In order to easily understand the behavior of the different materials concerning their interaction with an incident electromagnetic wave with specific wavelength, the material is modeled as a transmission line having a characteristic impedance Z0. 

If we have two materials with different Z0, then  by definition the reflection coefficient roh= (Z02-Zo1)/ Z02+Z01), iF  the two impedances are equal the reflection coefficient will be zero the two materials are matched. The characteristic impedance depends on the dielectric constant, the permeability and the resistivity of the material on both sides of the interface. . In case of metal air interface because of the high conductivity of the metal its characteristic impedance will be very different from that of the air and total reflection occurs from a relatively thin metal thickness. One can said if the metal is an ideal conductor it will make the electric field component equals zero. This is as if you short circuit the transmission line. The wave will be bounced back.

In case of partial reflection the refracted part will be partly absorbed in the material and the rest will be transmitted out of the materiel. The absorption depends on the interaction between the photos of the wave and the electrons in the material. It depends specifically on the energy band structure of the material,

wish you success

http://hkpho.phys.ust.hk/Training/Experiments/O6_Microwave.pdf

http://www.emu.dk/sites/default/files/physics_of_microwave_oven.pdf

http://users.wpi.edu/~vadim/IMS-2011.pdf

http://www.microwaves101.com/encyclopedias/radar-cross-section-physics

http://www.nikhef.nl/~h73/kn1c/praktikum/phywe/LEP/Experim/4_5_09.pdf

In order to easily understand the behavior of the different materials concerning their interaction with an incident electromagnetic wave with specific wavelength, the material is modeled as a transmission line having a characteristic impedance Z0. 

If we have two materials with different Z0, then  by definition the reflection coefficient roh= (Z02-Zo1)/ Z02+Z01), iF  the two impedances are equal the reflection coefficient will be zero the two materials are matched. The characteristic impedance depends on the dielectric constant, the permeability and the resistivity of the material on both sides of the interface. . In case of metal air interface because of the high conductivity of the metal its characteristic impedance will be very different from that of the air and total reflection occurs from a relatively thin metal thickness. One can said if the metal is an ideal conductor it will make the electric field component equals zero. This is as if you short circuit the transmission line. The wave will be bounced back.

In case of partial reflection the refracted part will be partly absorbed in the material and the rest will be transmitted out of the materiel. The absorption depends on the interaction between the photos of the wave and the electrons in the material. It depends specifically on the energy band structure of the material,

wish you success

Dear Sir,

we have a transmission = 0 .

And the equation is:

transmission + reflection = 0

All the best

F. Gräbner

"And the equation is:

transmission + reflection + refraction + absorption + adsorption + scattering + diffraction = 0"???

Isn't it = 1 or 100%? 

Thank you sir Frank Gräbner, Abdelhalim Zekry, and Gyorgy Banhegyi for your replies!

 The energy balance of the incident electromagnetic rays is that 

The incident energy= The reflected energy +refracted energy,

The refracted energy = The absorbed energy + the transmitted energy,

So, finally the conservation rule can be put in the form:

The incident energy=The reflected energy+The absorbed energy + the transmitted energy

DIVIDING BOTH SIDES OF THE EQUATION BY THE INCIDENT ENERGY , ONE FINALLY GETS:

THE REFLECTANCE + ABSORBANCE +TRANSMITTANCE=1,

In case of metals, both absorbance and transmittance are equal to zero and we have total reflection with reflectance equals to one.

wish you success.

metal cannot support electric field inside E=P=D=0

Dear Suhas Dinesh,

let me suggest a point for consideration. At optical frequencies metal is generally much more reflective than glass. However, that does not mean glass is not reflective, as one can ascertain when trying to look through a window at night with the lights on. This is interpreted by saying that light is partly reflected and partly refracted by glass, whereas it is totally reflected by metal. Amplification of light intensity by metal is excluded. But, do metal and glass show the same performance at MW or in the UV, too?

This experimental question is of theoretical relevance because Kirchhoff's law of thermal radiation is assumed to be valid for electromagnetic radiation. This law states that, when dealing with black-body radiation at thermodynamic equilibrium, an absorption/re-emission process is akin to reflection. According to Wikipedia (https://en.wikipedia.org/wiki/Kirchhoff%27s_law_of_thermal_radiation): 'Kirchhoff's law has as a corollary: the emissivity cannot exceed one (because the absorptivity cannot, by conservation of energy.)'

The behavior of metals was thoroughly investigated by Rubens and Hagen (and by Quincke) around 1900. If you can read German, I suggest that you have a look at them, and in particular at “Die Absorption ultravioletter, sichtbarer und ultraroter Strahlen in dünnen Metallschichten”. It deals specifically with film depth effects.

If a metal completely reflects the incident EM energy by impedance mismatch, E = o inside a conductor (short circuit), how does an EM wave penetrate inside a conductor? I mean, what about the skin depth stuff?

I believe it is the induction process that generates surface current (by Electric fields) and eddy currents (by magnetic fields), isn't it?

Getting confused!