GATE
– 2019
58)
A Reversed refraction survey was done over a two layered medium with
the interface between them dipping at an angle of 150 .
The velocities in the upper and lower medium are V1
and V2 respectively
with V2>V1..If
the critical angle is 45^0then which one of the following is correct?
(Vu and Vd are up
dip and down dip velocities).
A) V1 = V 2= VU
B) VU > Vd> V1
C) V1> Vd < Vu
D) Vu < Vd < V1
(Thanks to Chandrasekhar, ANU)
Solution:
The
apparent velocities for the refracted arrival when shooting in the
Down dip and Up dip are as follows
For Down dip V_{d}=
\frac{V_{1}}{sin(\theta_{c}+ \alpha)}
Where
Critical angle is \theta_{c} = 450
Dip of the bed is \alpha
= 150
Case(I):
For Down dip V_{d}=
\frac{V_{1}}{sin(\theta_{c}+ \alpha)}
V_{d}=
\frac{V_{1}}{sin(45+ 15)}
V_{d}=
\frac{V_{1}}{sin(60)}
V_{d}= 1.154 V_{1}
V_{d}>V_{1}
Case(2):
For Updip V_{u}=
\frac{V_{1}}{sin(\theta_{c}-\alpha)}
V_{u}=
\frac{V_{1}}{sin(45-15)}
V_{u}=
\frac{V_{1}}{sin(30)}
V_{u}= 2 V_{1}
V_{u}>V_{1}.
Hence
V_{u}>
V_{d}>V_{1}
A S-wave propagating through the earth shows the time differnce of 1.0 sec between its arrival on transverse and radial components an averaging velocity of 4km/sec and anisotropy is 8% then thickness of the layer is?
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