Question

A 2.0 -cm-high object is placed 12cm from a convex lens perpendicular to its principal axis.The lens forms a real image whose size is 1.5cm .Find the power of the lens.

Answer With Full Explanation

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[ Hint :- Answer is 175/9 D ]​

Answer 1

$\frak{Given}\:\begin{cases} \:\quad \sf Height \:of \:the \:object\:,\:h_o \:=\:\pmb{\frak{ 2 \: cm }} \\ \:\quad \sf Height \:of \:the \:image\:,\:h_i \:=\:\pmb{\frak{ 1.5 \: cm }} \\ \:\quad \sf Distance \:of \:the \:object\:,\:u \:=\:\pmb{\frak{ 12 \: cm }} \end{cases}$

Need To Find : The Power of the Lens [ Convex Lens ] ?

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━⠀

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

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

⠀⠀⠀

Where ,

• u = Distance of the Object &
• v = Position of image .

$\dag\underline{\frak{ Putting \:known \:Values \:in \:Given \:Formula \:\::\:}}$

$\qquad \twoheadrightarrow \:\sf Magnification \::\:\:\dfrac{ h_{image}}{h_{object}}\:=\:\dfrac{\:v}{u}\:\\\\ \qquad \twoheadrightarrow \:\sf \:\dfrac{ h_{image}}{h_{object}}\:=\:\dfrac{\:v}{u}\:\\\\ \qquad \twoheadrightarrow \:\sf \:\dfrac{-1.5}{2}\:=\:\dfrac{\:v}{-12}\:\\\\ \qquad \twoheadrightarrow \:\sf \:v\:=\dfrac{-1.5\:\times \: (- 12 )}{2}\:\:\\\\ \qquad \twoheadrightarrow \:\sf \:v\:=-1.5\:\times \: (- 6 )\:\:\\\\ \qquad \twoheadrightarrow \underline {\pmb{\boxed {\frak{\purple {\: \:v\:=\:9\:cm\:}}}}}\:\:\bigstar \:$

$\therefore \:\underline {\sf \:Position \:of \:an \:image \:is \:,\:v\:\pmb{\sf 9\:cm\:}.\:}$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━⠀

$\qquad \bigstar \:\underline {\pmb{\sf { { From \:\:Lens\:\:Formulae\:\::}}}}$

$\qquad \star \:\: \pmb{\underline {\boxed {\pink{\sf \:\dfrac{1}{f}\:=\:\dfrac{1}{v}\:-\:\dfrac{1}{u} }}}}$

Where ,

• f = Focal Length of lens ,
• u = Distance of the Object &
• v = Position of image .

$\dag\underline{\frak{ Putting \:known \:Values \:in \:Given \:Formula \:\::\:}}$

$\qquad \twoheadrightarrow \sf \:\dfrac{1}{f}\:=\:\dfrac{1}{v}\:-\:\dfrac{1}{u}\:\\\\ \qquad \twoheadrightarrow \sf \:\dfrac{1}{f}\:=\:\dfrac{1}{9}\:-\:\dfrac{1}{-12}\:\\\\ \qquad \twoheadrightarrow \sf \:\dfrac{1}{f}\:=\:\dfrac{1}{9}\:-\:\dfrac{1}{12}\:\\\\ \qquad \twoheadrightarrow \sf \:\dfrac{1}{f}\:=\:\dfrac{4 + 3 }{36}\:\:\\\\ \qquad \twoheadrightarrow \sf \:f\:=\:\dfrac{36 }{7}\:\:\\\\ \qquad \twoheadrightarrow \underline {\pmb{\boxed {\frak{\purple {\: \:f\:=\:\dfrac{36}{7}\:\:cm\:}}}}}\:\:\bigstar \:$

$\therefore \:\underline {\sf \:Focal\:Length \:of \:a\:mirror \: \:,\:f\:is\:\:\pmb{\sf 36/7\:cm\:}.\:}$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━⠀

$\qquad \bigstar \:\underline {\pmb{\sf { { From \:\:Power \:\:Formulae\:\::}}}}$

$\qquad \star \:\: \pmb{\underline {\boxed {\pink{\sf \:Power\:=\:\dfrac{1}{f\:(\:in\:m)}\:}}}}$

Where ,

• f = Focal Length of lens in metres .

$\dag\underline{\frak{ Putting \:known \:Values \:in \:Given \:Formula \:\::\:}}$

$\qquad \twoheadrightarrow \sf \:Power\:=\:\dfrac{1}{f\:(\:in\:m)}\:\:$

$\qquad \twoheadrightarrow \sf \:Power\:=\:\Bigg[ \:\:\dfrac{1}{\dfrac{36}{700}\:}\:\:\:\:\Bigg] \:\quad \qquad \because \:\bigg\lgroup 1\:m\:=\:100\:cm\:\bigg\rgroup\\\\ \qquad \twoheadrightarrow \sf \:Power\:=\:\Bigg[ \dfrac{1}{36\:}\:\times 700 \Bigg] \\\\ \qquad \twoheadrightarrow \sf \:Power\:=\: \dfrac{700}{36\:}\: \\\\ \qquad \twoheadrightarrow \sf \:Power\:=\: \dfrac{175}{\:9\:}\: \\\\ \qquad \twoheadrightarrow \underline {\pmb{\boxed {\frak{\purple {\: \:Power\:=\:\dfrac{175}{9}\:\:D\:}}}}}\:\:\bigstar \:$

$\therefore \:\underline {\sf \:Power \:of \:a\:lens \: \:\:is\:\:\pmb{\sf 175/9\:D\:}.\:}$