Question

The focal length of the convex lens is $15cm$ and that of the concave lens is $fcm$ . An object is placed $40cm$ away from the convex lens, now if the concave lens is kept adjacent to the convex lens the image is shifted $96cm$ away from the previous point. Find the value of $f$ .

Answer 1

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
$\therefore \frac{1}{15}=\frac{1}{v}+\frac{1}{40}$
$\therefore \frac{1}{v}=\frac{1}{15}-\frac{1}{40}$
$\therefore v=24cm$ When concave lens is placed,
$v^{'}=\left(\right.96+24\left.\right)=120cm$
$\frac{1}{f}=\frac{1}{v^{'}}-\frac{1}{u}$ where, $f=$ focal length of combination
$\therefore \frac{1}{f}=\frac{1}{120}+\frac{1}{40}\Rightarrow f=30cm$ Also,
$\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}$
$\therefore \frac{1}{30}=\frac{1}{15}+\frac{1}{f_{2}}$
$\therefore f_{2}=-30cm$ As negative sign indicates the lens is concave, neglecting it.
$\therefore $ Focal length $=30cm$