Silicon photonics
Which correctly describes what happens when the circumference of a ring resonator is shortened?
- AThe spacing between adjacent resonant wavelengths (FSR) becomes wider
- BThe spacing between adjacent resonant wavelengths (FSR) becomes narrower
- CResonance no longer occurs and all wavelengths pass through
Show answer and explanation
Answer: A. The spacing between adjacent resonant wavelengths (FSR) becomes wider
The FSR (free spectral range) is the spacing between adjacent resonant wavelengths, approximately FSR ≈ λ² / (ng × L) (ng is the group index, L the circumference), so it is inversely proportional to the circumference. For example, with λ = 1.55 µm, ng ≈ 4.2 and L ≈ 63 µm (radius about 10 µm), the FSR is about 9 nm. Halving the circumference roughly doubles the FSR, making it easier in WDM, which handles many wavelengths, to keep one ring from also responding to another wavelength. However, sharper bends increase bending loss, so there is a limit to how small the ring can be.