An object is placed at a distance of 4 cm from a concave lens of focal length 12 cm. Find the position and nature of the image.

here

object distance, u = -4 cm

focal length, f = -12 cm

so, if

$\frac{1}{f}$ = $\frac{1}{v}$ - $\frac{1}{u}$

then

$\frac{1}{v}$ = $\frac{1}{f}$ - $\frac{1}{u}$

$\frac{1}{v}$ = $\frac{-1}{4}$ - $\frac{1}{12}$

$\frac{1}{v}$ = -0.333

Thus, the image distance will be, v = -3cm

The image will be erect, diminished and virtual.