concave len

Since the image formed by a concave lens is always virtual, and is situated at position between the object and the lens, the image distance v must be shorter than the object distance u.

Because magnification m = v/u, m is thus smaller than unity (one). In other words, the size of the image is always smaller than that of the object.

When the object distance increases, the image distance also increases but would not go beyond the principal focus (remember that when the object is at infinity, the virtual image distance equals to the focal length of the lens, the magnification is minimum). That is to say, the increase of v would not be as much as the increase of u. This leads to reduction of m when u increases. The size of image becomes smaller.

This phenomenon can be proved by the thin lens formula,
1/u + 1/v = 1/f
For a concave lens, v is -ve, f is -ve
hence, 1/u - 1/v = 1/-f
Multiply both sides by v
v/u - 1 = -v/f
i.e. m = 1 - v/f (using magnification m = v/u)

Since for a concave lens, v never exceed f, m is always smaller than one. The image size is always smaller than the object size.

Now, if the object distance increases, the image distance also increases, say to a new value v', where v' > v
we have the new magnification m' = 1 - v'/f
Because v' > v, then (1-v'/f) < (1-v/f), i.e. m' < m
The magnification is reduced, that means the image size becomes smaller.