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6.1 Diffraction at a Canonical Conic Surface 123

region 0 ≤ ϑ ≤ ,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

sin

π

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

1

 

 

 

 

 

f (1)(2)

=

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n

 

cos

π

 

 

 

cos

π + ϑ

cos

π

 

cos

π

2ω

 

ϑ

 

 

 

 

 

 

 

 

 

 

 

 

n

 

 

n

 

 

n

 

 

+ n

+

 

 

 

 

 

 

 

 

 

 

 

 

sin ω

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(6.28)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

cos ω

cos

+

ϑ )

 

 

 

 

 

 

 

 

 

 

and

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

sin

π

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

1

 

 

 

 

g(1)(2)

=

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

π + ϑ

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n

 

 

cos

π

 

 

 

cos

+ cos

π

 

cos

π

2ω

 

ϑ

 

 

 

 

 

 

 

 

 

 

 

 

n

 

 

n

 

 

n

 

 

+ n

+

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

sin+ ϑ )

 

,

 

 

 

 

 

 

 

 

 

 

 

 

(6.29)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

cos ω

cos

+

ϑ )

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

and for the region π ω ϑ π the corresponding expressions are

 

 

 

 

 

 

 

sin

π

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

1

 

 

 

 

f (1)(2)

=

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

π ϑ

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n

 

 

cos

π

 

 

 

cos

cos

π

 

cos

π

2ω

 

ϑ

 

 

 

 

 

 

 

 

 

 

 

 

n

 

 

n

 

 

n

 

 

n

 

 

 

 

 

 

 

 

 

 

 

 

sin ω

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(6.30)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

cos ω

cos

+

ϑ )

 

 

 

 

 

 

 

 

 

 

and

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

sin

π

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

 

 

 

 

 

1

 

 

 

 

g(1)(2)

=

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

π ϑ

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n

 

 

cos

π

 

 

 

cos

+ cos

π

 

cos

π

2ω

 

ϑ

 

 

 

 

 

 

 

 

 

 

 

 

n

 

 

n

 

 

n

 

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

sin+ ϑ )

.

 

 

 

 

 

 

 

 

 

 

(6.31)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

cos ω

cos

+

ϑ )

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Some terms in functions f (1) and g(1) are singular in certain directions ϑ ; however, such singularities cancel each other and these functions are always finite. For the forward direction ϑ = 0 (that is, the shadow boundary behind the body),

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

sin

π

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

π

 

 

 

 

1

 

 

f

(1)

(1) = f

(1)

(2) = −

 

 

 

 

 

 

n

n

 

 

cot

+

 

cot ω

(6.32)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

cos

π

 

cos

π + 2ω

 

 

2n

n

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n

 

 

 

 

 

 

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

and

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

sin

π

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

π

 

1

 

 

 

 

 

 

(1)

(1) = g

(1)

(2) =

 

 

 

 

 

n

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

g

 

 

 

 

 

 

 

 

 

 

 

cot

 

 

 

 

cot ω.

(6.33)

 

 

 

 

 

cos

π

 

cos

π + 2ω

 

2n

n

 

2

 

 

 

 

 

 

 

 

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

TEAM LinG


124 Chapter 6 Axially Symmetric Scattering of Acoustic Waves at Bodies of Revolution

For the specular direction ϑ = 2ω (corresponding to the ray reflected from the conic surface),

 

 

 

 

 

 

 

 

 

 

1

 

sin

π

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

 

π

 

1

 

 

f

(1)

(1) =

 

 

 

 

 

 

n

n

 

 

 

 

+

 

cot

+

cot ω

(6.34)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

cos

π

 

cos

π − 2ω

 

 

2n

n

 

2

 

 

 

 

 

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

and

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

sin

π

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

 

 

π

 

1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

g

(1)

(1) =

 

 

 

 

 

 

 

n

 

n

 

 

 

 

cot

+

cot ω.

(6.35)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

π

 

 

 

π

2ω

 

2n

n

 

2

 

 

 

 

cos

 

 

 

cos

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

n

 

 

 

 

 

 

n

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Notice also that for the directions ϑ = 0 and ϑ = π , the relationships

f (1)(1) = f (1)(2)

and

g(1)(1) = g(1)(2)

(6.36)

are valid, due to the symmetry of the problem.

6.1.3Focal Fields

Focal fields for acoustic and electromagnetic waves generated by the nonuniform sources

j(1) are different, due to the vector nature of electromagnetic fields.

Given the incident wave (6.1), every point at the z-axis is a focal point for diffracted edge waves (Fig. 6.2). In the far zone, the position of any focal points with z > 0 is determined by the coordinate ϑ = 0, and the points with z < 0 are characterized by the coordinate ϑ = π . For the focal points, the general field expressions (6.8) and (6.9) are simplified. Under the condition a ξeff , they can be written as

a eikR

 

ξeff

Js(1)(kξ , ϕ0)e±ikξ cos ω

us(1) = −u0

 

 

 

 

2

R

0

ξeff

 

 

 

+Js(1)(kξ , α ϕ0)e ikξ cos dξ

0

and

 

 

 

 

 

ika eikR

sin ω

ξeff

 

uh(1) = −u0

 

 

 

 

Js(1)(kξ , ϕ0)e±ikξ

2

 

R

0

± sin

 

ξeff

Js(1)(kξ , α ϕ0)e ikξ cos dξ

0

dξ

(6.37)

cos ω dξ

. (6.38)

TEAM LinG


6.1 Diffraction at a Canonical Conic Surface 125

The upper (lower) sign here relates to ϑ = 0 (ϑ = π ). In terms of the local coordinates they look the same for both directions:

 

 

a eikR

ξeff

 

 

 

 

 

us(1) = −u0

 

 

 

 

 

 

 

 

 

Js(1)(kξ , ϕ0)eikξ cos ϕ1 dξ

 

 

2

 

 

R

0

 

 

 

+

ξeff

Js(1)(kξ , α ϕ0)eikξ cosϕ1) dξ .

 

(6.39)

0

 

 

 

 

uh(1) = −u0

ika eikR

sin ϕ1

ξeff

Js(1)(kξ , ϕ0)eikξ cos ϕ1 dξ

 

 

2

 

 

R

 

0

 

 

 

 

 

 

 

 

 

 

 

 

 

ξeff

 

 

,

 

+ sinϕ1)

0

Js(1)(kξ , α ϕ0)eikξ cosϕ1) dξ

(6.40)

with ϕ1 = ω(ϕ1 = π + ω) when ϑ = π (ϑ = 0). Then we use Equations (4.29) and (4.30) and find the following expressions for the focal field:

us(1) = u0af (1)(1)

 

eikR

 

(6.41)

 

R

and

 

 

 

 

 

uh(1) = u0ag(1)(1)

eikR

 

 

 

.

(6.42)

 

R

 

 

 

 

 

 

 

It is seen that the focal field is ka times higher in magnitude than the ray fields (6.21) and (6.22). In conjunction with Equations (6.41) and (6.42), we recall relationships (6.36), which are valid for the focal points.

6.1.4Bessel Interpolations for the Field us,h(1)

In the previous sections, we derived the ray and focal asymptotic expressions for the field us,h(1). The ray asymptotics (6.21) to (6.24) are valid under the condition ka sin ϑ 1 (i.e., away from the focal line), and the asymptotic expressions (6.41) and (6.42) determine the field directly on the focal line (where sin ϑ = 0). Now our goal is to construct such approximations, that would continuously describe the diffracted field both in the ray and focal regions. This can be done utilizing the Bessel functions J0(ka sin ϑ ) and J1(ka sin ϑ ).

For the large argument (ka sin ϑ 1), they can be approximated by (Gradshteyn and Ryzhik, 1994)

1

%eika sin ϑ i

π

+ eika sin ϑ +i

π

&

 

 

J0(ka sin ϑ )

 

4

4

 

(6.43)

2π ka sin ϑ

 

 

 

 

and

 

 

 

 

 

 

 

 

1

%eika sin ϑ i

3π

 

3π

& .

 

J1(ka sin ϑ )

 

 

+ eika sin ϑ +i

 

 

 

(6.44)

 

4

 

4

2π ka sin ϑ

 

 

 

TEAM LinG


126 Chapter 6 Axially Symmetric Scattering of Acoustic Waves at Bodies of Revolution

It follows from these expressions that

 

 

eika sin ϑ +i π4

 

1

[J0(ka sin ϑ ) i J1(ka sin ϑ )]

 

 

 

 

 

 

 

 

 

 

 

 

(6.45)

 

 

2

 

 

2π ka sin ϑ

and

 

 

 

 

 

 

 

 

 

eika sin ϑ i π4

 

1

 

[J0(ka sin ϑ ) + i J1(ka sin ϑ )].

 

 

 

 

 

 

 

 

 

 

(6.46)

 

2

 

 

2π ka sin ϑ

 

With these observations, the ray asymptotics (6.21) and (6.22) can be written as

 

 

 

 

a eikR

0 f (1)(1)[J0(ka sin ϑ ) i J1(ka sin ϑ )]

 

us(1) = u0

 

 

 

 

 

2

R

 

 

 

 

 

+ f 1)(2)[J0(ka sin ϑ ) + i J1(ka sin ϑ )]1

(6.47)

 

 

 

 

a eikR

0g(1)(1)[J0(ka sin ϑ ) i J1(ka sin ϑ )]

 

uh(1) = u0

 

 

 

 

 

2

R

 

 

 

 

 

+ g(1)(2)[J0(ka sin ϑ ) + i J1(ka sin ϑ )]1 .

(6.48)

Now we analytically continue these expressions into the focal region. If we take into account Equation (6.36), as well as the relationships J0(0) = 1 and J1(0) = 0, one can see that expressions (6.47) and (6.48) exactly transform into the focal asymptotics (6.41) and (6.42), when ϑ → 0 and ϑ π . This means that the above formulas (6.47) and (6.48) can be considered as the appropriate approximations for the diffracted field in the regions 0 ≤ ϑ ≤ and π ω ϑ π .

In the region ≤ ϑ π ω, where the stationary point 2 is not visible, the modified versions of Equations (6.47) and (6.48),

us(1) = u0

a

(1)(1)[J0(ka sin ϑ ) i J1(ka sin ϑ )]

eikR

(6.49)

 

 

f

 

 

 

2

R

 

 

and

 

 

 

 

 

 

 

 

 

 

 

 

 

(1)

= u0

a

(1)

(1)[J0

(ka sin ϑ ) i J1(ka sin ϑ )]

eikR

,

(6.50)

uh

 

g

 

 

 

2

 

 

R

represent the analytical continuation of the ray asymptotics (6.23) and (6.24) into the entire region ≤ ϑ π ω.

In the next sections, the above results found for the canonical problem are applied for calculation of the field scattered at certain bodies of revolution.

6.2SCATTERING AT A DISK

The geometry of the problem is shown in Figure 6.6. The incident plane wave is given by Equation (6.1) and propagates in the positive direction of the z-axis. Because of

TEAM LinG


6.2 Scattering at a Disk 127

Figure 6.6 An incident wave propagates in the direction along the z-axis and undergoes diffraction at a circular disk of radius a. Edge points 1 and 2 are in the plane x = 0.

that, and due to the axial symmetry of the disk, the scattered field is the same in all meridian planes ϕ = const. (0 ≤ ϕ ≤ 2π ). In addition, the scattered field is symmetric with respect to the disk plane: us(z) = us(z), uh(z) = −uh(z). Therefore, it is sufficient to investigate the field only in the plane ϕ = π/2 for the directions 0 ≤

ϑ π/2.

6.2.1Physical Optics Approximation

The relationships us(0) = Eϕ(0), uh(0) = Hϕ(0) are valid for the disk diffraction problem

under the conditions usinc = Eϕinc, uh(0) = Hϕ(0).

Acoustic Waves

In this approximation, the scattered field is considered as the radiation generated by the uniform component (1.31) of the scattering sources, which are induced on the illuminated side (z = −0) of the disk. This field is determined by Equations (1.16)

TEAM LinG

128 Chapter 6 Axially Symmetric Scattering of Acoustic Waves at Bodies of Revolution

and (1.17), where one should set

ϑ

=

π

,

ϕ

 

π

,

 

cos

 

sin ϑ sin ϕ ,

m

n

z

2

= 2

 

=

 

 

 

 

 

 

 

 

 

 

 

 

 

ˆ

· ˆ = − r

js = js(0) = −2iku0,

 

 

jh = jh(0) = 2u0 .

 

 

 

 

 

With these settings, the field is described by

 

 

 

 

 

 

 

 

 

 

(0)

 

 

 

ik

 

eikR

a

 

2π

 

ikr

sin ϑ sin ϕ

 

 

 

 

 

 

us

= u0

 

 

 

0 r

dr

0

e

 

 

 

dϕ

 

 

 

 

 

2π

R

 

 

 

 

and

≈ − cos ϑ ,

(6.51)

(6.52)

(6.53)

 

u(0)

=

u

 

ik

cos ϑ

eikR

a r dr

2π eikr sin ϑ sin ϕ dϕ .

(6.54)

0 2π

 

 

h

 

 

R

0

0

 

According to Gradshteyn and Ryzhik (1994),

 

 

 

2π

2π

e

 

dϕ = J0(x),

J0(x)xdx = x J1(x),

(6.55)

 

0

 

1

 

 

 

 

ix sin ϕ

 

 

 

where J0 and J1 are the Bessel functions. With these relationships, one can represent Equations (6.53) and (6.54) as

us(0) = u0

ia

eikR

 

 

 

 

J1(ka sin ϑ )

 

 

 

 

(6.56)

sin ϑ

R

 

 

and

 

 

 

 

 

uh(0) = u0

ia

 

 

eikR

 

 

cos ϑ J1(ka sin ϑ )

 

.

(6.57)

sin ϑ

R

For small arguments (ka sin ϑ 1), one can replace J1(ka sin ϑ ) by (ka sin ϑ )/2. Then we set ϑ = 0, π and find the field on the focal line:

us(0)

= u0

ika2 eikR

 

for ϑ = 0 and ϑ = π

 

 

 

 

 

 

 

 

,

 

(6.58)

2

R

 

 

and

 

 

 

 

 

 

 

 

 

 

 

 

 

(0)

 

 

ika2 eikR

 

0

 

 

uh

= ±u0

 

 

 

 

 

,

for ϑ = π

.

(6.59)

2

 

 

 

R

In the directions away from the focal line where ka sin ϑ 1, the Bessel function in Equations (6.56) and (6.57) can be replaced by its asymptotic expression (6.44).

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