Calculation of true thickness of the bed

CSIR_NET_2018_ DECEMBER

115) A P-wave is incident at an angle of 30⁰ upon a horizontal interface separating two media which are both poissions solid. If the P-wave velocity of the first layer is 3 km/sec and the critical angle is 60⁰ , then the angle of reflection of the shear wave is 

 

A) $\arcsin(\frac{1}{6})$ 

B) $\arcsin(\frac{1}{4})$

C) $\arcsin(\frac{1}{3})$ 

 D) $\arcsin(\frac{1}{2})$

(Thanks to Chandrasekhar, ANU)

Solution:

Here we use two formulas 

 

1. $\frac{V_{P}}{V_{S}}= \sqrt{\frac{2(1-\nu)}{(1-2\nu)}}$
   
   
2. $\frac{Sin Ip}{Vp1}=\frac{Sin Is}{Vs1}=\frac{Sin Rp}{Vp2}=\frac{Sin Rs}{Vs2}$

Poissions solid value ($\nu$) =0.25

Velocity of the P-wave (V )= 3 km/s

 

$\frac{V_{P}}{V_{S}}= \sqrt{\frac{2(1-\nu)}{(1-2\nu)}}$

$\frac{3}{V_{S}}=\sqrt{\frac{2(1-0.25)}{(1-2*0.25)}}$

$\frac{3}{V_{S}}=\sqrt{\frac{2*0.75}{0.5}}$

$\frac{3}{V_{S}}=\sqrt{3}$

$V_{S}=\frac{3}{\sqrt{3}}$

$V_{S}=\sqrt{3}km/sec$

i.e $V_{s_{1}}=\sqrt{3}km/sec$

now we use the 2^nd formula

$\frac{Sin Ip}{Vp1}=\frac{Sin Is}{Vs1}=\frac{Sin Rp}{Vp2}=\frac{Sin Rs}{Vs2}$

Ip - incident  and reflected angles  of P wave

Is - reflected angle of S-wave

Vp1 - P-wave velocity of the upper layer

Vs1 - S wave velocity of the upper layer

Vp2 - P wave velocity of the lower layer

Vs2 - S wave velocity of the lower layer

 

from this we can write $\frac{Sin Ip}{Vp1}=\frac{Sin Is}{Vs1}$

$Sin Is=Sin Ip(\frac{Vs_{1}}{Vp_{1}})$

$Sin Is=Sin 30(\frac{\sqrt{3}}{3})$

$Sin Is=\frac{1}{2}(\frac{\sqrt{3}}{3})$

$Sin Is=\frac{1}{3.4}$

$I_{s}=\arcsin(\frac{1}{3})$

References : Fundamentals of Geophysics , William Lowrie