Geometrical Optics: Infinite Rays Normally in ray optics, we draw a parallel line from the top of the image to the lens and stop when this line intersects an angled line (drawn from the height of the real object) and intersects. However, why do we stop? We can draw infinite rays from this object and they should be able to go out for infinitely far distances. Thus, we should be able to get a huge image (albeit a bit darker).

 A: (I wanted to put this as a comment, but I don't have enough reputation)
The different rays have to intersect at a point for an image to be formed there, as you said. What happens is that when you look at the rays, they seem to be coming from that point where they intersect, which is the location of the image.
If you put a paper at some other arbitrary point, you won't get the image! You'll get some sort of hazy formation that looks similar to the image, but it's blurry because that's not where the image is, it's just that it might be close enough that the rays arrange in similar patterns as they do for the image. I encourage you to try this out yourself.
Let's say you have a light source. It's rays extend to infinity. But if you randomly took a point where one of the rays meet, it doesn't mean the source is at that point, does it? The mirror just changes where the source appears to be.
A: An image is formed when there is a one to one correspondence between a point on the object and a point in space where all light rays emitted from that point on the object meet up. In other words, if we find that different light rays from the same point of the object end up meeting up at different points in space after passing through the lens, then we don't have an image there (or at least it will be pretty blurry).
This is why you typically see one arrow representing the object and then we look at light rays coming from the top. We pick rays that are easy to use and find where those rays intersect after going through the lens. This then means that the rest of the object must be in focus since it is represented by just a line that is perpendicular to the optical axis.
The problem with going out farther than where the image is formed is that all of the rays from that point on the object won't all go to that farther point. Hence no image will be out there.
A: I have added some more rays onto your diagram and omitted the arrowhead to make the diagram clearer. I have also ignored the effects of spherical aberration.  
Every point on the object produces light which travels along rays and after reflection from the mirror meet at a corresponding point on the image - neighbouring points on the object are neighbouring points on the image. 
 
All the light travelling along the red rays which originate from point $A$ on the object and after reflection from the mirror arrive at point $B$ on the image having taken the same time to travel from $A$ to $B$.
This means that the path lengths of all rays which travel from $A$ to $B$ is the same.
Waves leaving point $A$ in phase arrive at point $B$ in phase.
All the light travelling along the blue rays which originate from point $C$ on the object and after reflection from the mirror arrive at point $D$ on the image having taken the same time to travel from $C$ to $D$.
This means that the path lengths of all rays which travel from $C$ to $D$ is the same.
Waves leaving point $C$ in phase arrive at point $D$ in phase.  
Light which leaves every point on the object after reflection meets at a corresponding point on the image.
The path length from point $A$ to point $D$ after "reflection" from the mirror is not constant and so all the light which originates from point $A$ would not arrive in phase at point $D$ so no "image" of point "A" is formed at point $D$.
