Explanation

(c) Let as shown, 1 and 2 are positions of objects and images in two different situations.
Object
[image:9_image121.emf]
It is given
$\left| \frac{v₁}{u₁} \right| = 2\left| \frac{v₂}{u₂} \right|$
Here, $u₁ = \ - 15\ cm,\ u₂ = \ - 20\ cm$
$\therefore v₁ = 2v₂\ \times \frac{u₁}{u₂} = 2v₂\ \times \frac{15}{20} = \frac{3}{2}v₂$
$now,\ \ \frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
$\therefore\frac{1}{f} = \frac{1}{v₁} - \frac{1}{u₁}\ and\frac{1}{f} = \frac{1}{v₂} - \frac{1}{u₂}$
$so,\ \ \ \frac{1}{v_{1}} - \frac{1}{u_{1}} = \frac{1}{v_{2}} - \frac{1}{u_{2}}$
$\Longrightarrow \frac{2}{2v₂} + \frac{1}{15} = \frac{1}{v₂} + \frac{1}{20}$
$\Longrightarrow v = 20\ cm$
$\therefore\frac{v₁}{u₁} = 2\frac{v₂}{u₂} = 2\ \times \frac{20}{20} = 2$
$\Longrightarrow v₁ = 2u₁ = 2 \times 15 = 30\ cm$
Therefore,$\frac{1}{f} = \frac{1}{v₁} - \frac{1}{u_{1}} = \frac{1}{15} + \frac{1}{30} = \frac{3}{30}$
$\therefore f = 10cm = 0.10\ m$