The behavior of decreasing refractive index n for wavelengths greater than 700nm is an important result, because it corresponds transmission increasing of ITO /glass at near-infrared wavelength
Dear Nadir,
the reflectivity (here vertical incidence) is given by R = ((n2 - n1)/(n2+n1))2.
Here n2 and n1 describe the refractive indices at the boundary of the 2 media 1 and 2 for the corresponding wavelength. If the difference of the refractive indices decreases, decreases R and T increases. A low reflectivity means a high transmittance (the absorption is described by the imaginary part of the refractive index and must be discussed independently).
If the question was, why n decreases with decreasing wavelength, the resonances which can occur must be discussed for the corresponding material.
With Regard
R. Mitdank
Let me correct the upper formulae: I omitted a minus:
R = ((n2 - n1)/(n2 + n1))2
Dear professor,
I'm not an expert in ITO, however an increasing transmitance generally means the decreasing refractive index. I would expect that in ITO (as a wide-gap semiconductor) the normal dispersion will take part. Consequently, at near infrared wavelengths the free-carrier absorption will occur.
Best regards,
TH
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