Calculation of true thickness of the bed

^{1.A seismic wave emanating from a shot point, when incident at angle 30 degrees at a point P on a refraction, gets refracted at an angle 60 degrees. If the wave is incident 52m away from P, it would be critically refracted.(Thanks to Neeraja-AU)}

^{Find the}

^{a) critical angle}

^2)depth

^{3) cross over distance.}

a) critical angle

$$\frac{sini}{sinr}=\frac{V_{1}}{V_2}$$

$$\frac{sin30^o}{sin60^o}=\frac{V_{1}}{V_2}$$

$$\frac{V1}{V2}=\frac{1/2}{\sqrt3/2}=\frac{1}{\sqrt3}$$

$$\frac{sini}{sinr}=\frac{V_{1}}{V_2}$$

$$\frac{sini_c}{sin90^o}=\frac{1}{\sqrt3}$$

$$sini_c=\frac{1}{\sqrt{3}}$$

$$\cos i_c=\sqrt{(1-(\frac{1}{\surd3})^2)}$$

$$=\sqrt{1-\frac{1}{3}}$$

$$=\sqrt{\frac{3-1}{3}}$$

$$=\sqrt{\frac{2}{3}}$$

$$\tan i_c=\frac{sini_c}{cosi_c}=\frac{1/\sqrt{3}}{\sqrt{2}/\sqrt{3}}=\frac{1}{\sqrt{2}}$$

$$i_c=\tan^{-1}(\frac{1}{\sqrt{2}})$$

b) The depth of the refractor is about:

$$\triangle SPB$$

$$\tan30^o=\frac{x}{h}$$

$$\frac{1}{\sqrt{3}}=\frac{x}{h}$$

$$h=x\sqrt{3}$$

$$x=\frac{h}{\sqrt{3}}$$

$$\triangle SDC$$

$$\tan i_c=\frac{x+52}{h}$$

$$\frac{1}{\sqrt{2}}=\frac{x+52}{h}$$

$$\frac{h}{\sqrt{2}}={x+52}$$

$$\frac{h}{\sqrt{2}}=\frac{h}{\sqrt{3}}+52$$

$$h[\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}]=52$$

$$h[\frac{\sqrt{3}-\sqrt{2}}{\sqrt{6}}]=52$$

$$h=400m$$

c) cross over distance

$$z=\frac{x_{crs}}{2}\sqrt{\frac{V_2-V_1}{V_2+V_1}}$$

$$x_{crs}=2z\sqrt{\frac{V_2+V_1}{V_2-V_1}}$$

$$x_{crs}=2z\sqrt{\frac{V_2+V_1}{V_2-V_1}}$$

$$=2\times400\sqrt{\frac{1+\frac{V_1}{V_2}}{1-\frac{V_1}{V_2}}}$$

$$=800\sqrt{\frac{1+\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}}}$$

$$=800\sqrt{\frac{\sqrt{3}+1}{\sqrt{3}-1}}$$

$$=800\times1.9311$$

x_crs=1540m