An object is placed 10 cm in front of a convex lens of focal length 15 cm. The image produced will be
$1/15 = 1/v – 1/(-10)$
$1/15 = 1/v + 1/10$
$1/v = 1/15 – 1/10$
$1/v = (2 – 3)/30$
$1/v = -1/30$
$v = -30$ cm
The image distance $v$ is -30 cm. A negative image distance for a lens indicates that the image is formed on the same side as the object, which means it is a virtual image.
The magnification ($m$) is given by $m = v/u$:
$m = (-30) / (-10) = +3$
The positive magnification indicates the image is erect (virtual images are always erect). The magnitude of magnification $|m| = 3$ is greater than 1, indicating that the image is magnified (larger than the object).
Therefore, the image is virtual and magnified.