What are optic fibres/cables in simple words?

I guess you could think of it that way. The more technical description is that fiber optic cables are a type of waveguide (https://en.wikipedia.org/wiki/Waveguide) for visible light.

Yes, but they are low for good cables. I think I've seen cables where the loss is around 1dB/km.

Of course. Telecommunication companies use these for internet traffic.

 

I'm not sure what you mean by saving a percent loss. Do you mean how much light intensity will be lost over a 2500km cable?

 

For the ones used in actual applications, not quite so. It's true that a hollow cable having reflective inner surface can also transmit light from one end to the other but such scheme proves to suffer from high amount of loss per unit distance (I either read it somewhere or heard it from someone else, too bad it was long time ago I can't remember the source). In reality, generally speaking optical fibers take the form of a coaxial dielectric and flexible tube tube. It is coax because it consists of at least two 'layers'. The innermost tube is called core and the outer one is called cladding. One of the condition for it to be able to guide light is to have the core having higher refractive index than the cladding. In this way, rays hitting the core-cladding interface will undergo total internal reflection provided the incoming angle is bigger than the critical angle between the two media. As the others have mentioned, there is a wealth of information on optical fiber and a forum thread is by no means sufficient to even cover the basic, you will need an introductory textbook on this topic at the very least. As subfields of optics, optical fibers and lasers pose the same level of breadth and development.

 

Over 2500km, using the example given, 2.5dB of the signal is lost. The fraction ##f## of power transmitted is then
$$f=10^{-2.5/10}=0.56$$
So the fraction of power lost is ##44\%##.

 

I don't know, you will probably have to do some googling to find the answer. Since this question is not directly related to the original question though, it is probably more appropriate in a separate thread.

 

I'd start by looking up candela def, then lumens. Assuming a perfect burn you can measure the light output by heat temp during combustion. Prob take two pages of equations to get the answer.

 

Yeah it is interesting how commercial applications of technology so easily overtake industrial applications, and then, hopefully, lead to rapid advances in said tech.