Currently I am reading about super lens and came across these two waves, propagating and evanescent. If a negative index material is used as a lens then both propagating and evanescent can be passed comparing to a conventional lens in which only propagating can be passed. What is the difference between these two waves on image formation?

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    $\begingroup$ I'm skeptical of claims "both propagating and evanescent can be passed". Of course it is in theory true, but then again this is always true in theory, even for conventional materials. The practical limit is that the further axially you move away axially from the source, the evanescent wave grows exponentially weaker and therefore the signal to noise ratio becomes crap. This will equally limit negative and positive refractive index propagation. There are some superlenses that use the evanescent wave to allow smaller than wavelength features to be projected short distances through ... $\endgroup$ Commented Oct 9, 2013 at 13:04
  • $\begingroup$ ...curved mediums such that they are no longer smaller than wavelength scale variations (analogously to a quantum tunneling effect), whence they can be successfully resolved by conventional optics. These I believe will work if we can find the right materials and building methods for them. $\endgroup$ Commented Oct 9, 2013 at 13:06

2 Answers 2


The difference between propagating and evanescent is most obvious when looking at the intensity functions of the wave. You are probably familiar with a propagating wave, as its intensity is oscillating like a sine or cosine wave; this oscillation is what causes them to propagate or travel. An evanescent wave has an exponential wave function, typically an exponentially decaying curve if referring to intensity of light. This causes the wave to not really be able to travel through space and through a conventional lens.

I'm not really familiar with the super lenses you speak of, but I believe that the lenses are able to make evanescent waves visible due to the complex component of a evanescent wave function.

This comes into effect with reflection and refraction of light, particularly total internal reflection. If you were to have a block of glass and shine light into it which undergoes total internal reflection, no light would be visible where the light would have come out of the material if it would have went straight through; however, if another block of glass were put at this location with an extremely small distance between the two blocks (on the order of picometers, I believe), the light would come out of the second block of glass where a complex wave was created from the original electromagnetic wave in the space between the blocks, then was able to be changed back to a real wave in the second block of glass.


Well, there might be a bit of confusion of terms going on here; between "waves" and "fields".

When you have ordinary TIR going on at the boundary between a higher refractive index medium, and a lower index medium; with the beam incident from the high index side of the boundary, classical EM theory says there is no energy propagation in the low index medium; but there is an EM "field" in the low index medium, that drops exponentially with distance from the interface, in a space of the order of the wavelength. There isn't supposed to be any propagating wave, as a result of this evanescent field.

But, if a third medium is brought up to within the range of this evanescent field, and the refractive index is such that the TIRed beam would not be TIRed if media 1 and 3 were in optical contact, rather than media 1 and 2, then there will be an attenuated beam transmitted into the third medium. This process is known as "frustrated total internal reflection." The amplitude of the transmitted beam, in medium 3 is a sensitive function of the 1 - 3 spacing.

Beam splitter cubes can be obtained, with prescribed split ratios, that are set by controlled FTIR. Naturally, the split is wavelength dependent, so such splitter cubes, are typically used only in laser applications, here the wavelength is fixed. They aren't broad spectrum devices.


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