Why does high reflectivity correspond to an absorption peak?

[clanx]

i was reading papers and also doing my own simulations for a reflectivity experiment. i noticed though that reflectivity peak say at 3eV will correspond to a absorption peak (in the complex dielectric) at about 3eV too.

I'm just confused here because we all know that R + T +A = 1. But yet when R is high, it corresponds to a absorption peak too? what is wrong with my concept?

 

[Claude Bile]

What you observe is probably a manifestation of the Kramers-Kronig relations, that dictate that an absorption peak (i.e. a peak in the imaginary part of the refractive index) corresponds to an inflection point in the real refractive index (reflectivity).

Claude.

 

[clanx]

yes I understand that. but what about the physical picture? say if we happen to have this material with 100% reflection, shouldn't the absorption be 0?

 

[DrDu]

Some more information would be helpful, e.g., kind of material ...

 

[clanx]

the 100% reflection is just a hypothetical case. for the general case I'm referring to can be found on many scholarly articles. here is one I found on Google http://www.scielo.br/img/fbpe/bjp/v29n4/a06f1.gif
you can see clearly that a reflection peak corresponds to a complex dielectric peak at the same energy.

 

[DrDu]

What makes you think that a large imaginary part of the dielectric constant corresponds necessarily to large absorption?

 

[DrDu]

This effect is quite easy to observe btw. Make a line with a foil marker on a black sheet of paper or look at some crystals of a dye. The colour you see is the complementary colour of the one seen in transmission.
Reflection is always high when there is a large difference between the refractive indices (whether real or imaginary isn't important). If the imaginary part is high, the light won't enter deeply into the material. However, absorption is nevertheless low, as most light is reflected. In former times, this was exploited to generate monochromatic light in the far IR. It is called Reststrahlen method.

 

[clanx]

hmm. because the imag dielectric can also be written as 2nk where k is the extinction coefficient, and the absorption coefficient is directly proportional to the extinction coefficient? is there something wrong with concept?

 

[UltrafastPED]

You will find a nice introduction here:
http://www.tf.uni-kiel.de/matwis/amat/elmat_en/kap_3/backbone/r3_7_2.html

This goes on for several pages; follow the links at the bottom.

 

[DrDu]

hmm. because the imag dielectric can also be written as 2nk where k is the extinction coefficient, and the absorption coefficient is directly proportional to the extinction coefficient? is there something wrong with concept?

The problem is that this Beer's law only holds if reflection is negligible.
While k describes still extinction, i.e. the weakening of a ray propagating within the medium, if 2nk is high, most light won't even enter the medium because it gets reflected.
This has especially counterintuitive consequences in the case of metals. There, at low frequencies, n is small but k is high. Hence absorption is low although waves are strongly damped. Most of the light is reflected. If absorption rises, reflectivity becomes smaller and thin sheets of metal (like e.g. gold or copper) become partially transparent in the frequency region where they absorb.