In the last section where we considered N-wave interference by division of wavefront from N slits, Fabry-Perot etalons and interferometers cause interference by division of amplitude. These configurations are used prolifically in optics, and in fact many other fields have phenomena that can be understood in much the same way as the Fabry-Perot (for example, tunneling barrier conditions in a finite square well in Quantum Mechanics, or how a capacitor charges). The difference between the etalon and interferometer is that the etalon is basically just a dielectric slab with parallel faces, or two parallel stationary mirrors with high reflectivity. The interferometer, on the other hand, consists of two highly reflecting mirrors placed parallel to one another with a variable distance “d” between them. Fabry-Perot interferometers in particular have found a particularly important application in solid state laser cavities, so you may see a revisit to this if you head over to the “Lasers” portion of this website!

**Table of Contents:**

- Frequency Domain: Partial Wave Approach
- Frequency Domain: Scattering Matrix Approach
- Analysis of Transmitted Intensity
- Time Domain: Linear Response Theory

**FREQUENCY DOMAIN: Partial Wave Approach**

The basic layout of a Fabry-Perot interferometer (we will be considering varying distances, so it is an interferometer for our purposes) is shown in the figure below.

We will denote the distance between the mirrors as “d”, and we will assume lossless mirrors such that the transmission coefficient is given as:

\begin{equation} t_{1,2} = \sqrt{1-r_{1,2}^2} \end{equation}

I said before that the mirrors would have very high reflectivities, which would then mean that almost no light would couple into the interferometer and most of it would be reflected. This is still true; I promise I didn’t lie to you! The crazy thing about Fabry-Perot interferometers is that most of the light is reflected *until* a resonance condition is met. At resonance, we have perfect transmission (assuming that the mirrors are perfect mirrors, which is almost never the case because perfect, as we have all learned by now, is not very realistic; therefore, the transmission will be slightly lower, but still very close to, the total input field)! This happens because the light is allowed to form a standing wave in between the mirrors, so the entire field is transmitted out. At resonance, then, since all of the light is transmitted then there should be no light that is reflected, so the fields propagating back towards the source would all be destructively interfering with the first reflected wave.

Now that we have a general idea about how Fabry-Perot interferometers work, we can begin our partial wave analysis.

(1) Incident wave $\rightarrow~E_{inc}$

(2) Transmit through first mirror $\rightarrow~it_1E_{inc}$

(3) Propagate between mirrors $\rightarrow~it_1e^{\delta/2}E_inc$

(4) transmit through the second mirror $\rightarrow~(it_1)(it_2)e^{-i\delta/2}E_{inc}$, which reduces to $-t_1t_2e^{-i\delta/2}E_{inc}$

where we have

\begin{equation} \frac{\delta}{2} = \frac{n\omega{d}}{c}cos{\theta_t} \end{equation}

**FREQUENCY DOMAIN: SCATTERING MATRIX APPROACH**

Now the second method of obtaining the transmitted field is via the scattering matrices of the mirrors. The scattering matrix is given by:

\begin{equation} S_{1,2} = \begin{bmatrix} r_{1,2} & it_{1,2} \\ it_{1,2} & r_{1,2} \end{bmatrix}\end{equation}

which then has the corresponding transfer matrix:

\begin{equation} M_{1,2} = \frac{1}{it_{1,2}}\begin{bmatrix} -1 & r_{1,2} \\ -r_{1,2} & 1 \end{bmatrix}\end{equation}

So we then have that the total Fabry-Perot can be described by the matrix product:

\begin{equation} \frac{1}{it_{2}}\begin{bmatrix} -1 & r_{2} \\ -r_{2} & 1 \end{bmatrix}\begin{bmatrix} e^{-i\delta/2} & 0 \\ 0 & e^{-i\delta/2} \end{bmatrix} \frac{1}{it_{1}}\begin{bmatrix} -1 & r_{1} \\ -r_{1} & 1 \end{bmatrix} \end{equation}

We recall that $1/M_{2,2}$ in the transfer matrix multiplication gives the transmitted field. As such, the matrix multiplication above yields for the $M_{2,2}$ position:

\begin{equation} M_{2,2} = \frac{-1}{t_1t_2}[-r_1r_2e^{-i\delta/2}+e^{i\delta/2}] \end{equation}

Therefore, upon calculation of $1/M_{2,2}$ we have the same result as the partial wave approach for the transmitted intensity, namely:

\begin{equation} I_{trans} = |\frac{-t_1t_2e^{-i\delta/2}}{1-r_1r_2e^{-i\delta}}|^2I_{inc} \end{equation}

**ANALYSIS OF TRANSMITTED INTENSITY**

To analyze our equation for the transmitted intensity,

\begin{equation} I_{trans} = |\frac{-t_1t_2e^{-i\delta/2}}{1-r_1r_2e^{-i\delta}}|^2I_{inc} \end{equation}

we will consider the case where we have a symmetric Fabry-Perot, namely where $r_1=r_2=r$ and $t_1 = t_2 = t = \sqrt{1-r^2}$. Then we can simplify the transmitted intensity equation as follows:

\begin{equation} I_{trans} = \frac{t^4}{|1-r^2e^{-i\delta}|}^2I_{inc} \end{equation}

Then, evaluating the denominator:

\begin{align} |1-r^2e^{-i\delta}|^2 & = (1-Re^{-i\delta})(1-Re^{i\delta}) \\ & = 1 – R(e^{i\delta}+e^{-i\delta})+R^2 \\ & = 1 – 2Rcos{\delta} + R^2 \\ & = 1 – 2Rcos\delta + R^2 \\ & = 1 – 2R(1-2sin^2\frac{\delta}{2})+R^2 \\ & = (1-R)^2 + 4Rsin^2\frac{\delta}{2} \end{equation}

The transmitted intensity is thus:

\begin{equation} I_{trans} = \frac{T^2}{(1-R)^2[1+\frac{4R}{(1-R)^2}sin^2{\delta}{2}}I_{inc} \end{equation}

We can then define the “contrast” of the Fabry-Perot as:

\begin{equation} F = \frac{4R}{(1-R)^2} \end{equation}

We can then again rewrite the transmitted intensity, this time as a ratio to the incident intensity, as follows:

\begin{equation} \frac{I_{trans}}{I_{inc}} = \frac{1}{1+Fsin^2\frac{\delta}{2}} \end{equation}

Note that when

\begin{equation} \delta = \frac{2nd\omega}{c}cos(\theta_t) = 0, 2\pi, 4\pi … \end{equation}

then we have that:

\begin{equation} sin^2\frac{\delta}{2} = 0 \end{equation}

which means that

\begin{equation} \frac{I_{trans}}{I_{inc}} = 1 \end{equation}

which means, as we said, that all of the light is transmitted and constructively interferences at resonance. Conversely, when $\delta = \pi, 3\pi, 5\pi, …$, then we have:

\begin{equation} sin^2\frac{\delta}{2} = 1 \end{equation}

and

\begin{equation} \frac{I_{trans}}{I_{inc}} = \frac{1}{1+F} \end{equation}

Here we can see that if the reflectivity of the mirrors is large, then this corresponds to a high contrast F, and thus the transmitted intensity would be much lower than the incident intensity. This corresponds to a high finesse, which we will define shortly.

The Fabry-Perot transmission function is periodic, with a maximum each time the wave satisfies the condition:

\begin{equation} \frac{2\omega}{c}(ndcos\theta_t) = 2\pi{m} \end{equation}

Then, the difference in frequency between two modes i.e. for $\Delta{m}=1$ corresponds to the free spectral range of the Fabry-Perot, which is given as:

\begin{equation} \frac{2\Delta\omega_{FSR}}{c}(ndcos(\theta_t)) = 2\pi \end{equation}

We can note that the “effective optical thickness” is given by $ndcos\theta_t$, which we can denote “L” and thus rewrite the free spectral range in terms of frequency as:

\begin{equation} \Delta\nu_{FSR} = \frac{c}{2L} \end{equation}

The “free spectral range” is often referred to as the *axial mode spacing*, particularly when talking about cavity resonator modes.

**TIME DOMAIN DESCRIPTION: LINEAR RESPONSE THEORY**