Any kind of electrical power transfer has a typical ratio of voltage to current. For example, you can deliver 100 watts by 1 amp at 100 volts, or 10 amps at 10 volts, or any other product that comes to 100. It's convenient to express the ratio of volts to amps as a number of ohms (since dimensionally that's all ohms are anyway). Power sources, loads, and even transmission lines all have characteristic impedance, which tells you something about what will happen when things are connected together.
Impedance matching is the selection of components with identical impedance, or the addition of impedance transforming components to cause a component with an undesired impedance to appear as though it has a more desirable one.
As Brian Carlton pointed out, when you have matched impedance, you achieve maximal power transfer. This is often desirable, but not always. The thing to remember is that maximal power transfer is achieved at the cost of dissipating equal power in the source and the load.
So for example, a case for not matching impedance is when you want to efficiently use the energy from a source. A 0.1 ohm load would get optimal power out of a battery with a 0.1 ohm internal resistance, but half the energy would be dissipated in the battery itself, which would be rather wasteful of the stored energy. (Not to mention that the terminal voltage would fall to half!) By purposely using a load much with much higher resistance than the battery, most of the stored energy ends up doing work in the load.
On the other hand, you DO want to match impedance when, for example, you have an audio amplifier stage that ideally wants to drive a 600 ohm load, but you only have a 3.2 ohm speaker. An ordinary transformer, having a 1:N voltage ratio, will conveniently give you a 1:N^2 impedance ratio. Another common case is in RF work, where as volting pointed out, you need to minimize reflections, because reflected energy can cause excessive power dissipation in your source.
Presumably your waveguide will have to be conductive in order to transport the signal. So using high resistivity Si isn't going to help you. What you need is lateral resistivity that is low (highly conductive) whilst vertical conduction is low (high resistivity).
In semiconductor technology you have two isolation mechanisms, 1) dielectric (i.e. insulator) isolation and the 2) Junction isolation. I think dielectric isolation is obvious in context, you have a thin layer of SiO2 to keep the Metal layer and the substrate separate. Junction isolation means that you would form a junction through implanting dopants and converting the substrate into a opposite polarity. What you would be building is an extended diode with most likely a N-type conductive region on a P-type substrate. Diodes have capacitance which is also voltage variable. Do in doing this you would get higher capacitance which because of it's non-linearity could cause distortion (depending upon how you operate it).
Modelling this means as a good first order approximation of using the dielectric of the Si substrate as you suggest, depending upon your parameters you may have to bring in the di-electric (SiO2) but the permittivity of the SI is so high that the E-Field laines will be sucked into that.
Best Answer
The reflection at the end of a transmission line is given by
\$ \Gamma = \dfrac{Z_L - Z_0}{Z_L + Z_0}\$
Where Z0 is the line's characteristic impedance and ZL is the load impedance.
So if you have a 55 Ohm line and you terminate it with 50 Ohms, you're looking at about 5% reflection.
If you have a 50 Ohm line, followed by a short length of 55 Ohm line, terminated with 50 Ohms, you're going to get about 5% reflection from the mismatch between the two lines, and 5% reflection from the end termination.
But...these two reflections will interact with each other. Depending on your operating frequency and the length of the mismatched line, they could interfere with each other constructively or destructively. You could have anywhere from 0 to +/-10% reflection from this structure.
But this is about the voltage signal (or travelling wave) reflection. If you're talking about "losses" you probably want to convert this to power terms. If you know the signal reflection, that would correspond to \$(1-|\Gamma|)^2\$ fraction of the power transmitted to the load. For 10% reflection, that's about 81% power delivery or 0.9 dB of loss.
To find out the exact reflection for your situation, you'd need to either run a simulation, or do a pen-and-paper analysis with a Smith Chart, which we've discussed a little bit in a previous question. The simulator is probably the easier way to go, though, and it will enable you to do things like sensitivity analysis more quickly.