Optics 086 · Waveguides, structured light, and modern optics

Microring & Whispering-Gallery Resonator

Why does a small ring beside a waveguide make the transmitted light dip at certain wavelengths? Light coupled in from the straight waveguide circles the ring and resonates whenever a whole number of wavelengths fits the effective circumference, so a smaller ring has its resonances farther apart. On resonance the ring builds up a strong circulating field. When the coupling to the waveguide equals the ring's own loss, transmission at resonance drops to zero; this is critical coupling.

Couple a ring resonator to a single bus waveguide and follow the light coherently. Set the radius and gap, then read the lossless coupling, the exact resonance width, the circulating power and the shift caused by index changes, and freeze complex spectra.

Coherent through power

T

$0.50$

Circulating / input power

B

$50\%$

Exact spectral quality

Q_{\rm spec}

$0.00\pi$

Model regime $\text{valid model regime}$

Model checked against analytic results

Microring & Whispering-Gallery Resonator

Actual ring centreline radius

R

30\,\mu\mathrm m

Actual edge gap; assumed coupling fit

g

0.33\,\mu\mathrm m

Uniform intrinsic power attenuation

\alpha_{\rm dB}

10\,\mathrm{dB/cm}

Vacuum laser wavelength

\lambda_0

1549.59\,\mathrm{nm}

Imposed effective-index shift at reference

\Delta n_{\rm eff}

0

Probe arclength from the coupler

s/L

25\%

A dark output port can hide a bright ring. Tune the laser, match the loss to the coupling, and watch the field passing straight by cancel the field returning from the ring. Remove the intrinsic loss and all the light is transmitted, yet the ring still stores light.

The ring carries one travelling mode, and the coupler is a lumped element. The gap dependence uses an explicit, assumed fit, and the propagation constant has linear dispersion. The whispering-gallery mode is not solved in 3D, the material is not measured, and the model does not predict fabrication results. Backscattering, thermal and nonlinear effects are left out.

Start with the essentials

Focus question: Why does a small ring beside a waveguide make the transmitted light dip at certain wavelengths?
One-sentence intuition: The ring can be bright while the output is dark. The energy balance is $T+(1-a^2)B=1$ . The power circulating in the ring is not the power missing from the output.

Core mathematical model

A lossless coupler

\begin{pmatrix}E_{\rm out}\\E_+\end{pmatrix}=\begin{pmatrix}t&i\kappa\\i\kappa&t\end{pmatrix}\begin{pmatrix}1\\ae^{i\phi}E_+\end{pmatrix},\quad t^2+\kappa^2=1

The fields are normalized so that their squared magnitudes give mode power relative to the input. The ring lies in the X–Z plane, and the forward wave leaves the coupling point at the bottom heading toward positive X. The time dependence uses a negative frequency exponential. There is one bus waveguide and no drop port.

Power buildup and loss in the ring

B=\frac{\kappa^2}{|1-tae^{i\phi}|^2},\quad P(s)/P_0=B a^{2s/L},\quad a=10^{-\alpha_{\rm dB}L_{\rm cm}/20}

The buildup factor is the circulating power just after the coupler. Uniform attenuation removes a fraction of that power on every round trip. At resonance, critical coupling happens when the self-coupling amplitude equals the round-trip amplitude. With no intrinsic loss, the steady-state through amplitude stays at one.

The assumed dispersion and gap model

\beta(\omega)=\frac{(n_0+\Delta n)\omega_0+n_g(\omega-\omega_0)}{c},\quad K(g)=0.12e^{-(g-0.2\,\mu\mathrm m)/(0.12\,\mu\mathrm m)}

The reference wavelength is 1550 nm, with phase index 2.4 and group index 4.2; these are teaching assumptions. The gap law is an imposed coupler calibration over a limited range, not the result of a Maxwell overlap calculation or a manufacturer curve. The index shift is a shift of the effective mode index at the reference wavelength, not a calibrated refractometer reading of the environment.

Exact linewidth and the energy lifetime from round-trip decay

\mathrm{FSR}=\frac1{T_r},\quad\Delta f=\frac{2}{\pi T_r}\arcsin\!\frac{1-at}{2\sqrt{at}},\quad\tau_U=-\frac{T_r}{2\ln(at)},\quad T_r=\frac{n_gL}{c}

The width is the full width at half maximum of the circulating power. The spectral quality factor is the resonance frequency divided by that exact width, and the decay quality factor is the angular frequency times the energy lifetime. The two agree for a narrow resonance, but not exactly for arbitrary loss. After the drive is switched off, the decay plot samples the power once per complete round trip.

Common difficulties

Does zero transmission mean zero field?

Typical misconceptionThe ring holds only the power that is missing from the output.

Better mental modelNo. Coherent feedback can build up circulating power far larger than the input. At critical coupling the outgoing amplitudes cancel, and the intrinsic loss dissipates the incoming energy. With no intrinsic loss there is no absorption dip.

Run the experiment

01
Scene 1: Coherent feedback
Tune the laser to resonance, then set the intrinsic loss to match the coupler. Look at how the complex fields cancel, and how the input power divides between output and loss on a linear scale.
What to observe: A larger radius gives a smaller free spectral range, and the nearest resonance can switch to a different integer order. The index-shift comparison stays on the same unperturbed order.
02
Scene 2: Lossless all-pass
Select the lossless all-pass ring and tune to resonance again. The circulating power grows, but the through power stays at one. This is not the absorption dip of critical coupling.
What to observe: The colors in the apparatus show the averaged travelling power on a fixed display scale. They are not photons, they do not resolve the radial mode, and the animation does not run at the real speed.
03
Scene 3: Effective-index perturbation
Acquire a spectrum, change the effective index, and keep the frozen record as a reference. Then change the radius and look at the real gap in the close-up view.
What to observe: The saved scan holds simulated complex detector data, and the plotted power is recomputed from those same samples.