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diff --git a/source/know/concept/fabry-perot-cavity/index.md b/source/know/concept/fabry-perot-cavity/index.md new file mode 100644 index 0000000..e2852d4 --- /dev/null +++ b/source/know/concept/fabry-perot-cavity/index.md @@ -0,0 +1,233 @@ +--- +title: "Fabry-Pérot cavity" +date: 2021-09-18 +categories: +- Physics +- Optics +- Laser theory +layout: "concept" +--- + +In its simplest form, a **Fabry-Pérot cavity** +is a region of light-transmitting medium surrounded by two mirrors, +which may transmit some of the incoming light. +Such a setup can be used as e.g. an interferometer or a laser cavity. + +Below, we calculate its quasinormal modes in 1D. +We divide the $x$-axis into three domains: left $L$, center $C$, and right $R$. +The cavity $C$ has length $\ell$ and is centered on $x = 0$. +Let $n_L$, $n_C$ and $n_R$ be the respective domains' refractive indices: + +<a href="cavity.png"> +<img src="cavity.png" style="width:70%"> +</a> + + +## Microscopic cavity + +In its simplest "microscopic" form, the reflection at the boundaries +is simply caused by the index differences there. +Consider this ansatz for the [electric field](/know/concept/electric-field/) $E_m(x)$, +where $m$ is the mode: + +$$\begin{aligned} + E_m(x) + = \begin{cases} + A_1 e^{- i k_m n_L x} & \mathrm{for}\; x < -\ell/2 \\ + A_2 e^{- i k_m n_C x} + A_3 e^{i k_m n_C x} & \mathrm{for}\; \!-\!\ell/2 < x < \ell/2 \\ + A_4 e^{i k_m n_R x} & \mathrm{for}\; x > \ell/2 + \end{cases} +\end{aligned}$$ + +The goal is to find the modes' wavenumbers $k_m$. +First, we demand that $E_m$ and its derivative $\idv{E_m}{x}$ +are continuous at the boundaries $x = \pm \ell/2$: + +$$\begin{aligned} + A_1 e^{i k_m n_L \ell/2} + &= A_2 e^{i k_m n_C \ell/2} + A_3 e^{- i k_m n_C \ell/2} + \\ + A_4 e^{i k_m n_R \ell/2} + &= A_2 e^{- i k_m n_C \ell/2} + A_3 e^{i k_m n_C \ell/2} +\end{aligned}$$ +$$\begin{aligned} + - i k_m n_L A_1 e^{i k_m n_L \ell/2} + &= - i k_m n_C A_2 e^{i k_m n_C \ell/2} + i k_m n_C A_3 e^{- i k_m n_C \ell/2} + \\ + i k_m n_R A_4 e^{i k_m n_R \ell/2} + &= - i k_m n_C A_2 e^{- i k_m n_C \ell/2} + i k_m n_C A_3 e^{i k_m n_C \ell/2} +\end{aligned}$$ + +Rearranging the four equations above yields the following linear system: + +$$\begin{aligned} + 0 + &= A_1 - A_2 e^{i k_m (n_C - n_L) \ell/2} - A_3 e^{- i k_m (n_C + n_L) \ell/2} + \\ + 0 + &= A_2 e^{- i k_m (n_C + n_R) \ell/2} + A_3 e^{i k_m (n_C - n_R) \ell/2} - A_4 + \\ + 0 + &= n_L A_1 + n_C \big( A_3 e^{- i k_m (n_C + n_L) \ell/2} - A_2 e^{i k_m (n_C - n_L) \ell/2} \big) + \\ + 0 + &= n_C \big( A_3 e^{i k_m (n_C - n_R) \ell/2} - A_2 e^{- i k_m (n_C + n_R) \ell/2} \big) - n_R A_4 +\end{aligned}$$ + +Which can be rewritten in matrix form as follows, with the system matrix on the left: + +$$\begin{aligned} + \begin{bmatrix} + 1 & -e^{i k_m (n_C - n_L) \ell/2} & -e^{- i k_m (n_C + n_L) \ell/2} & 0 \\ + 0 & e^{- i k_m (n_C + n_R) \ell/2} & e^{i k_m (n_C - n_R) \ell/2} & -1 \\ + n_L & -n_C e^{i k_m (n_C - n_L) \ell/2} & n_C e^{- i k_m (n_C + n_L) \ell/2} & 0 \\ + 0 & -n_C e^{- i k_m (n_C + n_R) \ell/2} & n_C e^{i k_m (n_C - n_R) \ell/2} & -n_R + \end{bmatrix} + \cdot + \begin{bmatrix} + A_1 \\ A_2 \\ A_3 \\ A_4 + \end{bmatrix} + = + \begin{bmatrix} + 0 \\ 0 \\ 0 \\ 0 + \end{bmatrix} +\end{aligned}$$ + +We want non-trivial solutions, where we +cannot simply satisfy the system by setting $A_1$, $A_2$, $A_3$ and +$A_4$; this constraint will give us an equation for $k_m$. Therefore, we +demand that the system matrix is singular, i.e. its determinant is zero: + +$$\begin{aligned} + 0 = + &- n_C (n_L + n_R) \big( e^{i k_m (2 n_C - n_L - n_R) \ell/2} + e^{- i k_m (2 n_C + n_L + n_R) \ell/2} \big) + \\ + &+ (n_C^2 + n_L n_R) \big( e^{i k_m (2 n_C - n_L - n_R) \ell/2} - e^{- i k_m (2 n_C + n_L + n_R) \ell/2} \big) +\end{aligned}$$ + +We multiply by $e^{i k_m (n_L + n_R) \ell / 2}$ and +decompose the exponentials into sines and cosines: + +$$\begin{aligned} + 0 + = i 2 (n_C^2 + n_L n_R) \sin(k_m n_C \ell) + - 2 n_C (n_L + n_R) \cos(k_m n_C \ell) +\end{aligned}$$ + +Finally, some further rearranging gives a convenient transcendental equation: + +$$\begin{aligned} + \boxed{ + 0 + = \tan(k_m n_C \ell) + i \frac{n_C (n_L + n_R)}{n_C^2 + n_L n_R} + } +\end{aligned}$$ + +Thanks to linearity, we can choose one of the amplitudes +$A_1$, $A_2$, $A_3$ or $A_4$ freely, +and then the others are determined by $k_m$ and the field's continuity. + + +## Macroscopic cavity + +Next, consider a "macroscopic" Fabry-Pérot cavity +with complex mirror structures at boundaries, e.g. Bragg reflectors. +If the cavity is large enough, we can neglect the mirrors' thicknesses, +and just use their reflection coefficients $r_L$ and $r_R$. +We use the same ansatz: + +$$\begin{aligned} + E_m(x) + = + \begin{cases} + A_1 e^{-i k_m n_L x} & \mathrm{for}\; x < -\ell/2 \\ + A_2 e^{-i k_m n_C x} + A_3 e^{i k_m n_C x} & \mathrm{for}\; \!-\!\ell/2 < x < \ell/2 \\ + A_4 e^{i k_m n_R x} & \mathrm{for}\; \ell/2 < x + \end{cases} +\end{aligned}$$ + +On the left, $A_3$ is the reflection of $A_2$, +and on the right, $A_2$ is the reflection of $A_3$, +where the reflected amplitudes are determined +by the coefficients $r_L$ and $r_R$, respectively: + +$$\begin{aligned} + A_3 e^{- i k_m n_C \ell/2} + &= r_L A_2 e^{i k_m n_C \ell/2} + \\ + A_2 e^{-i k_m n_C \ell/2} + &= r_R A_3 e^{i k_m n_C \ell/2} +\end{aligned}$$ + +These equations might seem to contradict each other. +We recast them into matrix form: + +$$\begin{aligned} + \begin{bmatrix} + 1 & - r_R e^{i k_m n_C \ell} \\ + - r_L e^{i k_m n_C \ell} & 1 + \end{bmatrix} + \cdot + \begin{bmatrix} + A_2 \\ A_3 + \end{bmatrix} + = + \begin{bmatrix} + 0 \\ 0 + \end{bmatrix} +\end{aligned}$$ + +Again, we demand that the determinant is zero, in order to get non-trivial solutions: + +$$\begin{aligned} + 0 + &= 1 - r_L r_R e^{i 2 k_m n_C \ell} +\end{aligned}$$ + +Isolating this for $k_m$ yields the following modes, +where $m$ is an arbitrary integer: + +$$\begin{aligned} + \boxed{ + k_m + = - \frac{\ln(r_L r_R) + i 2 \pi m}{i 2 n_C \ell} + } +\end{aligned}$$ + +These $k_m$ satisfy the matrix equation above. +Thanks to linearity, we can choose one of $A_2$ or $A_3$, +and then the other is determined by the corresponding reflection equation. + +Finally, we look at the light transmitted through the mirrors, +according to $1 \!-\! r_L$ and $1 \!-\! r_R$: + +$$\begin{aligned} + A_1 e^{i k_m n_L \ell/2} + &= (1 - r_L) A_2 e^{i k_m n_C \ell/2} + \\ + A_4 e^{i k_m n_R \ell/2} + &= (1 - r_R) A_3 e^{i k_m n_C \ell/2} +\end{aligned}$$ + +We simply isolate for $A_1$ and $A_4$ respectively, +yielding the following amplitudes: + +$$\begin{aligned} + A_1 + &= (1 - r_L) A_2 e^{i k_m (n_C - n_L) \ell/2} + \\ + A_4 + &= (1 - r_R) A_3 e^{i k_m (n_C - n_R) \ell/2} +\end{aligned}$$ + +Note that we have not demanded continuity of the electric field. +This is because the mirrors are infinitely thin "magic" planes; +had we instead used the full mirror structure, +then we would have demanded continuity, as you maybe expected. + + + +## References +1. P.T. Kristensen, K. Herrmann, F. Intravaia, K. Busch, + [Modeling electromagnetic resonators using quasinormal modes](https://doi.org/10.1364/AOP.377940), + 2020, Optical Society of America. |