$A$ concave lens of focal length $15 \,cm$ forms an image $10 \,cm$ from the lens. How far is the object placed from the lens? Draw the ray diagram.

(N/A) Focal length of concave lens,$f = -15 \,cm$ (by sign convention).
Image distance,$v = -10 \,cm$ (since the image formed by a concave lens is virtual and on the same side as the object).
According to the lens formula:
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
Substituting the values:
$\frac{1}{-10} - \frac{1}{u} = \frac{1}{-15}$
$-\frac{1}{u} = -\frac{1}{15} + \frac{1}{10}$
$-\frac{1}{u} = \frac{-2 + 3}{30} = \frac{1}{30}$
$u = -30 \,cm$
The negative sign indicates that the object is placed $30 \,cm$ in front of the lens.