Question

If an object of 7 centimetre height is placed at a distance of 12 cm from a convex lens of focal length 8 cm find the position nature and height of the image

Answer 1

$\Large\underline{\underline{\sf{\color{Red}{GIVEN}}}}$

$\bullet\ \sf{Size\ of\ the\ object(h_o)=\ }\bold{7\ cm}$

$\bullet\ \sf{Object\ distance(u)=\ }\bold{-12\ cm}$

$\bullet\ \sf{Focal\ length(f)=\ }\bold{8 \ cm}$

$\Large\underline{\underline{\sf{\color{Red}{TO\ FIND}}}}$

$\longmapsto \sf{The\ }\bold{position,\ nature\ }\sf{and\ }\bold{height\ of\ the\ image(h_i)}\sf{.}$

$\Large\underline{\underline{\sf{\color{Red}{SOLUTION}}}}$

$\textsf{We know,}$

$\boxed{\sf{\dag\ \frac{1}{v}-\frac{1}{u}=\frac{1}{f} }}$

$\sf{Putting\ the\ known\ values:-}$

$\displaystyle{\sf{\dashrightarrow \frac{1}{v}-\frac{1}{-12}=\frac{1}{8} }}\\\\\displaystyle{\sf{\dashrightarrow \frac{1}{v}+\frac{1}{12}=\frac{1}{8} }}\\\\\displaystyle{\sf{\dashrightarrow \frac{1}{v}=\frac{1}{8}-\frac{1}{12} }}\\\\\displaystyle{\sf{\dashrightarrow \frac{1}{v}=\frac{3-2}{24} }}\\\\\displaystyle{\sf{\dashrightarrow \frac{1}{v}=\frac{1}{24} }}\\\\\boxed{\boxed{\displaystyle{\sf{\dashrightarrow v=24\ cm }}}}$

$\underline{\bold{Hence,\ the\ image\ is\ formed\ 24\ cm\ on\ the\ right\ side}}\\\underline{\bold{ of\ the\ mirror\ (As\ v\ is\ positive).}}$

$\rule{190}3$

$\sf{We\ also\ know\ that,}$

$\boxed{\displaystyle{\sf{\dag\ \frac{v}{u}=\frac{h_i}{h_o} }}}$

$\sf{Putting\ the\ known\ values :-}$

$\displaystyle{\sf{\dashrightarrow \frac{24}{-12}=\frac{h_i}{7} }}\\\\\displaystyle{\sf{\dashrightarrow -2=\frac{h_i}{7} }}\\\\\boxed{\boxed{\displaystyle{\sf{\dashrightarrow h_i=-14\ cm}}}}$

$\underline{\bold{Hence,\ the\ size\ of\ the\ image\ is\ 14\ cm,\ pointing}}\\\underline{\bold{downwards.}}$

$\rule{190}3$

$\sf{We\ also\ know\ that,}$

$\boxed{\sf{\dag\ Magnification(m)=\frac{v}{u} }}$

$\sf{Putting\ the\ known\ values :-}$

$\sf{\dashrightarrow m=-2}\\\sf{(From\ above)}$

$\bold{\underline{Hence,\ the\ image\ is\ real\ and\ inverted\ as\ the}}\\\bold{\underline{magnification\ is\ negative.}}$

$\mathbb{CONCLUSION :--}$

$\bullet\ \sf{Image\ is\ formed\ 24\ cm\ behind\ the\ mirror.}$

$\bullet\ \sf{Position-Beyond\ 2F_2.}$

$\bullet\ \sf{Nature-Real\ and\ inverted.}$

$\bullet\ \sf{Height\ of\ the\ image(h_i)=-14\ cm.}$

Answer 2

Given

SIze of the object = 7 cm

Object distance = -12 cm

Focal length = 8 cm

To find :-

Position, nature and height of the image

FORMULA TO BE USED

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

$\implies \dfrac{1}{8}=\dfrac{1}{v} - \dfrac{1}{12}$

$\implies \dfrac{1}{v} = \dfrac{1}{8} - \dfrac{1}{12}$

$\implies \dfrac{1}{v} = \dfrac{1}{24}$

$\implies v = 24 \ cm$

$\boxed{\boxed{\bold{Image \ is \ formed \ on \ the \ right \ side \ of \ the \ mirror.}}}}}}$

                                                     

$\dfrac{v}{u} = \dfrac{h_i}{h_o}$

$\implies \dfrac{24}{-12} = \dfrac{h_i}{7}$

$\implies h_i = -14 \ cm$

$\boxed{\boxed{\bold{The \ image \ is \ enlarged}}}}}}$

                                                   

$m = \dfrac{v}{u}$

$= \dfrac{24}{-12}$

$= -2$

$\boxed{\boxed{\bold{Image \ is \ real \ and \ Inverted}}}}}}$