Question

An object is placed $40cm$ away from a convex lens of focal length $15cm$ and a sharp image is formed on a screen. Now a concave lens is placed in contact with the convex lens. The screen now has to be moved away from the lens system by $96cm$ to get a sharp image again. What is the focal length of the concave lens (in $cm$ )?

Answer 1

Answer: 30

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$
$\therefore \frac{1}{15}=\frac{1}{v}+\frac{1}{40}$
$\therefore \frac{1}{v}=\frac{1}{15}-\frac{1}{40}$
$\therefore v=24cm.$
When concave lens is placed,
$v^{\prime }=(96+24)=120cm$
$\frac{1}{f}=\frac{1}{v^{\prime }}-\frac{1}{u}$
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 $=30cm$