Why can we see the image produced by a converging lens?

[Johnny B.]

1. The problem statement, all variables and given/known data
Diverging lenses produce a "virtual" image, which can be seen but can't be projected, and converging lenses produce a "real" image which can be projected but not seen.
How come we can see the image produced by a converging lens, which is supposed to be a real image?

2. Relevant equations
None, I don't think I'll need to use any equation for this.

3. The attempt at a solution
I'm clueless... why is it that real images can in theory not be seen? I know that when I look through a magnifying glass, which is a converging lens, I can definitely see something. Is it because the image is "projected" in my retina?
If I trace the rays of light through the lens I get something like this:
http://upload.wikimedia.org/wikipedia/commons/6/69/Converging_lens_object_far.png
meaning the image is inverted and shrunken, but I can't get any further than that.

Thanks in advance!

[rl.bhat]

Whenever diverging rays from any object or from an image fall on our eye, we can see them. Even in your ray diagram if you keep your eye beyond the image you can see an inverted image

[Johnny B.]

Thanks...
So, if I arrange the lens like this

EYE ------- LENS ------- OBJECT

as when looking through a magnifying glass, the image forms between my eye and the lens and I can see it because there are rays going through my eye.
Have I gotten it right?

What's a "diverging" ray btw? Are all rays diverging?

[rl.bhat]

Thanks...
So, if I arrange the lens like this

EYE ------- LENS ------- OBJECT

as when looking through a magnifying glass, the image forms between my eye and the lens and I can see it because there are rays going through my eye.
Have I gotten it right?
What's a "diverging" ray btw? Are all rays diverging?
You are right.
Rays from all the real objects are diverging. After refracting from a lens it may converge or diverge depending on the nature of the lens and position of the object.