Question

An object and a screen are mounted on an optical bench and a converging lens is placed between them so that a sharp image is received on the screen. The linear magnification of the image is $2.5$. The lens is now moved $30cm$ nearer the screen and a sharp image is again formed on the screen. The focal length of the lens is:

• A. $14.0cm$
• B. $14.3cm$
• C. $14.6cm$
• D. $14.9cm$

Answer 1

The correct option is B $14.3cm$

By lens formula,

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

Given, Magnification $m=2.5=v/u$

$v=2.5u.........................\left(i\right)$

Position 1,sitive

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}....................\left(ii\right)$

Position 2,

Object on left, image on right, focal length is po

Image distance decreases and object distance increases by 30 cm

$\frac{1}{v-30}+\frac{1}{u+30}=\frac{1}{f}......................\left(iii\right)$

Substituting $\left(i\right)$ in $\left(ii\right),$

$\frac{1}{2.5u}+\frac{1}{u}=\frac{1}{f}$

$⟹u=7f/\mathrm{5......................}\left(iv\right)$

Equating $\left(ii\right)\&\left(iii\right),$

$\frac{1}{v}+\frac{1}{u}=\frac{1}{v-30}+\frac{1}{u+30}............\left(v\right)$

Substituting $\left(i\right)\&\left(iv\right)\text{in}\left(v\right),$

$\frac{1}{2.5\times \left(7/5\right)f}+\frac{1}{\left(7/5\right)f}=\frac{1}{2.5\times \left(7/5\right)f-30}+\frac{1}{\left(7/5\right)f+30}$

Solving, $f=100/7=14.3cm$