Ring Resonators

Question

If optical field will propagate into the following ring resonator. Optical field will propagate into the ring and propagate to the coupling region. I didn't get where and how the interference will take place and what we will get from the output at the wave guide ?

Answer 1

I am assuming for simplicity that either the waveguides are one-moded or, if not, the input to the system is in one mode alone and the system design is such that coupling between modes is negligible.

Usually with resonant ring systems like this one used as interferometers, we are using them to sense changes in the ring's optical phase delay.

Actually, if the ring is lossless (we can make it so practically if the ring radius is big enough) and if the coupled region is lossless (i.e. all of the optical power stays in the one bound mode of each waveguide), then we don't get interference for this system: as you can see, under the conditions I state, all the energy going into the input waveguide must come out (at steady state, after the resonant cavity has been "filled" with energy). Instead, you get a very sensitive dependence of the output phase on the ring's phase delay. So, you can put this system inside a fibre optical Mach-Zehnder interferometer to convert this sensitive phase shift to an amplitude shift that is highly sensitive to the ring's optical path length.

When the ring itself has a small amount of loss, you get a sensitive dependence of the output amplitude on the ring's optical path length.

Let's do some simple analysis. The coupled region between the two waveguides is analogous to a bulk optic partly silvered mirror. If the coupled region is mirror-symmetric about a transverse plane in the middle of the coupled region, you can adapt the analysis in my answer here to show that the most general transfer matrix relating the two input complex mode amplitudes $\vec{a} = \left(\begin{array}{c}a_1\\a_2\end{array}\right)$ to the two output complex mode amplitudes $\vec{b} = \left(\begin{array}{c}b_1\\b_2\end{array}\right)$ is $\vec{b} = U \vec{a}$ where:

$$U = e^{i\,\beta}\,\left(\begin{array}{cc}\cos\frac{\alpha}{2}&i\,\sin\frac{\alpha}{2}\\i\,\sin\frac{\alpha}{2}&\cos\frac{\alpha}{2}\end{array}\right);\quad \alpha,\beta\in\mathbb{R}\qquad(1)$$

and the angle $\alpha$ sets the coupler's split ratio. Now, if we link one coupler output through the ring with phase delay $\phi$ and with a small attenuation $a < 1$ such that $a = 1-\epsilon$ where $0<\epsilon\ll 1$ then

$$a_2 = a\,e^{i\,\phi}\,b_2\qquad(2)$$

On putting (2) into (1) we get:

$$\frac{b_1}{a_1} = \frac{e^{i \beta } \left(-\cos \left(\frac{\alpha }{2}\right)+a\,e^{i (\beta +\phi )}\right)}{-1+a \,\cos \left(\frac{\alpha }{2}\right)\,e^{i (\beta +\phi )}}$$

As we foresaw, if $a=1$ (no ring attenuation) then the magnitude of this quantity is always unity and the phase is:

$$\pi -\phi + 2\arctan\left(\frac{\sin(\beta+\phi)}{\sin(\beta+\phi) - \cos\left(\frac{\alpha }{2}\right)}\right)$$

which you can make arbitrarily close to having a step of $2 \pi$ radians by having $\cos(\alpha/2)$ arbitrarily near to 1. I plot this phase below for $\beta = \pi$ and with $\alpha = 1$ (green plot), $\alpha = 0.5$ (blue plot) and $\alpha = 0.1$ (red plot).

as you can see you get very sensitive dependence of the output phase on the phase to ring delay, which, as I said, can be converted to a sensitive amplitude dependence by including the resonant ring system in one arm of an optical fibre Mach-Zehnder interferometer.

But there is in practice no need to use the Mach-Zehnder interferometer. If we match the loss in the ring to the coupler split ratio so that $a =\cos(\alpha/2)$ then the output intensity has a perfect null, moreover, as we make $a$ nearer and nearer to 1, the notch gets very sharp. I plot this intensity below for $\beta = \pi$ and with $a =\cos(\alpha/2) = 0.9$ (green plot), $a =\cos(\alpha/2) = 0.95$ (blue plot) and $a =\cos(\alpha/2) = 0.99$ (red plot).