# Refraction seismology - travel time for wave

I am taking an introductory class in seismology, but have some difficulties understanding the logic behind the formula used to calculate the time it takes for a refracted wave to return to the surface from which it was sent out originally.

So, say we have a seismic wave sent out from a horizontal layer. We also assume that the interfaces below are horizontal, and parallel to the top layer. If the wave hits the first interface at the so-called critical angle, it will be reflected back to the surface with the same angle at which it hit the interface. Now, the formula given in the book states that in order to calculate travel time, we use:

$$t = \frac{2 h_1 cos(i_c)}{v_1}$$

where $h_1$ is the vertical height of the layer, $i_c$ is the critical angle, and $v_1$ is the velocity of the wave as it travels through the layer.

As mentioned, I don't quite see the logic of this formula. I know that time = distance/velocity. Surely the shortest possible distance to travel back and forth between two horizontal layers, separated with distance $h_1$ is simply $2 h_1$. So by that logic, the shortest possible time a wave could possibly use, would be:

$$t = \frac{2 h_1}{v_1}$$

But in the given formula we multiply the numerator with $cos(i_c)$, which would yield an even shorter value for distance. And I just can not see how this makes sense at all. Now, if we were to use the cosine-term in the denominator, then the formula would make perfect sense to me. But why isn't it so? Any help here would be greatly appreciated!

• I think, and I remark that I think (I need to review my optics notes) that you can derive that formula via Snell's law and imposing minimum condition, so that the sine in Snell gives you a cosine in the derivative – Jorge Lavín Jan 24 '13 at 0:37
• Thanks. I would really, really be grateful if you could provide me with the derivation. – user12277 Jan 24 '13 at 8:19