Calculation of true thickness of the bed

 

1. A P-wave incident on the boundary at an angle of 30 degrees between media of velocities V₁ and V₂ . The angle of P-wave incidence i_p and reflected P & S - waves are i’_p ,i’_s and refracted P & S – waves are r_p and r_s respectively .

Velocity of reflected P-wave is (Vp1) = 4000m/s

Velocity of reflected S-wave is (Vs1) = 2500m/s

Velocity of refracted P-wave is (Vp2) = 5000m/s

Velocity of refracted S-wave is (Vs2) = 3500m/s respectively.

Solution:
 
Snell's law:
$$\frac{sini_{p}}{V_{1p}}= \frac{sini^{!}_{p}}{V_{1p}} =\frac{sini^{!}_{s}}{V_{1s}}=\frac{sinr_{p}}{V_{2p}}=\frac{sinr_{s}}{V_{2s}}$$

Velocity of reflected P-wave is (Vp1) = 4000m/s

Velocity of reflected S-wave is (Vs1) = 2500m/s

Velocity of refracted P-wave is (Vp2) = 5000m/s

Velocity of refracted S-wave is (Vs2) = 3500m/s

$i_{P}$- incident  angle of P-wave
$i^{!}_{p}$ - reflected  P-wave
$i^{!}_{s}$ - reflected  s-wave
$r_{p}$- refracted p-wave
$r_{s}$- refracted s-wave

Reflected P-wave angle is:

$$\frac{sini_{p}}{V_{1p}}= \frac{sini^{!}_{p}}{V_{1p}}$$

$$\frac{sin30^{0}}{4000}= \frac{sini^{!}_{p}}{4000}$$

$$\frac{sini^{!}_{p}}{4000}=\frac{sin30^{0}}{4000}$$

$${sini^{!}_{p}}=\frac{sin30^{0}}{4000} \times4000$$

$${sini^{!}_{p}}={sin30^{0}}$$

$${i^{!}_{p}}={30^{0}}$$

Refracted P-wave angle is

$$\frac{sini_{p}}{V_{1p}}= \frac{sinr_{p}}{V_{2p}}$$

$$\frac{sin30^{0}}{4000}= \frac{sinr_{p}}{5000}$$

$$\frac{sinr_{p}}{5000}=\frac{sin30^{0}}{4000}$$

$${sinr_{p}}=\frac{sin30^{0}}{4000} \times5000$$

$${sinr_{p}}={sin30^{0}}\times1.25$$

$${sinr_{p}}=0.5 \times1.25$$

$${sinr_{p}}=0.625$$

$$r=\arcsin{(0.625)}$$

$$r=38.68^{0}$$

Reflected S-wave angle is:

$$\frac{sini_{p}}{V_{1p}}= \frac{sini^{!}_{s}}{V_{1s}}$$

$$\frac{sin30^{0}}{4000}= \frac{sini^{!}_{s}}{2500}$$

$$\frac{sini^{!}_{s}}{2500}=\frac{sin30^{0}}{4000}$$

$${sini^{!}_{s}}=\frac{sin30^{0}}{4000} \times2500$$

$${sini^{!}_{s}}=0.31$$

$$i_{s}^{!}=\arcsin{(0.31)}$$

$$i_{s}^{!}=18^{0}$$

Refracted S-wave angle is

$$\frac{sini_{p}}{V_{1p}}= \frac{sinr_{s}}{V_{2s}}$$

$$\frac{sin30^{0}}{4000}= \frac{sinr_{s}}{3500}$$

$$\frac{sinr_{s}}{3500}=\frac{sin30^{0}}{4000}$$

$${sinr_{s}}=\frac{sin30^{0}}{4000} \times3500$$

$${sinr_{s}}={sin30^{0}}\times0.875$$

$${sinr_{s}}=0.5 \times0.875$$

$${sinr_{s}}=0.4375$$

$$r_{s}=\arcsin{(0.4375)}$$

$$r_{s}=25.9^{0}$$