Here, object size, `h_(1) = 4.5 cm`

object distance, `u = -12 cm`

focal length, `f = +15 cm`

image distance, `v = ?` (to be calculated)

magnification,`m = ?`

As `1/u + 1/v = 1/f` , `1/v = 1/f -1/u`

Putting `u = -12 cm` and `f = + 15 cm` , we get

` 1/v = 1/15 - 1/(-12) = (4+5)/60` or `v = 60/9 = 6.7 cm`

i.e., image is formed `6.7 cm` behind the convex mirror. It must be virtual and erect.

If `h_(2)` is size of the image, then `m = h_(2)/h_(1) = (-v)/u` or `m = h_(2)/h_(1) =(-(6.7))/(-12) = 0.558`

i.e., `h_(2) = 0.558 h_(1) = 0.558 xx 4.5 = 2.5 cm`

As the needle is moved farther from the mirror, image moves away from the the mirror till it is at focus F of the mirror. The size of the image goes on decreasing.