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GATE 2019 58

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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)}


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}












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