Question

A cancave mirror produces 10 cm long image of an object of height 2 cm. What is the magnificent produced ?​

Answer 1

Given :-

Height of Image = I = 10 cm

Height of Object = O = 2 cm

As per relation we know that,

m = I/O

m = 10/2

{tex}\bf{m\:=\:5}[/tex]

Hence,

The magnification = m = 5

From this we can conclude that image so formed is larger than the object.

Answer 2

$\pmb{\frak{ Given \:}}\begin{cases} \:\quad \sf Height \:of\:the\:image\:,\:h_i \:=\:\pmb{\frak{ 10\:cm\:}}\\ \:\quad \sf Height \:of\:the\:object \:,\:h_o \:=\:\pmb{\frak{ 2\:cm\:}}\end{cases}$

Need To Find : The Magnification of the image .

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━⠀

$\qquad \bigstar \:\underline {\pmb{\sf \: From \:Magnification \:Formula \::\:}}$

$\qquad \star\:\pmb{\underline {\boxed {\pink{\sf \: Magnification \:=\:\dfrac{ h_{image}}{h_{object}}\:\:}}}}$

$\pmb{\sf{ Where \:}}\begin{cases} \:\quad \sf \:h_{image} \:=\:\pmb{\frak{ 10\:cm\:}}\\ \:\quad \sf h_{object} \:=\:\pmb{\frak{ 2\:cm\:}}\end{cases}$

$\qquad \dag\:\underline {\frak{ Substituting \:known \:Values \:in \:Formula \:\::\:}}$

$\twoheadrightarrow \sf Magnification \:=\:\dfrac{ h_{image}}{h_{object}}\:\:\\\\\\ \twoheadrightarrow \sf Magnification \:=\:\dfrac{ 10}{2}\:\:\\\\\\ \twoheadrightarrow \sf Magnification \:=\:\cancel{\dfrac{10}{2}}\:\:\\\\\\\twoheadrightarrow \pmb {\underline {\boxed {\purple {\:\frak{ \:Magnification \:\:=\:5\:}}}}}\:\bigstar \:$

∴ Hence , The Magnification Produced is 5 .