GATE-2018 (24) :Wyllie – Time average equation

CSIR-NET 2015

90. In a refraction survey, the velocities inferred for the upper and the lower layers are 3000m/s and 5000m/s respectively. If the cross over distance is 4000m, the depth of the refractor is

A) 2000m              B) 1000m           C) 4000m            D) 6000m

Solution:

The intersection of direct wave and refracted wave is called “Crossover distance”

At the crossover distance, the refraction and direct waves have equal travel times.

If the incident wave hits at the critical angle, the critically refracted head wave travels along the boundary of layer1 and layer 2

Velocity of first layer (V₁) = 3000 m/s

Velocity of second layer (V₂) = 5000 m/s

Cross over distance (X_cr) = 4000 m

X_cr = 2*Z √((V₂+V₁)/(V₂-V₁)) --------------------(1)

Z= (X_cr/2)* √((V2-V1) / ( V2+ V1))

Z= (4000/2)* √ (5000 -3000) /(5000+3000)

Z= 2000 * √ ((2000)/(8000))

Z= 2000* (1/2)

Z= 1000 m

Therefore Depth (Z) = 1000 m