You got it. We can't make optical transceivers as good as electrical ones. Not as small or power-efficient.
They require significantly different fabrication processes, and we don't know how to fab them into the same chip as electrical ones. I mean: you can either have photonics, or performant digital (or analog) electronics.
We've gotten really, really good at making small electronics, per the latest tech coming out of Intel & TSMC. We are... not that good at making photonics.
I wonder what the latency for switching medium is these days too (for the super small transceivers). To my understanding optical is better for attenuation than electric (less noise, and thus easier to shove more frequencies and higher frequencies on the same pipe), and can be faster (both medium dependent, neither yet approaching the upper bound of c).
I'm imaging the latency incurred by the transceiver is eventually offset from the gains in the signal path (for signal paths relevant to circuit boards and ICs)
The big issue is really 1. Photonic waveguides are much larger than electronic ones (due to the wavelength) 2. You loose dynamic range and in EO conversion (shot noise is significant at optical frequencies) 3. Co integration of optics and photonics components is nontrivial due to the different materials and processes. 4. Power efficiency of EO conversion is also not that great.
Where photonics shines is for transmission of high frequencies (i.e. a lot of data) over long distances and being immune to EM interference. So there is certainly a tradeoff for at what transmission distances to go optical and as data rates keep going up the tradeoff length has become shorter and shorter. Intel, Nvidia, AMD et al. All do research into optical interconnect.
If so, does that matter at all here? Dunno if that holds up for such kind of devices and/or at these scales (much shorter distance, but also much higher speed).
Which brings the question: why operating wavelengths are smaller but “waveguides” are bigger in optical fiber communication. In fact, fiber itself is a waveguide and its diameter is tens of micro meters.
I'm not sure what kind of refractive indices are possible in much smaller photonic circuits, particularly if it's not practical to develop and run everything in a permanent vacuum.
You get to this result if you take the electromagntic wave equation - a partial differential equation - and solve that for your transmission line configuration.
The proper analogy in the realm of electrical waveguides is the hollow waveguide. The hollow waveguide supports TE- and TM-modes but not TEM modes just like a dielectric conductor. The size is also a function of the dielectric constant ε.
What we mostly use are TEM waveguides like microstrips or coaxial cables. The difference between electrical waveguides that supports TEM modes and waveguides that supports TE/TM modes is that the former has two independent potential planes and the latter only one. Also TEM waveguides do not have a lower cutoff frequency. A TEM wave with any frequency can propagate on any microstrip configuration.
This is not true for TE/TM waves.
What's important to understand is that for microstrips/coaxial cables the power isn't transferred in the metal but in the space (dielectric) around the metal - see Poynting vector. So what happens if you have a second conductor in that space? You get crosstalk! So TEM transmission lines do not contain the wave like hollow waveguides or optical fibers (edit: ok coaxial cables do, microstrips don't)
Now the question, how big is the microstrip? Is it just the width of the signal conductor? No, it is not.
Edit: The width of the metal lines in a chip is given by the current it must carry - current density requirement, electro-migration issues. Power lines are wide because they have to supply power to the circuit but logic traces in CMOS technology only carry negligible amount of current. In circuits like RF power amplifiers with bipolar transistors the trace width is much larger because it has to carry a much larger current. But again, microstrip lines do not have a lower cutoff frequency.
I'm not an expert on integrated electronic circuits, but I guess the difference could matter depending on application.