Exploring the Optical Phenomenon of Total Internal Reflection in Salt Solutions

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In summary, the conversation discusses a salt solution with increasing refractive index causing a laser to bend in a clockwise direction. However, there is also a question about the possibility of the laser bending down again and the lack of reflections. The explanation is that the index of refraction changes continuously and Snell's law must be used to describe the propagation. This results in a differential equation that explains the trajectory of the laser, including the backwards bend at the top due to total internal reflection. Despite transmission dominating over reflection near the top, the shallow angle of incidence still causes a change in the laser's trajectory.
  • #1
Anton Alice
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Hello,

please take a look at the following picture:
laser.png


So we have a salt solution, with increasing refractive index, as you go down the solution.
The Laser is steadily refracted at the layers from high-to-low index and therefore bends in clockwise direction.
But how is it possible for the laser to bend down again? And also, shouldn't we also see reflections? Because at any point, there is a layer of high-to-low index transition. And at any such point there should also be a reflection. And as the angle of incidence get higher, as the laser path gets more horizontal, the reflection should dominate.

EDIT:

Does it bend down, because of total internal reflection at the peak of the curve?
 
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  • #2
Since the index of refraction changes continuously, you have to use Snell's law to describe the propagation one infinitesimal distance at a time. If you work it out, you can get a differential equation that will describe the trajectory of a beam of light. That it bends back downwards at the top is due to total internal reflection.

That being said, as the difference in index of refraction on two sides of an interface goes to zero, transmission dominates over reflection (the reflection coefficient becomes insignificant compared to the transmission coefficient). Though this would seem to contradict the idea of total internal reflection causing the backwards bend at the top, the beam near the top has a very shallow angle of incidence on either side of the peak, so that very small differences in refractive index can still amount to a change in the beam's trajectory.
 
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yes, thank you.
 

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