Geoscience Reference
In-Depth Information
Table 4.2
The width of the Fresnel zone for
specified depths and signal wavelengths
Wavelength (km)
Depth (km)
0.10
0.20
0.30
0.40
0.50
2
0.63
0.90
1.11
1.28
1.44
5
1.00
1.42
1.74
2.01
2.25
10
1.42
2.00
2.45
2.84
3.17
30
2.45
3.46
4.24
4.90
5.48
50
3.16
4.47
5.48
6.33
7.08
unlikely to be resolved as being due to a fault since it would appear as a slight time
error. Thus, a 20-Hz signal could resolve faults with throws in excess of about 30 m
in 4.8-km s
−
1
material. However, in 6.4-km s
−
1
material, the same 20-Hz signal
could only resolve faults with throws of more than 40 m; in 7.2-km s
−
1
material,
the minimum resolvable throw would be 45 m. A 30-Hz signal would mean
that faults of 20, 27 and 30 m, respectively, would be resolvable. Clearly, shorter
wavelengths (higher frequencies) give better resolution of structures and greater
detail than are obtainable from long wavelengths. Unfortunately, high-frequency
vibrations do not travel as well through rock as low-frequency vibrations do
(Section 8.1.3), so high frequencies are generally absent from reflections from
deep structures.
The lateral extent that a reflector must have in order for it to be detectable
is also controlled by wavelength. To understand the lateral resolution obtainable
with a given source signal, we need to introduce the concept of the
Fresnel
zone
. The first Fresnel zone is that part of a reflecting interface which returns
energy back to the receiver within half a cycle of the first reflection (Fig. 4.50).
In plan view, the first Fresnel zone is circular; the second Fresnel zone is a
ring (concentric with the first Fresnel zone) from which the reflected energy
is delayed by one-half to one cycle; and so on for the third, fourth and
n
th
Fresnel zones. For an interface to be identified as such, it must be at least as
wide as the first Fresnel zone (usually referred to as
the
Fresnel zone). If a
reflecting zone is narrower than the first Fresnel zone, it effectively appears to be a
diffractor.
The width of the Fresnel zone
w
can be calculated by applying Pythagoras'
theorem to triangle SOQ of Fig. 4.50(b):
d
+
4
2
w
2
2
=
d
2
+
(4.100)
On rearranging this, we obtain
2
4
=
2
d
λ
+
λ
w
2
(4.101)
