Here, object size, `h_(1) = 2.0 cm`, focal length of convex lens, `f = 10 cm`
object distance, `u -15 cm`, Image distance, `v = ?`, Image size, `h_(2) = ?`
As `1/v - 1/u = 1/f, 1/v = 1/f + 1/u = 1/10 - 1/15 = 1/30`
or `v = 30 cm`.
As v is positive, the image formed is on the right side of the lens. It must be real and inverted.
As linear magnification, `m = (h_(2))/(h_(1)) = v/u, (h_(2))/(2.0) = 30/(-15) = -2`
or ` h_(2) = -4.0 cm`.
Negative sign of m and `h_(2)` show that the image is inverted.
Thus a real, inverted image enlarged 2 times (i.e. `4.0 cm` tall) is formed at a distance of `30 cm` on the right side of the lens.