CSIR-NET 2014
101. Assume that the P-wave velocities in the overlying and the underlying layers are 6 km/s and 7 km/s respectively. Now, if we consider that the thickness of the overlying layer is 5km, then the direct wave and the head wave will have the same travel time at distance of
A) 5√13 km B) 42km C)65km D)10√13 km
Solution:
The interception of
direct wave and refracted wave is called “Crossover distance”.
At crossover
distance the refraction and direct waves have equal travel times.
The direct wave
propagates along the atmosphere -upper layer (called layer 1)
boundary.
If the incident wave
hits at the critical angle, the critically refracted head wave
travels along the boundary of layer 1 and layer 2.
X crossover =
2*Z √((V2+V1)/(V2-V1))
--------------------(1)
Velocity
of first layer (V1)
= 6 km/s
Velocity
of second layer (V2)
= 7 km/s
Depth
(Z) = 5 km
X crossover
=2*5 √((7+6)
/ (7-6))
X crossover
=10
√(13/1)
X crossover
=10
√13
Therefore
Cross over distance = 10√13
km
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