Instrument aperture sizes on Hubble Telescope When looking at the diagram of the focal plane of the Hubble Space Telescope I noticed that the aperture sizes of most of the scientific instruments are very small. They occupy a very tiny fraction of the focal plane.

I understand that for a spectrograph (like the COS) you need a very small aperture to look at point sources. But why the Advanced Cmaera for Surveys or Wide Field Camera 3 also have so small apertures? Most of the focal plane is not used and it must be very tedious for the guidance system to move the target object exactly into the small aperture of the selected instrument. Even though the WFC3 is called "wide field", its field of view is less than 3 arc minutes.

What were the reasons behind selecting these aperture sizes? Is HST designed to observe details of small targets? Are the sizes optimized for majority of observations?

On the other hand the HST is taking images of large objects like nebulae and usually it is necessary to stitch a number of individual frames to create the whole image of a bigger object. For example the Carina nebula (stitched from 48 frames) or the Crab nebula (24 frames).
EDIT: You can't have a very wide field, and simultaneously a very high resolving power, in an instrument with a large focal ratio (or any instrument, really), unless you use sensors with an unrealistic pixel count. Details below:

The linear size of the focal plane is large - one document says it's the size of a "dinner plate". But the instrument is an f/24, which is a large focal ratio. At that ratio, a big linear size in the focal plane doesn't mean much in terms of angular size.

From what I've seen, the Wide Field Camera has a total field of 4.7 arcmin, or 280 arcsec according to this document:

http://asd.gsfc.nasa.gov/archive/hubble/a_pdf/news/SM2-MediaGuide.pdf

Anyway, yes, that's a small field, if you compare it with ground-based amateur astrographs, which can have a field as wide as several degrees of arc.

But the Hubble telescope has a very high resolving power - 0.05 arcsec. Let's take your information and assume the WFC has a total field of 3 arcmin.

3 arcmin / 0.05 arcsec = 3600 pixels in theory, as a linear size of the WFC

If you wanted the WFC to have a bigger total field, you'd have to either sacrifice resolving power, or use a sensor with an enormous pixel count, which is not feasible.

The WFC uses several sensors, each one of them 800 x 800 pixels.



TLDR: Wide field, or great resolving power - pick one. They chose resolving power, and they compensate the narrow field by doing multiple passes.