Chapter 11, Problem 65P

To determine

To Find: The angle of refraction.

Answer to Problem 65P

  $44°$

Explanation of Solution

Given Information:

The angle of incidence ${\mathrm{\phi }}_1=38°$

Specify gravity 1 $\mathrm{S}{\mathrm{G}}_1=3.6$

Specify gravity 2 $\mathrm{S}{\mathrm{G}}_2=2.8$

Concept used:

  $\frac{{\mathrm{v}}_1}{{\mathrm{v}}_2}=\frac{\sin ({\mathrm{\phi }}_2)}{\sin ({\mathrm{\phi }}_1)}=\sqrt{\frac{\mathrm{E}/{\mathrm{\rho }}_2}{\mathrm{E}/{\mathrm{\rho }}_1}}$ $=\sqrt{\frac{\mathrm{S}{\mathrm{G}}_2{\mathrm{\rho }}_{\mathrm{W}}}{\mathrm{S}{\mathrm{G}}_1{\mathrm{\rho }}_{\mathrm{W}}}=}$ $\sqrt{\frac{\mathrm{S}{\mathrm{G}}_1}{\mathrm{S}{\mathrm{G}}_2}}$

So,

  $\mathrm{\phi }={\sin }^{-1}\left(\sin {\mathrm{\phi }}_1\sqrt{\frac{\mathrm{S}{\mathrm{G}}_1}{\mathrm{S}{\mathrm{G}}_2}}\right)$

Here, ${\mathrm{v}}_1$ is speed of wave in rock 1,

  ${\mathrm{v}}_2$ is speed of wave in rock 2,

  ${\mathrm{\theta }}_1$ is the angle of incidence,

  $\mathrm{E}$ is elastic modulus for both the rock,

  $\mathrm{S}\mathrm{P}$ is specific gravity of rock 1,

  $\mathrm{S}\mathrm{P}$ is specific gravity of rock 2,

  ${\mathrm{\rho }}_{\mathrm{w}}$ is density of water,

  ${\mathrm{\rho }}_1$ is density of rock 1,

  ${\mathrm{\rho }}_2$ is density of rock 2.

Calculation:

  $\mathrm{\phi }={\sin }^{-1}\left(\sin {\mathrm{\phi }}_1\sqrt{\frac{\mathrm{S}{\mathrm{G}}_1}{\mathrm{S}{\mathrm{G}}_2}}\right)$

then,

  ${\mathrm{\phi }}_2={\sin }^{-1}\left(\sin 38°\sqrt{\frac{3.6}{2.8}}\right)$

  ${\mathrm{\phi }}_2=44°$

Conclusion:

Angle of refraction of wave is ${\mathrm{\phi }}_2=44°$ .