Question

A 2 cm candle is placed in front of convex mirror having focal length of 15 cm . Image is formed 10 cm away from the mirror . Find object distance ?
a) 6 cm
b) -30 cm
c) 30 cm
d) 6 cm
With explanation​

Answer 1

Answer :-

Distance of the object is -30 cm from the mirror . [Option.b]

Explanation :-

We have :-

→ Height of the object = 2 cm

→ Focal length of the mirror = 15 cm

→ Position of the image = 10 cm

________________________________

Since the mirror is convex, we have :-

• f = + 15 cm

• v = + 10 cm

Now, according to mirror formula :-

1/v + 1/u = 1/f

⇒ 1/u = 1/f - 1/v

⇒ 1/u = 1/15 - 1/10

⇒ 1/u = (1 - 2)/30

⇒ 1/u = -1/30

⇒ -u = 30

⇒ u = - 30 cm

Note : Since the object, by convention is always placed to the left of the mirror, it's x-coordinate 'u' is always negative.

Answer 2

Given : Height of candle is ( h ) 2 cm , Focal Length of the mirror is ( f ) 15 cm , Distance between mirror and image is ( v ) 10 cm & The mirror is Convex Mirror .

Exigency To Find : The object distance ( u ) .

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⠀⠀Given that ,

⠀⠀⠀⠀⠀⠀⠀⠀▪︎⠀⠀The mirror is CONVEX MIRROR .

Therefore,

⠀⠀⠀⠀⠀⠀⠀⠀▪︎⠀ Focal Length of the mirror is ( f ) 15 cm ,

⠀⠀⠀⠀⠀⠀⠀⠀▪︎⠀ Distance between mirror and image is ( v ) 10 cm

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

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

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

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

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

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

$\qquad:\implies \sf \dfrac{1}{10} + \dfrac{1}{u}\:=\:\dfrac{1}{15}$

$\qquad:\implies \sf \dfrac{1}{u}\:=\:\dfrac{1}{15} - \dfrac{1}{10}$

$\qquad:\implies \sf \dfrac{1}{u}\:=\: \dfrac{1- 2}{30}$

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

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

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

⠀⠀⠀⠀⠀⠀⠀⠀▪︎⠀Here , u denotes object distance which is - 30 cm .

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