# Reflection transfer function of a Fabry Perot cavity

I am trying to calculate the transfer function of a Fabry Perot optical cavity. All the resources I have looked up, do the following. They start with an incident light field $$e^{i\omega t}$$ and write down the reflected light field. The reflected light field will have contributions from : 1) The field that gets reflected from the first mirror. 2) The field that gets transmitted into the cavity, undergoes one roundtrip inside the cavity and then gets transmitted 3) Field that undergoes two roundtrips before getting out and so on.

Now when they write down the phase shifts(wrt to the incident field) undergone by these different components, they consider a phase shift of $$\pi$$ for the first case because it gets reflected from the mirror. However, for the later cases they only consider the phase shift picked up by the light due to the extra distance it covers(corresponding to different number of round trips). ie a phase shift of $$\frac{2\pi}{\lambda}*{\Delta x}$$, where $$\Delta x$$ is $$2L$$ for one round trip and $$4L$$ for two round trips etc, L being the length of the cavity. They do not take into account the phase picked up by reflection from the mirros. For a single round trip, the phase due to reflection from a mirror would be $$\pi$$. For multiple round trips, it would be $$3\pi$$, $$5\pi$$, etc, but they would all amount to $$\pi$$.

To calculate the transfer function you do $$F = \frac{I_{reflected}}{I_{incident}}$$. If I take the extra phase of $$\pi$$ for all the contributions, I get a different answer.

I am confused why would they not consider the phase shift of $$\pi$$ for all the terms in the reflected field.

You should now see that there is always an odd number of reflections, but if your etxra phase from each reflection is $$\pi$$, then the phase you pick up from 1 reflection is the same as from 3 ($$3\pi=full\,period+\pi$$), or from any odd number, so there is no need to account for the extras.