Question

A convex lens produces a double size real image when an object is placed at a distance of 18 cm from it. Where should the object be placed to produce a triple size real image ?

Answer 1

Given $h_1=2h_0$, $h_1=\frac{v}{u}$

$\Rightarrow \frac{v}{u}=2$

$\Rightarrow v=2u=36\text{cm}$

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

$\frac{1}{36}+\frac{1}{18}=\frac{1}{f}$

$\Rightarrow \frac{1}{f}=\frac{3}{36}$

$\Rightarrow f=12\text{cm}$

Now, $h_i=3h_0$

$\frac{h_i}{h_0}=\frac{v}{u}$

$\Rightarrow \frac{v}{u}=3\Rightarrow v=3u$

$\frac{1}{3u}+\frac{1}{4}=\frac{1}{12}\Rightarrow 3u=40$

$u=16\text{cm}$