Need an explanation: Optics

For microwave range we use hollow metallic waveguide
but for optical range we go for dielectric wave guides.
why cannt we use hollow metallic waveguide for light wave transmission

I referred many books, all they say is at optical range, in metallic waveguides signal incurs heavy attenuation due to the fact that in this range the refractive index of the metal becomes frequency dependent and complex.

But this is still a vague explanation to me.

Can any body kindly explain in detail?

Thanks

Interesting question.

I would suspect that moding (excitation of numerous high order modes) at light frequencies would be pronounced and that a metallic waveguide operating close to the fundamental mode would just be too small. I don't know if it is even practical to consider. Consider metallic guide at mm wavelengths The opening is almost too small to see. WR3 for example cutoff is 173 GHz and shows 3 to 5 dB attenuation per foot. It looks like it makes a good load. Blue light is around 750 teraHz and red is a little lower around 430 teraHz.

I don't think the spirit of the question includes things like the "Solatube", which is a product for skylights that is sized similar to a heater duct and "guides" light from a hole on the roof to a ceiling opening.

Hi Azulykit,

I do not think that the problem for "optical" propagation in metallic waveguides is the size of them. As discussed in previous post, waveguides are, in theory, highpass filters. So, even a "big" waveguide can support an optical frequency. I know that if monomode propagation is what we need then we must have as smallest waveguide size as possible (not very practical as you said). But what about multimode propagation? I think that the problem is high attenuation of the very high frequencies in the metallic waveguides.
As you see, I am in a disagreement with you. Not personal........

I do feel that the problem is with the high attenuation.
but 'm not able to reason for this high attenuation.