Question

The focal length of a convex lens is 15 cm. An object is placed at a distance 25cm

away from the optic centre. Calculate the distance of the image from the object. Also

calculate magnification​

Answer 1

$\bf \underline{ \underline{\maltese\:Given} }$

$\sf\Rrightarrow The \: focal \: length \: of \: a \: convex \: lens \: (f)= 15 \: cm$

$\sf\Rrightarrow\sf An \: object \: distance \: from \: optic \: centre \: (u)= 25 \: cm$

$\bf \underline{ \underline{\maltese\:To \: find } }$

$\sf\Rrightarrow Distance \: of \: the \: image \: from \: the \: object = \: ?$

$\Rrightarrow\sf Magnification = \: ?$

$\bf \underline{ \underline{\maltese\:Solution } }$

$\implies \boxed{ \sf\dfrac{1}{f} = \: \dfrac{1}{v} - \dfrac{1}{u} }$

$\implies \sf \dfrac{1}{v} =\: \dfrac{1}{f} + \dfrac{1}{u}$

$\implies \sf \dfrac{1}{v} =\: \dfrac{1}{15} + \dfrac{1}{ - 25}$

$\implies \sf \dfrac{1}{v} =\: \dfrac{1}{15} - \dfrac{1}{ 25}$

$\implies \sf \dfrac{1}{v} =\: \dfrac{5}{75} - \dfrac{3}{75}$

$\implies \sf \dfrac{1}{v} =\: \dfrac{5 - 3}{75}$

$\implies \sf \dfrac{1}{v} =\ \dfrac{2}{75}$

$\implies \sf v = \cancel\dfrac{75}{2}$

$\implies \sf v = 37.5 \: cm$

$\sf \Rrightarrow \underline{Hence, \:image \: distance \: (h_i)= 37.5 \ \: cm}$

$\bf \underline{Now},$

$\sf We\: are \: said \: to \: magnify..$

$\sf Magnification = \dfrac{Height \: of \: image \: \:(h_i)}{H eight \: of \: object \:\: (h_o)}$

$\sf\implies m = \dfrac{37.5}{25}$

$\sf\implies m = 1.5$

$\sf\Rrightarrow \underline{Hence, \:Magnification \: (m)= 1.5}$