Calculation of true thickness of the bed

CSIR-NET JUNE 2018

100)The travel time line segments in an up-dip seismic refraction survey have gradients of 0.6666 sec/km and 0.3333 sec/km respectively. If the critical angle is 60⁰, the dip angle of the refractor is a 
(Thanks to Suresh Sir ANU) 
(Thanks to Neeraja AU)

a)10⁰ 
b)15^o 
c)30⁰ 
d)45⁰

sol:

Gradients are given ,

$\frac{dt_1}{dx_1}=0.6666sec/km$

$\frac{dt_2}{dx_2}=0.3333sec/km$

 Velocities is nothing but inverse of the Gradients .

$V_1=\frac{1}{0.6666}=1.5015 km/sec$

$V_2=\frac{1}{0.3333}=3.0003 km/sec$

$\sin(i_c-\theta)=\frac{V_1}{V_2}----------(1)$

Critical angle is given 

$I_c=60^o$

$V_1=1.5km/s$

$V_2=3km/s$ 

What is the dip angle$(\theta)=?$

$\sin(60^o-\theta)=\frac{1.5}{3}$

$\sin(60^o-\theta)=0.5$

$\sin(60^o-\theta)=\frac{1}{2}$

$\sin(60^o-\theta)=sin 30^o$

$60^o-\theta=\sin^{-1}\sin30^o$

$60^o-\theta=30^o$

$\theta=30^o$