Question

An object is placed at a distance of 3 cm from a concave lens of focal length 12 cm. Find the position, size and nature of image​

Answer 1

u=-3cm

f=-12cm

$Lens ~Formula=~\: \boxed{ \frac{1}{f} = \frac{1}{v} - \frac{1}{u}}$

$\frac{1}{v} = (\frac{1}{ - 12}) - (\frac{1}{ - 3} )$

$\frac{1}{v} = \frac{ - 1}{12} + \frac{1}{3}$

$\frac{1}{v} = \frac{ - 1 + 4}{12}$

$\frac{1}{v} = \frac{3}{12}$

$v = \frac{12}{3}$

$\underline { \underline {v = 4cm}}$

Position = 4cm

Nature= Virtual,erect & magnified.

Answer 2

$\sf \: u=-3cm \\ \\ </p><p> \sf \: f=-12cm \\ \\ </p><p> \sf \: \begin{gathered}Lens ~Formula=~\: \boxed{ \frac{1}{f} = \frac{1}{v} - \frac{1}{u}} \\ \\\end{gathered} \\ </p><p> \sf \: \frac{1}{v} = (\frac{1}{ - 12}) - (\frac{1}{ - 3} ) \\ \\ </p><p> \sf \: \frac{1}{v} = \frac{ - 1}{12} + \frac{1}{3} \\ \\ </p><p> \sf \: \frac{1}{v} = \frac{ - 1 + 4}{12} \\ \\ </p><p> \sf \: \frac{1}{v} = \frac{3}{12} \\ \\ </p><p> \sf \: v = \frac{12}{3} \\ \\ </p><p> \sf \: \underline { \underline {v = 4cm}} \\ \\ </p><p> \sf \: Position = 4cm \\ \\ </p><p> \sf \: Nature= Virtual,erect \: and \: magnified.$