Physics - Geometrical Optics Question with Solution | TestHub

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.