What exactly gives a larger field of view to the donated "spy" telescopes that NASA may send to Mars? How much larger?

Question

This answer to Could one of the interstellar probes discover Planet IX by accident? links to Space.com's NASA May Launch Donated Spy Satellite Telescope to Mars which says:

An unexpected gift

The two donated telescopes were apparently built for a National Reconnaissance Office program called Future Imagery Architecture, which was terminated in 2005.

NASA announced in June 2012 that it had acquired the instruments, which are designed to have a much wider field of view than Hubble, despite sporting Hubble-like 8-foot-wide (2.4 meters) main mirrors.

In November, the space agency asked scientists to suggest potential uses for the NRO scopes, which are basically just primary and secondary mirrors, with no instruments attached. More than 60 serious proposals came flooding in, 33 of which — including MOST — were presented in early February at the Study on Applications of Large Space Optics (SALSO) workshop in Huntsville, Ala.

I don't understand the use of "despite" there, but my question is:

Question(s):

1. What exactly gives a larger field of view to the donated "spy" telescopes that NASA may send to Mars? I'm assuming that are Cassegrain-like telescopes. It's possible that sources might be available from Was Hubble really related to spy satellites?
2. How much larger are we talking about here; 50%? a factor of ten?

Answer 1

The first mission to use one of the NRO contributed mirrors is the Nancy Grace Roman Space Telescope (previously WFIRST). 1 It has a wider field of view (FOV) for three reasons. First, the donated mirrors do have a shorter focal length than Hubble's. 1 Also, the primary systems are different. The Hubble primary is a 2-mirror Richey-Cretien design (similar to a Cassigrain) 2, while the Roman telescope primary is a 3-mirror anastigmat 3. 3-mirror anastigmat systems are specifically designed for wide FOV applications. [4] If the Mars telescope uses the WFIRST/Roman design, it will have a much larger FOV than Hubble's.

The actual FOV for the Roman telescope is 45 by 23 arc minutes. [5] The high resolution portion of the Hubble telescope FOV is about 18 arc minutes (see image below). For a comparison between the Hubble Wide Field 3 Camera (WFC3) and the Roman telescope, see ref [6]. Although this is quite dramatic, it's not quite fair as the FOV of the WFC3 (2.7 by 2.7 arc minutes) [7] is only a fraction of the total field of view of the telescope. I couldn't find a good link to explain the overall FOV of the Hubble, but I have an image I made a while back. As you can see, the manner used to split the image between five separate instruments, means the FOV of each instrument is only a fraction of the total FOV of the telescope itself. This is the third reason why the Roman telescope will be able to create much larger FOV images.

1 https://en.wikipedia.org/wiki/Nancy_Grace_Roman_Space_Telescope

2 https://asd.gsfc.nasa.gov/archive/hubble/technology/optics.html

3 https://roman.gsfc.nasa.gov/images/stsci/roman-mission-operations-tools.pdf

4 https://en.wikipedia.org/wiki/Three-mirror_anastigmat

5 https://www.stsci.edu/files/live/sites/www/files/home/roman/_documents/roman-science-sheet.pdf

6 https://svs.gsfc.nasa.gov/13672

7 https://en.wikipedia.org/wiki/Wide_Field_Camera_3

Answer 2

The answer is that the two telescopes have different focal lengths and that focal length and mirror size are parameters that can be independently adjusted (there are obviously other tradeoffs to do with resolution and focal ratio).

Here's a description of why.

Here is a rather terrible diagram of a simplified Newtonian reflector – I've simplified it just by unfolding the optics so there is no (flat) secondary mirror to make it easier to see what is going on:

In this picture the focal length – the distance from the mirror at which the image is in focus – is $l$, and the mirror radius is $r$. Two rays are shown coming in, incident at the same point on edge of the mirror. The first ray comes in parallel with the centreline of the mirror and is reflected down through an angle $\theta$. From elementary trignometry.

$$\tan\theta = \frac{r}{l}$$

The second rat comes in at an angle $\phi$ to the centreline, and is therefore reflected downwards through an angle $\theta + \phi$ (to convince yourself of this you need to draw fiddly pictures of the bit of mirror where the reflection happens).

So the question is: how far below the first ray does the second ray end up? This distance is shown as $y$ on the diagram. Well you can do a little geometry to show that

$$ \begin{align} \tan(\theta + \phi) &= \frac{r + y}{l}\\ \frac{\tan\theta + \tan\phi}{1 - \tan\theta\tan\phi} &= \frac{r}{l} + \frac{y}{l}&&\text{using double-angle formula for $\tan$}\\ \tan\theta + \tan\phi &\approx \frac{r}{l} + \frac{y}{l}&&\text{assuming $\theta$, $\phi$ small}\\ \end{align} $$

or, in other words

$$\phi\approx \frac{y}{l}$$

For small angles.

What this means is that, for a sensor with a size of $Y$, the maximum angle it can get light from is

$$\phi_\text{max} \approx \frac{Y}{l}$$

Or in other words the field of view of the telescope goes inversely as the focal length, completely independent of the radius of the mirror.

Note again that I've just treated light as rays in all this: nothing here tells you anything about the resolution of the device, just how much it can see.