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I was reading that for an incident angle greater than the critical angle, there will be a total internal reflection. The cosine of the refraction angle is therefore an imaginary number. If we make a straight forward derivation we get an exponential attenuation factor for the refracted wave. And this wave (named Evanescent wave) only propagates along the boundary. When we replace the cosine of the refraction angle on the Fresnel's formula, the amplitude of the reflected wave is the same of the incident wave, i.e the real part of Fresnel's reflection coefficient is equal to the unity. More calculations gives that there is no flow energy across the boundary and I guess this is why the total amplitude is reflected backward, i.e without losing energy. My questions are here: What is the physical meaning of the evanescent wave? If there are no flow energy why is there a propagation factor (along the boundary) for this wave? If this wave is not propagating how we can understand the existence of this wave? If there is no flow energy across the boundary but there is energy, why is the amplitude of reflected wave the same of the incident wave? is there no loss of energy?

thanks a lot.

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In electromagnetism, the evanescent field is sometimes referred to as the reactive field. That is because it represents stored energy. Physically one can think of the process as follows. The light comes in, encounters the boundary and sets up a reactive field, which temporarily stores the energy, and then finally the energy propagates away in the reflected field. The reason why there needs to be such a reactive field is because the boundary conditions require it.

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The evanescent wave is a form of light tunneling. It does propagate perpendicular to it. If a sufficiently thin sheet of air is bordered by glass of both sides, then the extinction at incidence beyond the limiting angle is not complete, and some light will come through on the other side.

It should be noted that an evanescent wave does propagate, along the interface.

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