Question

An earthquake P wave passes across a boundary in rock where its velocity increases from 6.5 km/s to 8.0 km/s. If it strikes this boundary at 30o, what is the angle of refraction?

Answer 1

Answer:

37.98°

Explanation:

Data provided in the question:

Initial velocity of the P wave, v₁ = 6.5 km/s          

Final velocity of the P wave, v₂ = 8.0 Km/s            

Angle of strike at the boundary, s = 30°

Now                              

let the angle of refraction be 'r'

we know,

$\frac{\sin(r)}{\sin(s)}=\frac{v_2}{v_1}$                      

therefore,

$\frac{\sin(r)}{\sin(30)}=\frac{8}{6.5}$

or

sin(r) = 1.231 × sin(30)

or

sin(r) = 0.6154

or

r = 37.98°

Answer 2

Answer:

$\theta = 37.97 degree$

Explanation:

Given data:

velocity $v_1 = 6.5 km/s$

$v_2 = 8.0 km/s$

Angle of striking = 30 degree

angle of refraction is assumed to be X

we know that

$\frac{sin\theta_2}{sin\theta_1} =\frac{v_2}{v_1}$

$\frac{sin\theta_2}{sin 30} =\frac{8}{6.5}$

$\frac{sin\theta_2}{0.5} =\frac{8}{6.5}$

$sin\theta_2 = \frac{8}{6.5} \times 0.5$

$\theta = sin^{-1} 0.615$

$\theta = 37.97 degree$