Physics - Geometrical Optics Question with Solution | TestHub

Lens formula is given by

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u} \dots \left(\mathrm{i}\right)$

Where$f$is focal length of lens,$v$is image distance and$u$is object distance.

Given$f=10 \mathrm{c}\mathrm{m}$(as lens is converging)

$u=\mathrm{ }-8\mathrm{ }\mathrm{c}\mathrm{m}$(as object is placed on left side of the lens)

Substituting these values in Eq. (i) , we get

$\frac{1}{10}=\frac{1}{v}-\frac{1}{-8}$

$\Rightarrow \frac{1}{v}=\frac{1}{10}-\frac{1}{8}$

$\Rightarrow \frac{1}{v}=\frac{8-10}{80}$

$\therefore v=\frac{80}{-2}=-40 \mathrm{c}\mathrm{m}$

Hence, magnification produced by the lens

$m=\frac{v}{u}=\frac{-40}{-8}=5$