Question: Why does Nancy Grace Roman = 100 × Hubble? Why does the new space telescopes wide field camera so much wider than the old one's?
The telescopes have very roughly the same wavelengths and apertures, so does the new one have one hundred times more pixels at the same resolution, or 100 times larger pixels? Is there better correction optics that allows for a flatter field for a flat array, or a curved focal plane array, or other optical correction?
Hubble was designed in the late 1980s, when electronic detectors were small (the original Wide Field/Planetary Camera had two different 2x2 arrays of 800x800-pixel CCDs); it also had to carry multiple different instruments, each taking up nominal space in the focal plane. So it was optimized to use small fields of view to go with the small detectors.
WFIRST/Nancy Grace Roman is much more tightly focused on large-field surveys, so it can devote most of the field of view to the IR camera. The reason it is now possible to do this is probably a combination of three things:
Large, sensitive, low-noise IR detectors are now available; the current plan is to use a total of 18 individual 4k x 4k detectors -- a huge advance over the original mid-1990s 256 x 256 NICMOS detector installed in HST in 1997, and over the single 1k x 1k detector in the infrared channel of WFC3 (installed in 2009);
We have lots of experience in building array mosaics with lots of individual detectors (the planned $\sim 300$ megapixel mosaic is relatively modest compared with the gigapixel arrays in Pan-STARRS1 and LSST/Vera Rubin);
Computer technology has advanced to the point where we can now store all the data such an instrument would take and transmit it to the ground in a reasonable time. Hubble can currently take and transmit between 2 and 3 gigabytes of data per day; Roman is forecast to take and transmit about 500 times as much data.