Calculation of true thickness of the bed

Gate_2018

23) Assume a flat earth with crustal thickness of 35 km and average crustal and mantle P-wave velocities of 6.4 km/sec and 8.1 km/sec respectively. The minimum distance from the epicenter of the near surface earthquake at which P_n wave are observed is ________km.

(Thanks to Chandrasekhar, ANU)

Solution:

$P_n$ - Doubly refracted wave (equivalent to a head wave) that travelled partly in the upper mantle.

The interception of reflected wave and refracted wave is called "Critical distance", here both the waves have equal travel times.

The minimum distance is nothing but  “Critical Distance”.

The Critical distance $(X_{c})= 2Z \tan\theta_{c}$----(1)

To find out the $\tan\theta_{c}$ ,

From snell's law at critical angle is given that,

 

$\sin\theta_{c}=\frac{V_{1}}{V_{2}}$

$\cos^{2}\theta_{c}=1-\sin^{2}\theta_{c}$

$\cos\theta_{c}=\sqrt{1-\sin^{2}\theta_{c}}$

$\cos\theta_{c}=\sqrt{1-\frac{V_{2}^{2}}{V_{1}^{2}}}$

$\cos\theta_{c}=\sqrt{\frac{V_{2}^{2}-V_{1}^{2}}{V_{2}^{2}}}$

$\tan\theta_{c}=\frac{\sin\theta_{c}}{\cos\theta_{c}}$

$\tan\theta_{c}=\frac{\frac{V_{1}}{V_{2}}}{\sqrt{\frac{V_{2}^{2}-V_{1}^{2}}{V_{2}^{2}}}}$

$\tan\theta_{c}=\frac{V_{1}}{\sqrt{V_{2}^{2}-V_{1}^{2}}}$

Now substitute the given values in the equation (1) we get 

 

$X_{c}= 2Z\frac{V_{1}}{\sqrt{V_{2}^{2}-V_{1}^{2}}}$

$X_{c}= 2\times 35\times\frac{6.4}{\sqrt{(8.1)^{2}-(6.4)^{2}}}$

$X_{c}= \frac{448}{\sqrt{65.61-40.96}}$

$X_{c}= \frac{448}{\sqrt{24.64}}$

$X_{c}= \frac{448}{{4.9648}}$

$X_{c}= 90.23525 km$

Extra information:

 1. Minimum distance- Critical Distance

$X_{c}= critical distance=2Z\tan\theta_{c}=2Z\frac{V_{1}}{\sqrt{V_{2}^{2}-V_{2}^{1}}}$

2. Maximum distance - Cross over Distance

$X_{cr} = Cross over distance= 2Z\sqrt{\frac{V_{2}+V_{1}}{V_{2}-V_{1}}}$

3. Intercept time

$t_{i} = intercept travell time= 2Z\frac{\sqrt{V_{2}^{2}-V_{1}^{2}}}{V_{2}V_{1}}$.

Reference : Fundamentals of Geophysics by William Lowrie.