Solution:Virtual and erect.
Given, Power = -10 D,
Distance of object (u) = - 15 cm.
Now,
Focal length, f = 1/đ.
= 1/10 = - 0.1 m = -10 cm.
Because the focal length is negative, the given lens is concave lens and in the case of concave lens no matter where the object is placed other than infinity, the virtual and Erect image will be formed between the optical center and the focus of the concave lens.