Question

An object is placed $40\, cm$ away from a convex lens of focal length $15\, cm$ 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 $96\, cm$ to get a sharp image again. What is the focal length of the concave lens (in $cm$ )?

Answer 1

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