per unit volume in matter. This induces
dipole moment, which has the same harmonic time dependence as the incident electric field,
gives rise to a radiation field of its own. When we add together all contributions to the electric
field at a point, we again observe the characteristic interference which led us to the notion of
refractive index.
Electric dipole radiation from a thin layer of uniform polarization
The polarization oscillating at frequency π is written as
π(π±, π‘) = π(π±)πβπππ‘ππ₯ (3.14)
Consider then a thin sheet of material of uniform polarization and thickness πΏ (Fig 3.6).
Using Eq. 3.2, we integrate to find the electric field at a point which lies a distance π§ from the
sheet.
πΈπ₯(π§) =1
4πν0π2ππΏ β«
ππππ
π ππ (3.15)
Using cylindrical coordinates (π, π) where π 2 = π2 + π§2, we write the surface integral as
β«ππππ
π ππ = β« ππ β«
ππππ
π
ππππ₯(π)
0
2π
0
πππ
= β« ππ β«ππππ
π
π πππ₯(π)
π§
2π
0
π ππ
=1
ππβ« (ππππ πππ₯(π) β ππππ§)
2π
0
ππ
The term ππππ πππ₯(π) varies extremely rapidly with π πππ₯(π), taking on every possible phase if
π πππ₯(π) varies as much as a wavelength. Hence it averages out by the integral over π. This
leads us to the result
Fig 3.6 Dipole radiation by an induced
polarization in a sheet of thickness πΉ
6
πΈπ₯(π§) = βππ
2ν0ππΏππππ§ (3.16)
We assume that π is due to an incoming field
πΈπ₯πππ(π§, π‘) = πΈ0ππ(ππ§βππ‘) (3.17)
Then the dipole radiation field due to the induced polarization π = ππν0πΈ0 is written as
πΈπ₯πππ(π§, π‘) = β
1
2πππππΏ πΈ0ππ(ππ§βππ‘) (3.18)
Total field: superposition of incoming and induced dipole fields
Now we consider that the uniform medium of ππ fills the half-infinite space for π§ > 0. Then the
total field at a position π§ is given by a sum of three terms: the incident field, the contributions
from slices to the left of π§, and the contributions from slices to the right of π§. Thus we have
πΈπ₯(π§) = πΈ0ππππ§ β1
2ππππ β« πΈπ₯(π§β²)πππ(π§βπ§β²)
π§
0
ππ§β² β1
2ππππ β« πΈπ₯(π§β²)πβππ(π§βπ§β²)
β
π§
ππ§β² (3.19)
Differentiating πΈπ₯(π§) twice with respect to π§, we obtain the wave equation
π2πΈπ₯
ππ§2= βπ2πΈπ₯ β π2πππΈπ₯ = βπ2(1 + ππ)πΈπ₯ = βπ2 (
π
π)
2
πΈπ₯ (3.20)
where π = β1 + ππ is the index of refraction. Thus the total field is
πΈπ₯(π§) = πΈπ₯πππ(ππ
)π§ (3.21)
3.3. Diffraction
Diffraction traditionally involves apertures or obstacles whose dimensions are large compared
to a wavelength. The diffraction of the waves around the obstacles or through the apertures
gives rise to a consequent spreading of the waves. Simple arguments based on Fourier
transforms show that the angles of deflection of the waves are confined to the region π β€ π/π,
where π is the wavelength and π is a linear dimension of the aperture or obstacle. Although
diffraction effects are generally most pronounced for waves where π is roughly similar to the
dimensions of the diffracting objects, the various approximation of diffraction theory work best
for π βͺ π and fail badly for π~π or π > π.
Historically, diffraction patterns were classed as Fresnel or Fraunhofer diffraction, depending on
the relative geometry involved. If ππ < π2 or ππ~π2 where π is the distance from the aperture
7
or obstacle to the observation point, the image of the aperture or obstacle, although easily
recognizable, becomes structured with fringes around its periphery. The phenomenon observed
is known as Fresnel or near-field diffraction. The Fraunhofer or far-field diffraction applies if
ππ β« π2, where the projected pattern will have spread out considerably, bearing little or no
resemblance to the actual aperture or obstacle. We consider only the Fraunhofer limit here.
Fraunhofer diffraction by a hole in a plane
A plane wave of light passing through a small hole in a wall exhibits a remarkable interference
pattern on the far side. Referring to Fig. 3.7 we see that the radiation to the right of the wall is a
superposition of the incoming radiation and the radiation arising from the oscillating dipoles in
the wall. If we were to fill in the hole so as to make the wall complete, then nothing would
penetrate the wall. That is, the complete wall radiates just enough to completely cancel the
incoming plane wave to the right of it. Hence the radiation which appears when the hole is
open must be precisely canceled out by the radiation from the stopper as we cover up the hole.
This makes our calculation of the diffraction pattern very simple. We need only calculate the
radiation from the stopper itself. The radiation field to the right of the hole is equal and
opposite in amplitude to that which would be emitted by the stopper if it were radiating all by
itself with the same dipole moment per unit area as the rest of the screen.
Our job then consists of two parts. First, we must find out what the dipole moment per unit area
of our wall is in terms of the incoming radiation. Second, we can replace the combination of
incoming radiation and wall by the βequivalentβ stopper alone. Giving this stopper a dipole
moment per unit area which is equal and opposite to the rest of the wall, we calculate its
radiation pattern and have the answer to our problem.
Fig 3.7 An incoming plane wave strikes an
opaque wall with a hole in it. If the βstopperβ
were inserted into the hole, then no radiation
would penetrate. Hence the stopper radiates
just enough to cancel the radiation which
would be transmitted through the hole. The
diffraction pattern on the right is obtained by
taking the negative of the electric field which
would be radiated by the stopper.
8
Let ππ be the dipole moment per unit area of our wall. If the wall were complete, then the field
contributed by its oscillating charges (see Eq. 3.16) is
πΈπ₯(π§, π‘) = βππ
2ν0ππ ππ(ππ§βππ‘) (3.22)
with the incoming radiation
πΈπ₯πππ(π§, π‘) = πΈ0ππ(ππ§βππ‘) (3.23)
The in order that the total radiation to the right of the wall be zero, we must have
ππ =2ν0πΈ0
ππ (3.24)
We thus replace the incoming radiation and the wall with its hole by the stopper having dipole
moment per unit area equal to βππ . If we let π±β² refer to a point on the aperture and π± be the
place at which we wish to know the electric field, then, using Eqs. 3.2 and 3.24, we obtain
πΈπ₯(π±, π‘) β πππΈ0
2ππβπππ‘ β«
πππ|π±βπ±β²|
|π± β π±β²|ππβ² (3.25)
The surface integral is over the aperture area. We assume here that the diffraction angle is
relatively small so that we neglect the projection of the dipole moment normal to the
propagation direction.
Diffraction by a rectangular aperture
We consider a rectangular aperture to extend from π₯ = βπ/2 to π₯ = +π/2 and from π¦ =
βπ/2 to π¦ = +π/2. The incoming radiation is polarized in the π₯ direction. We assume that
π, π β« π so that the diffraction angle is small and thus Eq. 3.25 is valid. We have then
πΈπ₯(π₯, π¦, π§, π‘) =πππΈ0
2ππβπππ‘ β« β«
πππ|π±βπ±β²|
|π± β π±β²|
π2
βπ2
π2
βπ2
ππ₯β²ππ¦β² (3.26)
We apply the condition π, π βͺ π
|π± β π±β²| β π β π§ β π±β² (3.27)
where π§ is a unit vector in the direction of π±. Then, Eq. 3.26 reduces to
πΈπ₯(π₯, π¦, π§) β πππΈ0
2π
ππππ
πβ« β« πβπππ§β π±β²
π2
βπ2
π2
βπ2
ππ₯β²ππ¦β² (3.28)
The integral splits into the product of an integral over π₯β² and one over π¦β². Thus we have
πΈπ₯(π₯, π¦, π§) =πππΈ0
2π
ππππ
π(β« πβππ(sin ππ₯)π₯β²
π2
βπ2
ππ₯β²) (β« πβππ(sin ππ¦)π¦β²
π2
βπ2
ππ¦β²) (3.29)
where
9
sin ππ₯ =π₯
π and sin ππ¦ =
π¦
π (3.30)
Integrating, we obtain
πΈπ₯(π₯, π¦, π§) =πππππΈ0ππππ
2ππ
sin (ππ2 sin ππ₯)
ππ2 sin ππ₯
sin (ππ2 sin ππ¦)
ππ2 sin ππ¦
(3.31)
Note that the directly forward-going diffracted field (π₯ = 0 and π¦ = 0),
πΈπ₯(0,0, π§) =ππππ
2πππΈ0ππππ (3.32)
Is (i) πΈπ₯(0,0, π§) β πΈ0ππ(ππ+π
2) and (ii) πΈπ₯(0,0, π§) β πβ1, πβ1.
Diffraction pattern
The time-averaged radiation power per unit solid angle is
ππ
πΞ©= π2π β π§ =
1
2πν0π2|π|2 = πΌ0
sin2 (ππ2 sin ππ₯)
(ππ2 sin ππ₯)
2
sin2 (ππ2 sin ππ¦)
(ππ2 sin ππ¦)
2 (3.33)
where
πΌ0 =πν0π2π2π2πΈ0
2
8π2 (3.34)
is the intensity per unit solid angle in forward direction. The intensity pattern is shown in Fig. 3.8.
Note that the minima occur at angles such that
sin ππ₯ =2ππ
ππ=
ππ
π or sin ππ¦ =
2ππ
ππ=
ππ
π,
(3.35)
for π, π = 1,2,3, β¦ .
Fig. 3.8 Intensity of diffracted radiation versus (ππ/π) π¬π’π§ π½π and (ππ/π) π¬π’π§ π½π for a rectangular
aperture
10
Diffraction by a circular aperture
We consider a circular aperture of radius π as shown in Fig. 3.9.
The incoming radiation polarized in the π₯ direction is normally incident on the aperture. The
field of diffracted radiation then is
πΈπ₯(π₯, π¦, π§, π‘) =πππΈ0
2ππβπππ‘ β« β«
πππ|π±βπ±β²|
|π± β π±β²|
π
0
2π
0
πβ²ππβ²ππβ² (3.36)
Using Eq. 3.27, we obtain
πΈπ₯(π₯, π¦, π§) β πππΈ0
2π
ππππ
πβ« β« πβπππ§β π±β²
π
0
2π
0
πβ²ππβ²ππβ² (3.37)
Since π§ β π±β² = sin π cos π β πβ² cos πβ² + sin π sin π β πβ² sin πβ² = πβ² sin π cos(π β πβ²),
πΈπ₯(π₯, π¦, π§) =πππΈ0
2π
ππππ
πβ« [β« πβπππβ² sin π cos(πβπβ²)
2π
0
ππβ²]π
0
πβ²ππβ² (3.38)
The angular integral can be transformed into
1
2πβ« πβπππβ² cos π½
2π
0
ππ½ = π½0(ππβ²) (3.39)
where π½ = π β πβ² and π = π sin π. Then the field is expressed as
πΈπ₯(π₯, π¦, π§) = πππΈ0
ππππ
πβ« π½0(ππβ²)
π
0
πβ²ππβ² (3.40)
Since π
ππ₯[π₯π½1(π₯)] = π₯π½0(π₯), the radial integral can be done directly:
Fig 3.9
11
πΈπ₯(π₯, π¦, π§) = πππΈ0
ππππ
π
1
π2[πππ½1(ππ)] =
ππππππ
ππ2πΈ0
π½1(ππ sin π)
ππ sin π
(3.41)
The time-averaged diffracted power per unit solid angle is
ππ
πΞ©= ππ
(ππ)2
4π|2π½1(ππ sin π)
ππ sin π|
2
(3.42)
where
ππ =1
2πν0πΈ0
2(ππ2) (3.43)
is the total power normally incident on the aperture. Figure 10.10 shows the diffraction pattern.
3.4. Babinetβs Principle of Complementary Screens
Babinetβs principle relates the diffraction fields of one diffracting screen to those of the
complementary screen. If the surface of the original screen is ππ and that of the
complementary screen is ππ, then ππ + ππ = π forms the entire screen as shown in Fig. 3.11.
Fig 3.10
Fig 3.11 A diffraction screen πΊπ and its
complementary screen πΊπ
12
If the surface integral in Eq. 3.25 is done over ππ and ππ, we have the following relation
between the diffraction fields πΈπ and πΈπ:
πΈπ + πΈπ = πΈ0 (3.44)
where πΈ0 is the unobstrcuted field. The principle implies that when πΈ0 = 0, πΈπ = βπΈπ, i.e.,
these disturbances are precisely equal in magnitude and π out of phase. We would, therefore,
observe exactly the same diifraction pattern with either ππ and ππ in place where πΈ0 is
negligible.
Fig. 3.12 Fresnel diffraction pattern of a rectangular (a) aperture and (b) obstacle calculated from
Babinet's principle. [K. M. Abedin et. al., Optics & Laser Tech. 39, 237 (2007)]
Top Related