An object is placed at $(i) \; 10 \; cm$,$(ii) \; 5 \; cm$ in front of a concave mirror of radius of curvature $15 \; cm$. Find the position,nature,and magnification of the image in each case.

(N/A) The focal length $f = -15 / 2 \; cm = -7.5 \; cm$.
$(i)$ The object distance $u = -10 \; cm$.
Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$:
$\frac{1}{v} + \frac{1}{-10} = \frac{1}{-7.5}$
$\frac{1}{v} = \frac{1}{10} - \frac{1}{7.5} = \frac{7.5 - 10}{75} = \frac{-2.5}{75} = -\frac{1}{30}$
So,$v = -30 \; cm$.
The image is $30 \; cm$ from the mirror on the same side as the object. Magnification $m = -\frac{v}{u} = -\frac{(-30)}{(-10)} = -3$.
The image is magnified,real,and inverted.
$(ii)$ The object distance $u = -5 \; cm$.
$\frac{1}{v} + \frac{1}{-5} = \frac{1}{-7.5}$
$\frac{1}{v} = \frac{1}{5} - \frac{1}{7.5} = \frac{7.5 - 5}{37.5} = \frac{2.5}{37.5} = \frac{1}{15}$
So,$v = 15 \; cm$.
The image is formed at $15 \; cm$ behind the mirror. Magnification $m = -\frac{v}{u} = -\frac{15}{(-5)} = 3$.
The image is magnified,virtual,and erect.