Friends, I have a basic doubt.
It is well known fact, when ZnO is doped with Al or Ga, the transparency in visible region increases. Similarly for the case of Indium doped Tin Oxide (ITO).
1) What is the possible mechanism/physics lying beneath for the increase in optical visible region transparency when respective dopants are added in ZnO/SnO?
Awaiting for your reply
Thanks in advance.
-Amir.
Amir, you have to say in what spectral range, near the bandgap? or in the visible region?
Amir, you have to say in what spectral range, near the bandgap? or in the visible region?
Amir, you have to say in what spectral range, near the bandgap? or in the visible region?
Amir, you have to say in what spectral range, near the bandgap? or in the visible region?
Amir, you have to say in what spectral range, near the bandgap? or in the visible region?
Amir, you have to say in what spectral range, near the bandgap? or in the visible region?
Amir, you have to say in what spectral range, near the bandgap? or in the visible region?
Amir, you have to say in what spectral range, near the bandgap? or in the visible region?
Dear Dr. Yong Zhang,
Many thanks for the reply.
The question is about the increase in optical visible region transparency. (I have just now edited the question for better understanding)
For eg., SnO shows less transparency whereas Indium doped SnO(ITO) shows higher visible region transparency. What factor enhances the optical transparency. I will be very much grateful if you can please provide the reason.
Thanks in advance
sorry for submitting my response multiple times :)
In the visible region (much below the bandgap), there is no absorption. However, the transmission is also affected by reflection. In general, the real part of the dielectric function is inversely proportional to the bandgap, so the higher bandgap material tends to have lower refractive index below the bandgap, thus less reflection. Adding Al probably increases the bandgap.
Dear Dr. Yong Zhang,
thanks for submitting the clarification. I am satisfied with your answer.
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