How the reflection and transmissions mechanism are different in metals, dielectric and plasma? I know that the density of free electrons is playing the role. Can anyone give an insight what is actually happening physically? Why the EM is partially transmitted and partly reflected through the dielectric without being absorbed but gets attenuated in the metal? What is the main difference between the role of valance electrons in dielectrics and free electrons in metals? Note:I am aware of the mathematical results, I want to know how all these happening inside.
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2 Answers
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In dielectrics, the EM field interacts with the dipoles by elastically deforming them (e.g. the sphericity on an electronic orbital around the nucleus), just like a spring. Oscillating a dipole creates a counter-field which interferes with the initial EM wave. The resulting is the light transport in the material as well as the reflection / refraction near the interface.
In (perfect) conductors, the EM field interact with the free electrons, which are way more mobile.
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1$\begingroup$ And also, in metals it is a collective response of the electron gas, dependent on the density. Unlike the Lorentz dipole oscillator in a dielectric, the response of a single free electron is not a good starting point for metals. $\endgroup$– user137289Feb 15, 2018 at 8:27
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Clarification on metals: inside perfect metals there cannot be an electric field, for the electrons have nothing that stops them from going where they are “pushed” to. This is why light is completely reflected at a perfect metal, and why the electric field of light has a phase jump at the boundary of a perfect metal, and as a result (following Maxwell’s equations) the magnetic field hasn’t.
