An object is placed at 5 cm distance in front of concave mirror of radius of curvature 15 cm.

What is the nature and magnification of the image?

3, virtual and erect

u = -5cm , f = -7.5 cm

$ \frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{-1}{7.5} + \frac{1}{5} = \frac{1}{15}$

$ v = 15cm$ 

Since v is positive , so image is virtual.

$m = -\frac{v}{u} = +3$

SO image is erect.