Question

the focal length of a concave mirror is 20 cm. Find the position of an object in front of the mirror so that the real image is two times the size of the object​

Answer 1

Answer:

The object is placed 30 cm before the mirror.

Explanation:

Given focal length= -20 cm (convex mirrors have -ve focal lengths)

Since image is real the magnification =

$\dfrac{v}{u} = 2$

$v = 2u$

Using mirror formula:

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

$- \dfrac{1}{20} = \dfrac{1}{u} + \dfrac{1}{2u} \\ - \dfrac{1}{20} = \dfrac{3}{2u} \\ u = - 30$

Therefore the object is placed 30 cm before the mirror.

Answer 2

Given : The focal length of a concave mirror is 20 cm & the real image is two times the size of the object .

Exigency To Find : Position of an object in front of mirror .

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⠀⠀Given that ,

》 The Focal Length of mirror is -20 cm [ as , Concave mirror focal length is always in negative '-ve' ]

》 The size of real image is two times the size of the object .

$\qquad:\implies \sf Size_{(\:Real\:image\:)} \:\:(\:or \:Magnification \:\:'M'\:)\: = \dfrac{v}{u} =\: 2 \:\\\\\:\:\qquad [\:\because \sf\: image \:\:is\:\bf\:real\:\:]$

$\qquad:\implies \sf Magnification \: = \dfrac{v}{u} =\: 2 \:$

$\qquad:\implies \sf \dfrac{v}{u} =\: 2 \:$

$\qquad:\implies \sf v =\: 2u \:$

$\qquad:\implies \sf v \:\:( \:or\:image \:Distance \:\:)\:=\: 2u \:$

$\qquad :\implies \pmb{\underline{\purple{\:v \:\:( \:or\:image \:Distance \:\:)\:=\: 2u\: }} }\:\:\bigstar$

⠀⠀⠀⠀⠀Finding the object distance :

$\dag\:\:\frak{ As,\:We\:know\:that\::}\\\\ \qquad\maltese\:\:\bf By \: Mirror\:Formula\:\::$

$\qquad \dag\:\:\bigg\lgroup \sf{ \qquad \dfrac{1}{f} \:\: =\:\:\dfrac{1}{v} \:+ \:\dfrac{1}{u}\qquad }\bigg\rgroup$

⠀⠀⠀⠀⠀Here , f is the focal length, v is the image distance & u is the object distance.

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

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

$\qquad \dashrightarrow \:\sf \qquad \dfrac{1}{(-20)} \:\: =\:\:\dfrac{1}{2u} \:+ \:\dfrac{1}{u}$

$\qquad \dashrightarrow \:\sf \qquad -\dfrac{1}{20} \:\: =\:\:\dfrac{1}{2u} \:+ \:\dfrac{1}{u}$

$\qquad \dashrightarrow \:\sf \qquad -\dfrac{1}{20} \:\: =\:\:\dfrac{1}{2u} \:+ \:\dfrac{1}{u}$

$\qquad \dashrightarrow \:\sf \qquad -\dfrac{1}{20} \:\: =\:\:\dfrac{3 }{2u} \:$

$\qquad \dashrightarrow \:\sf \qquad -2u(1) \:\: =\:\: 20 \times 3 \:$

$\qquad \dashrightarrow \:\sf \qquad -2u \:\: =\:\: 60 \:$

$\qquad \dashrightarrow \:\sf \qquad u \:\: =\:\: -30\: \:$

$\qquad \dashrightarrow \pmb{\underline{\purple{\:u \:\:( \:or\:object \:Distance \:\:)\:=\: - 30 \:cm\: }} }\:\:\bigstar$

⠀⠀⠀⠀⠀▪︎ Here u denotes position of an object ( or object distance ) which is -30 cm .

⠀⠀⠀⠀⠀$\therefore {\underline{ \sf \:Hence, \:The\:position \:of\:object \:is\:\bf -30\:cm\: }}$