A cube of side $$2\,m$$  is placed in front of a concave mirror of focal length $$1\,m$$  with its face $$P$$ at a distance of $$3\,m$$  and face $$Q$$ at a distance of $$5\,m$$  from the mirror. The distance between the image of face $$P$$ and $$Q$$ is

Solution :
$$\eqalign{ & {\text{For}}\,P:\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \cr & {\text{or}}\,\,\frac{1}{{{v_P}}} + \frac{1}{{ - 3}} = \frac{1}{{ - 1}} \cr & {\text{or}}\,\,{v_P} = - \frac{3}{2}m \cr & {\text{For}}\,Q:\frac{1}{{{v_Q}}} + \frac{1}{{ - 5}} = \frac{1}{{ - 1}} \cr & {\text{or}}\,\,{v_Q} = \frac{{ - 5}}{4}m \cr} $$
So horizontal distance between image of $$P$$ and $$Q$$ is $$0.25\,m$$