Calculation of true thickness of the bed

 96. A Thick Sedimentary formation, in Which the seismic wave Velocity V Increases with depth Z from $V_0$ at the surface to $V_1$ at the bottom according to the relation $V_1=V_0e^{\lambda z}$ . the Two way reflection travel time at a point closest to the short point is …..

(Special Thanks to Pawan Singh)

(Thanks to Ashok, AKNU)

Solution:

Given that $V=V_0e^{\lambda z}$

At  Z  =  0          $V = V_0$     ( surface )

At   Z =  Z          $V = V_1$     ( bottom )

 We know that two Way Trave Time is 

$dt = \frac{2dz}{v}$

 By integrating this equation we get

$t = 2\int_{0}^{z} \frac{1}{V_0e^{\lambda z}}dz$

$=\frac{2}{V_0}\int_{0}^{z}{e^{-\lambda z}}dz$

$= \frac{2}{V_0}[\frac{e^{-\lambda z}}{-\lambda}]^z_0$

$= \frac{2}{V_0\lambda}[{-e^{-\lambda z}}]^z_0$    

 since  ${e^{-\lambda z }}= \frac{V_0}{V_1}$

$= \frac{2}{V_0\lambda}[-\frac{V_0}{V}]^z_0$

$= \frac{2}{\lambda}[-\frac{1}{V(z)}+\frac{1}{V_0}]$

$= \frac{2}{\lambda}[-\frac{1}{V_1}+\frac{1}{V_0}]$ 

$t = \frac{2}{\lambda}[\frac{1}{V_0}-\frac{1}{V_1}]$