1
$\begingroup$

I'm very new to the topic of surface plasmons and I have been reading about different methods of exciting them. There is one method in which a prism is set up to allow phase matching of an incident light ray to the required wave vector for exciting a surface plasmon. The light goes through total internal reflection in the prism and then either tunnels through the metal (on which we want to excite the surface plasmon), or a lower indexed dielectric to excite the surface plasmon.

This might be a lack of understanding of attenuated total internal reflection, but apparently this field harbors no energy and 100% of the incident light is reflected. My question is how is energy conserved (how can it excite surface plasmons)? Is it because the energy of the attenuated field only time averages to 0? But then how can all the light be reflected?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

I think you may be misreading that the field holds no energy. All of the light can be reflected at steady state. When the light is first incident on the system, energy is stored in the plasmon part of the light-plasmon system. Once the steady state has been reached, the energy going in equals the energy coming out (less a trickle needed to make up for any nonideal loss in the metal). This is not only true of surface plasmons: its true of any electromagnetic system that can store energy before steady state is reached.

This is very like the establishment of the electromagnetic fields in a parallel LC resonant tank circuit, and indeed this is a good classical picture to keep in your head of the plasmon. The "tank circuit" has an energy input and an energy output. At first, a great deal of energy is "sunken" into the tank circuit whilst its fields are established. Once they are, though, the only energy needed per cycle is to make up for nonideal losses, so that the energy input equals the energy output.

Another idea is to think of a cylindrical water tank with an inlet near the top, and, diametically opposite, an outlet. When the water begins to fill the tank, no water comes out the outlet. However, at last the tank will be filled, and water will begin pouring out of the outlet at the same rate as it comes into the inlet (less a small volumetric rate is there is some other slow leak in the tank).

Improve this answer
$\endgroup$
2
  • $\begingroup$ So are you saying that all the light that is able to cross the boundary will eventually be a part of the reflected beam? But then what happens if I put another material with high dielectric constant very near to where the light becomes attenuated? That light won't be able to escape through the reflected incident beam. $\endgroup$
    – Atreyu
    Nov 25, 2013 at 5:36
  • $\begingroup$ Yes, but if you solve Maxwell's equation for a situation like this, the metal losslessly reflects any incoming wave. Have you looked at the article en.wikipedia.org/wiki/Surface_plasmon ? I mean, even the diagram on the top right gives you a good intuitive feel. There is quite a bit of energy stored in the metal at steady state. If you instead put a dielectric material near the evanescent field and your in TIR regime, then you've got quite a different situation. There is, for example, no skin depth confining the field. Have a look through the calcs in the Wiki page $\endgroup$
    – Selene Routley
    Nov 25, 2013 at 5:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.