Question

5. A Speherical mirror of focal length 20 cm will form a image on
screen placed at a distance 60 cm from the mirror. At What
distance the object must be placed from the mirror.​

Answer 1

Given :-

Focal Length of Mirror = f = - 20 cm

Image Distance = v = - 60 cm

We are required to find out the object distance. Considering the mirror as Concave.

Let the object distance = u

By using Mirror Formula.

1/u + 1/v = 1/f

1/u - 1/60 = -1/20

1/u = -1/20 + 1/60

1/u = (-3 + 1)/60

u = - 60/2

$\bf{u\:=-\:30\:cm}$

Hence,

The object distance = u = -30 cm

Answer 2

Given : A Speherical mirror of focal length ( f ) -20 cm & A image on screen placed at a distance ( v ) -60 cm from the mirror.

Exigency To Find : What distance the object ( u ) must be placed from the mirror ?

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━

❍ Let's Consider the mirror as concave mirror .

⠀⠀⠀⠀⠀Given that ,

⠀⠀⠀⠀⠀▪︎ Focal length ( f ) of the object is : - 20 cm

⠀⠀⠀⠀⠀▪︎ Image Distance ( v ) of the image is : - 60 cm .

Now ,

⠀⠀⠀⠀⠀Finding Object Distance ( u ) :⠀

$\dag\:\:\sf{ As,\:We\:know\:that\::}\\\\\qquad \bigstar\:\:\bf Formula\: of \: mirror\:\::$

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

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

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

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

$\qquad:\implies \sf \: \bigg( \dfrac{-1}{20} \bigg)\: = \bigg( \dfrac{-1}{60}\bigg) + \dfrac{1}{u}$

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

$\qquad:\implies \sf \: \bigg( \dfrac{-3 + 1}{60} \bigg)\: = \dfrac{1}{u}$

$\qquad:\implies \sf \: \bigg( \dfrac{-2}{60} \bigg)\: = \dfrac{1}{u}$

$\qquad:\implies \sf \: \dfrac{-2}{60} \: = \dfrac{1}{u}$

$\qquad:\implies \sf \: \cancel {\dfrac{-2}{60}} \: = \dfrac{1}{u}$

$\qquad:\implies \sf \: \dfrac{-1}{30} \: = \dfrac{1}{u}$

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

$\qquad:\implies \bf \: u\: = \:-30 \:$

$\qquad :\implies \frak{\underline{\purple{\:u\: = -30\: cm\: }} }\:\:\bigstar$

Therefore,

⠀⠀⠀⠀⠀⠀⠀⠀▪︎ u is the object distance is - 30 cm .

⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm {\:Hence\:,\:The \:Object \:Distance \:(\ u \ )\:is\:\bf{-30\:cm}}}}$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀