Image of the flame formed by the concave mirror on the wall will be real and inverted, hence magnification



                        $m=\frac{I}{O}=\frac{-1\text{.}5}{0·5}=-3$

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

∴                       $v = 3u$

Using               $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

                        $\frac{1}{3u}+\frac{1}{u}=-\frac{1}{1\text{.}5}$              $\left(f=-1\text{.}5\text{m}\right)$

                        $\frac{4}{3u}=-\frac{1}{1\text{.}5}$

∴                        $u=\frac{-4\times 1\text{.}5}{3}=-2\text{m}$

Thus the flame is at distance 2m from the mirror and                $v = 3u = –6 \mathrm{m}$

Hence the distance between wall and mirror is 6 m. It implies that the distance between the flame and the wall will be 4m.