I think the formula you show is not the travel time for a refracted ray. It is T0, the intercept of the time-distance curve with the x-axis for the refracted ray.
See this figure, and associated formula, and the difference should be apparent. Perhaps your text is in error or unclear.
Keep in mind that multiple ray paths can reach the receiver location - including a direct ray, reflected ray, and refracted ray. Which ray is the first arrival depends upon the underlying structure and distance from the source. The refracted ray is the first arrival at distances greater than XCROSS1.
This is one of the places where wave particle dualism gets some people in trouble. Many are taught that it means that light can be a wave and a particle, and that phrasing can lead to some confusion. I find it more intuitive to just rip the bandaid off quickly and say light is neither a wave nor a particle. It is something which, in some situations, can be well modeled as a wave, and in some situations can be well modeled as a particle, but it is its own thing (which can be well modeled in all known cases using a more complicated concept, a "wavefunction").
You can think of photons getting randomly reflected or transmitted on the boundary, but the truth is that the billiard-ball photon model really isn't very effective at describing what happens at this boundary. This is one of the regions where wave mechanics models the effects very well, while particle models don't do so well. If you use wave mechanics, the idea of a wave getting partially reflected and partially transmitted isn't difficult to believe at all. In fact, it's pretty easy to prove.
Thinking in wave terms at these boundaries also gives correct answers in peculiar situations where the particle model simply falls on its face. Consider the interesting case of an "evanescent wave."
In this setup, the laser and prism are set up at the correct angles to cause "total internal reflection." This means that, by the simple models, 100% of the light should bounce off the side of the prism and into the detector. Indeed, if the prism is in the open air, we do see 100% reflection (well, within the error bars of absorption). However, bring an object close to the prisim (but not touching) and things change. You end up seeing effects from the object, even though 100% of the light was supposed to be reflected!
If you think of light like photons, this is hard to explain. If you look at it as a wave governed by Maxwell's equations, you see that you would violate the law of conservation of energy if there was a "pure" reflection. Instead it creates a reflection and an "evanescent wave" which is outside the prism, and its strength falls off exponentially, which is really hard to explain with particles!
Of course, these too are all simplifications. The real answer to your question is that the wavefunction of the light interacts with the electromagnetic fields of the atoms in the prisim, and the result of that interaction leads to reflection, refraction, diffusion, absorption, and eveansecent waves. However, naturally those equations are a bit harder to understand, so we use the older, simpler models from before quantum mechanics. We just have to be sure to use the one which is most applicable in any given situation, because none of them are quite right.
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Refraction and reflection seismics serve different purposes. The refraction seismic method is normally used for localized near-surface investigations (site surveys). The reflection seismic method is typically used to investigate targets that are up to several kilometers deep. That's of course in the context of exploration seismics. You can't easily use reflection seismics for (very) shallow investigations because most reflections will be over-critical (i.e. beyond the critical incidence angle), in particular when you increase the source-receiver distance (offset).