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

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>V_1..If the critical angle is 45^0then which one of the following is correct? (V_u and V_d are up dip and down dip velocities).

A) V₁ = V ₂= V_U 

B) V_U > V_d> V₁  

C) V₁> V_d < V_u 

D) V_u < V_d < 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)}$

For Up dip $ V_{u}= \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

$V_{u}> V_{d}>V_{1}$