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

(𝑎 ≪ 𝜆), 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 𝜔 ( ∝ 𝑒−𝑖𝜔𝑡 ) as

𝐄𝑖 = 𝛜0𝐸0𝑒𝑖𝑘𝐧0⋅𝐱

𝐁𝑖 =1𝑐 𝐧0 × 𝐄𝑖

(3.1)

where 𝑘 = 𝜔/𝑐, 𝐧0 is a unit vector in the direction of incidence, and 𝛜0 is the incident

polarization vector. These fields induce an electric dipole moment 𝐩 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),

𝐄𝑠 = 𝛜𝐸𝑠𝑒𝑖𝑘𝐧⋅𝐱 =1

4𝜋휀0𝑘2 𝑒𝑖𝑘𝑟

𝑟(𝐧 × 𝐩) × 𝐧

𝐁𝑠 =1𝑐 𝐧 × 𝐄𝑠

(3.2)

where 𝐧 is a unit vector in the direction of observation, 𝛜 is the polarization vector, and 𝑟 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:

𝑑𝜎

𝑑Ω(𝐧, 𝛜; 𝐧0, 𝛜0) =

𝑟2 12 𝑐휀0|𝛜∗ ⋅ 𝐄𝑠|

12 𝑐휀0|𝛜𝟎

∗ ⋅ 𝐄𝑖| (3.3)

Substituting Eq. 3.2 into Eq. 3.3, we obtain

𝑑𝜎

𝑑Ω(𝐧, 𝛜; 𝐧0, 𝛜0) =

𝑘4

(4𝜋휀0𝐸0)2|𝛜∗ ⋅ 𝐩|2 (3.4)

The differential cross section varies with wave number as 𝑘4 (or in wavelength as 𝜆−4), which is

known as Rayleigh’s law.

Scattering by a dielectric sphere

We consider a small dielectric sphere of radius 𝑎 with a uniform isotropic dielectric constant

휀𝑟(𝜔). The electric dipole moment induced by the incident field is

𝐩 = 4𝜋휀0 (휀𝑟 − 1

휀𝑟 + 2) 𝑎3𝐄𝑖

(3.5)

Then the differential cross section is

𝑑𝜎

𝑑Ω(𝐧, 𝛜; 𝐧0, 𝛜0) = 𝑘4𝑎6 |

휀𝑟 − 1

휀𝑟 + 2|

2

|𝛜∗ ⋅ 𝛜𝟎|2 (3.6)

The scattered radiation is linearly polarized in the plane defined by 𝛜 and 𝐧 (Fig 3.2).

Typically the incident radiation is unpolarized, i.e., an equal incoherent mixture of 𝛜0 = 𝐞𝑥

(polarization parallel to the scattering plane) and 𝛜0 = 𝐞𝑦 (polarization perpendicular to the

scattering plane):

(i) for 𝛜0 = 𝐞𝑥, 𝛜∥ = cos 𝜃 𝐞𝑥 − sin 𝜃 𝐞𝑧

(ii) for 𝛜0 = 𝐞𝑦, 𝛜⊥ = 𝐞𝑦.

Therefore, we can write

𝑑𝜎

𝑑Ω=

𝑑𝜎∥

𝑑Ω+

𝑑𝜎⊥

𝑑Ω (3.7)

where

Fig 3.2 Polarization and propagation

vectors for the incident and scattered

radiation

3

𝑑𝜎∥

𝑑Ω=

𝑘4𝑎6

2 |휀𝑟 − 1휀𝑟 + 2|

2

cos2 𝜃

𝑑𝜎⊥

𝑑Ω=

𝑘4𝑎6

2 |휀𝑟 − 1휀𝑟 + 2|

2

(3.8)

Thus the total differential cross section is

𝑑𝜎

𝑑Ω= 𝑘4𝑎6 |

휀𝑟 − 1

휀𝑟 + 2|

2 1

2(1 + cos2 𝜃) (3.9)

and the total cross section is

𝜎 = ∫ (𝑑𝜎

𝑑Ω) 𝑑Ω =

8𝜋

3𝑘4𝑎6 |

휀𝑟 − 1

휀𝑟 + 2|

2

(3.10)

We can see that the scattered radiation is partially polarized. Its polarization Π(𝜃) is defined as

Π(𝜃) =

𝑑𝜎⊥

𝑑Ω−

𝑑𝜎∥

𝑑Ω𝑑𝜎⊥

𝑑Ω+

𝑑𝜎∥

𝑑Ω

=sin2 𝜃

1 + cos2 𝜃 (3.11)

Π(𝜃) ≈ 1, i.e., the scattered radiation is linearly polarized, around 𝜃 = 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 𝜔 (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 𝑟 from a simple linear array of 𝑁

parallel dipoles (𝐩𝑛 = 𝑝𝐞𝑥 , 𝑛 = 0,1, … 𝑁 − 1) spaced a distance 𝑑 (≪ 𝑟) apart (see Fig. 3.4).

The electric dipole field (Eq. 3.2) can be written as 𝐄 = 𝐸𝑥𝐞𝑥, where

𝐸𝑥(𝑟, 𝜃, 𝑡) =𝑝𝑘2

4𝜋휀0

𝑒𝑖(𝑘𝑟−𝜔𝑡)

𝑟∑ 𝑒−𝑖𝑘𝑛𝑑 sin 𝜃

𝑁−1

𝑛=0

=𝑝𝑘2

4𝜋휀0

𝑒𝑖(𝑘𝑟−𝜔𝑡)

𝑟

1 − 𝑒−𝑖𝑁𝛼

1 − 𝑒−𝑖𝛼 (3.12)

with 𝛼 ≡ 𝑘𝑑 sin 𝜃. Then we find the intensity

𝐼(𝑟, 𝜃) =1

2𝑐휀0|𝐸𝑥|2 =

𝑐𝑝2𝑘4

32𝜋2휀0𝑟2

sin2 𝑁𝛼2

sin2 𝛼2

(3.13)

The intensity pattern for sets of 𝑁 = 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 𝜶 = 𝒌𝒅 𝐬𝐢𝐧 𝜽 for sets of 𝑵 = 𝟐, 𝟒, 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 𝜔 is written as

𝐏(𝐱, 𝑡) = 𝑃(𝐱)𝑒−𝑖𝜔𝑡𝐞𝑥 (3.14)

Consider then a thin sheet of material of uniform polarization and thickness 𝛿 (Fig 3.6).

Using Eq. 3.2, we integrate to find the electric field at a point which lies a distance 𝑧 from the

sheet.

𝐸𝑥(𝑧) =1

4𝜋휀0𝑘2𝑃𝛿 ∫

𝑒𝑖𝑘𝑅

𝑅𝑑𝑆 (3.15)

Using cylindrical coordinates (𝑟, 𝜃) where 𝑅2 = 𝑟2 + 𝑧2, we write the surface integral as

∫𝑒𝑖𝑘𝑅

𝑅𝑑𝑆 = ∫ 𝑑𝜃 ∫

𝑒𝑖𝑘𝑅

𝑅

𝑟𝑚𝑎𝑥(𝜃)

0

2𝜋

0

𝑟𝑑𝑟

= ∫ 𝑑𝜃 ∫𝑒𝑖𝑘𝑅

𝑅

𝑅𝑚𝑎𝑥(𝜃)

𝑧

2𝜋

0

𝑅𝑑𝑅

=1

𝑖𝑘∫ (𝑒𝑖𝑘𝑅𝑚𝑎𝑥(𝜃) − 𝑒𝑖𝑘𝑧)

2𝜋

0

𝑑𝜃

The term 𝑒𝑖𝑘𝑅𝑚𝑎𝑥(𝜃) varies extremely rapidly with 𝑅𝑚𝑎𝑥(𝜃), taking on every possible phase if

𝑅𝑚𝑎𝑥(𝜃) varies as much as a wavelength. Hence it averages out by the integral over 𝜃. This

leads us to the result

Fig 3.6 Dipole radiation by an induced

polarization in a sheet of thickness 𝜹

6

𝐸𝑥(𝑧) = −𝑖𝑘

2휀0𝑃𝛿𝑒𝑖𝑘𝑧 (3.16)

We assume that 𝑃 is due to an incoming field

𝐸𝑥𝑖𝑛𝑐(𝑧, 𝑡) = 𝐸0𝑒𝑖(𝑘𝑧−𝜔𝑡) (3.17)

Then the dipole radiation field due to the induced polarization 𝑃 = 𝜒𝑒휀0𝐸0 is written as

𝐸𝑥𝑖𝑛𝑑(𝑧, 𝑡) = −

1

2𝑖𝑘𝜒𝑒𝛿 𝐸0𝑒𝑖(𝑘𝑧−𝜔𝑡) (3.18)

Total field: superposition of incoming and induced dipole fields

Now we consider that the uniform medium of 𝜒𝑒 fills the half-infinite space for 𝑧 > 0. Then the

total field at a position 𝑧 is given by a sum of three terms: the incident field, the contributions

from slices to the left of 𝑧, and the contributions from slices to the right of 𝑧. Thus we have

𝐸𝑥(𝑧) = 𝐸0𝑒𝑖𝑘𝑧 −1

2𝑖𝑘𝜒𝑒 ∫ 𝐸𝑥(𝑧′)𝑒𝑖𝑘(𝑧−𝑧′)

𝑧

0

𝑑𝑧′ −1

2𝑖𝑘𝜒𝑒 ∫ 𝐸𝑥(𝑧′)𝑒−𝑖𝑘(𝑧−𝑧′)

∞

𝑧

𝑑𝑧′ (3.19)

Differentiating 𝐸𝑥(𝑧) twice with respect to 𝑧, we obtain the wave equation

𝑑2𝐸𝑥

𝑑𝑧2= −𝑘2𝐸𝑥 − 𝑘2𝜒𝑒𝐸𝑥 = −𝑘2(1 + 𝜒𝑒)𝐸𝑥 = −𝑛2 (

𝜔

𝑐)

2

𝐸𝑥 (3.20)

where 𝑛 = √1 + 𝜒𝑒 is the index of refraction. Thus the total field is

𝐸𝑥(𝑧) = 𝐸𝑥𝑒𝑖𝑛(𝜔𝑐

)𝑧 (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 𝜃 ≤ 𝜆/𝑑,

where 𝜆 is the wavelength and 𝑑 is a linear dimension of the aperture or obstacle. Although

diffraction effects are generally most pronounced for waves where 𝜆 is roughly similar to the

dimensions of the diffracting objects, the various approximation of diffraction theory work best

for 𝜆 ≪ 𝑑 and fail badly for 𝜆~𝑑 or 𝜆 > 𝑑.

Historically, diffraction patterns were classed as Fresnel or Fraunhofer diffraction, depending on

the relative geometry involved. If 𝑟𝜆 < 𝑑2 or 𝑟𝜆~𝑑2 where 𝑟 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

𝑟𝜆 ≫ 𝑑2, 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 “equivalent” 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 “stopper”

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 𝑃𝑠 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

𝐸𝑥(𝑧, 𝑡) = −𝑖𝑘

2휀0𝑃𝑠𝑒𝑖(𝑘𝑧−𝜔𝑡) (3.22)

with the incoming radiation

𝐸𝑥𝑖𝑛𝑐(𝑧, 𝑡) = 𝐸0𝑒𝑖(𝑘𝑧−𝜔𝑡) (3.23)

The in order that the total radiation to the right of the wall be zero, we must have

𝑃𝑠 =2휀0𝐸0

𝑖𝑘 (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 −𝑃𝑠. If we let 𝐱′ refer to a point on the aperture and 𝐱 be the

place at which we wish to know the electric field, then, using Eqs. 3.2 and 3.24, we obtain

𝐸𝑥(𝐱, 𝑡) ≅𝑖𝑘𝐸0

2𝜋𝑒−𝑖𝜔𝑡 ∫

𝑒𝑖𝑘|𝐱−𝐱′|

|𝐱 − 𝐱′|𝑑𝑆′ (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 𝑥 = −𝑎/2 to 𝑥 = +𝑎/2 and from 𝑦 =

−𝑏/2 to 𝑦 = +𝑏/2. The incoming radiation is polarized in the 𝑥 direction. We assume that

𝑎, 𝑏 ≫ 𝜆 so that the diffraction angle is small and thus Eq. 3.25 is valid. We have then

𝐸𝑥(𝑥, 𝑦, 𝑧, 𝑡) =𝑖𝑘𝐸0

2𝜋𝑒−𝑖𝜔𝑡 ∫ ∫

𝑒𝑖𝑘|𝐱−𝐱′|

|𝐱 − 𝐱′|

𝑎2

−𝑎2

𝑏2

−𝑏2

𝑑𝑥′𝑑𝑦′ (3.26)

We apply the condition 𝑎, 𝑏 ≪ 𝑟

|𝐱 − 𝐱′| ≅ 𝑟 − 𝐧 ⋅ 𝐱′ (3.27)

where 𝐧 is a unit vector in the direction of 𝐱. Then, Eq. 3.26 reduces to

𝐸𝑥(𝑥, 𝑦, 𝑧) ≅𝑖𝑘𝐸0

2𝜋

𝑒𝑖𝑘𝑟

𝑟∫ ∫ 𝑒−𝑖𝑘𝐧⋅𝐱′

𝑎2

−𝑎2

𝑏2

−𝑏2

𝑑𝑥′𝑑𝑦′ (3.28)

The integral splits into the product of an integral over 𝑥′ and one over 𝑦′. Thus we have

𝐸𝑥(𝑥, 𝑦, 𝑧) =𝑖𝑘𝐸0

2𝜋

𝑒𝑖𝑘𝑟

𝑟(∫ 𝑒−𝑖𝑘(sin 𝜃𝑥)𝑥′

𝑎2

−𝑎2

𝑑𝑥′) (∫ 𝑒−𝑖𝑘(sin 𝜃𝑦)𝑦′

𝑏2

−𝑏2

𝑑𝑦′) (3.29)

where

9

sin 𝜃𝑥 =𝑥

𝑟 and sin 𝜃𝑦 =

𝑦

𝑟 (3.30)

Integrating, we obtain

𝐸𝑥(𝑥, 𝑦, 𝑧) =𝑖𝑎𝑏𝑘𝐸0𝑒𝑖𝑘𝑟

2𝜋𝑟

sin (𝑘𝑎2 sin 𝜃𝑥)

𝑘𝑎2 sin 𝜃𝑥

sin (𝑘𝑏2 sin 𝜃𝑦)

𝑘𝑏2 sin 𝜃𝑦

(3.31)

Note that the directly forward-going diffracted field (𝑥 = 0 and 𝑦 = 0),

𝐸𝑥(0,0, 𝑧) =𝑖𝑎𝑏𝑘

2𝜋𝑟𝐸0𝑒𝑖𝑘𝑟 (3.32)

Is (i) 𝐸𝑥(0,0, 𝑧) ∝ 𝐸0𝑒𝑖(𝑘𝑟+𝜋

2) and (ii) 𝐸𝑥(0,0, 𝑧) ∝ 𝜆−1, 𝑟−1.

Diffraction pattern

The time-averaged radiation power per unit solid angle is

𝑑𝑃

𝑑Ω= 𝑟2𝐒 ⋅ 𝐧 =

1

2𝑐휀0𝑟2|𝐄|2 = 𝐼0

sin2 (𝑘𝑎2 sin 𝜃𝑥)

(𝑘𝑎2 sin 𝜃𝑥)

2

sin2 (𝑘𝑏2 sin 𝜃𝑦)

(𝑘𝑏2 sin 𝜃𝑦)

2 (3.33)

where

𝐼0 =𝑐휀0𝑎2𝑏2𝑘2𝐸0

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 𝜃𝑥 =2𝑚𝜋

𝑘𝑎=

𝑚𝜆

𝑎 or sin 𝜃𝑦 =

2𝑛𝜋

𝑘𝑏=

𝑛𝜆

𝑏,

(3.35)

for 𝑚, 𝑛 = 1,2,3, … .

Fig. 3.8 Intensity of diffracted radiation versus (𝒌𝒂/𝟐) 𝐬𝐢𝐧 𝜽𝒙 and (𝒌𝒃/𝟐) 𝐬𝐢𝐧 𝜽𝒚 for a rectangular

aperture

10

Diffraction by a circular aperture

We consider a circular aperture of radius 𝑎 as shown in Fig. 3.9.

The incoming radiation polarized in the 𝑥 direction is normally incident on the aperture. The

field of diffracted radiation then is

𝐸𝑥(𝑥, 𝑦, 𝑧, 𝑡) =𝑖𝑘𝐸0

2𝜋𝑒−𝑖𝜔𝑡 ∫ ∫

𝑒𝑖𝑘|𝐱−𝐱′|

|𝐱 − 𝐱′|

𝑎

0

2𝜋

0

𝑟′𝑑𝑟′𝑑𝜙′ (3.36)

Using Eq. 3.27, we obtain

𝐸𝑥(𝑥, 𝑦, 𝑧) ≅𝑖𝑘𝐸0

2𝜋

𝑒𝑖𝑘𝑟

𝑟∫ ∫ 𝑒−𝑖𝑘𝐧⋅𝐱′

𝑎

0

2𝜋

0

𝑟′𝑑𝑟′𝑑𝜙′ (3.37)

Since 𝐧 ⋅ 𝐱′ = sin 𝜃 cos 𝜙 ⋅ 𝑟′ cos 𝜙′ + sin 𝜃 sin 𝜙 ⋅ 𝑟′ sin 𝜙′ = 𝑟′ sin 𝜃 cos(𝜙 − 𝜙′),

𝐸𝑥(𝑥, 𝑦, 𝑧) =𝑖𝑘𝐸0

2𝜋

𝑒𝑖𝑘𝑟

𝑟∫ [∫ 𝑒−𝑖𝑘𝑟′ sin 𝜃 cos(𝜙−𝜙′)

2𝜋

0

𝑑𝜙′]𝑎

0

𝑟′𝑑𝑟′ (3.38)

The angular integral can be transformed into

1

2𝜋∫ 𝑒−𝑖𝑏𝑟′ cos 𝛽

2𝜋

0

𝑑𝛽 = 𝐽0(𝑏𝑟′) (3.39)

where 𝛽 = 𝜙 − 𝜙′ and 𝑏 = 𝑘 sin 𝜃. Then the field is expressed as

𝐸𝑥(𝑥, 𝑦, 𝑧) = 𝑖𝑘𝐸0

𝑒𝑖𝑘𝑟

𝑟∫ 𝐽0(𝑏𝑟′)

𝑎

0

𝑟′𝑑𝑟′ (3.40)

Since 𝑑

𝑑𝑥[𝑥𝐽1(𝑥)] = 𝑥𝐽0(𝑥), the radial integral can be done directly:

Fig 3.9

11

𝐸𝑥(𝑥, 𝑦, 𝑧) = 𝑖𝑘𝐸0

𝑒𝑖𝑘𝑟

𝑟

1

𝑏2[𝑎𝑏𝐽1(𝑎𝑏)] =

𝑖𝑘𝑒𝑖𝑘𝑟

𝑟𝑎2𝐸0

𝐽1(𝑘𝑎 sin 𝜃)

𝑘𝑎 sin 𝜃

(3.41)

The time-averaged diffracted power per unit solid angle is

𝑑𝑃

𝑑Ω= 𝑃𝑖

(𝑘𝑎)2

4𝜋|2𝐽1(𝑘𝑎 sin 𝜃)

𝑘𝑎 sin 𝜃|

2

(3.42)

where

𝑃𝑖 =1

2𝑐휀0𝐸0

2(𝜋𝑎2) (3.43)

is the total power normally incident on the aperture. Figure 10.10 shows the diffraction pattern.

3.4. Babinet’s Principle of Complementary Screens

Babinet’s principle relates the diffraction fields of one diffracting screen to those of the

complementary screen. If the surface of the original screen is 𝑆𝑎 and that of the

complementary screen is 𝑆𝑏, then 𝑆𝑎 + 𝑆𝑏 = 𝑆 forms the entire screen as shown in Fig. 3.11.

Fig 3.10

Fig 3.11 A diffraction screen 𝑺𝒂 and its

complementary screen 𝑺𝒃

12

If the surface integral in Eq. 3.25 is done over 𝑆𝑎 and 𝑆𝑏, we have the following relation

between the diffraction fields 𝐸𝑎 and 𝐸𝑏:

𝐸𝑎 + 𝐸𝑏 = 𝐸0 (3.44)

where 𝐸0 is the unobstrcuted field. The principle implies that when 𝐸0 = 0, 𝐸𝑎 = −𝐸𝑏, 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 𝑆𝑎 and 𝑆𝑏 in place where 𝐸0 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)]