Question

A concave mirror of focal length 20cm forms an image having twice the size of object. For virtual position of the object,the position of object will be at

Answer 1

Answer:

Explanation:

Given,

Focal length, f = - 20 cm

To Find,

Object Distance, u = ?

Formula or method to be used,

Magnification formula, m = - v/u

Mirror formula, 1/f = 1/v + 1/u

Solution,

Putting all the values, we get

Magnification formula, m = - v/u

⇒ m = - v/u = 2

⇒ v = - 2u .... (i)

Now, putting eq (i)'s value in the mirror formula, we get

1/f = 1/v + 1/u

⇒ 1/- 20 = 1/- 2u + 1/u

⇒ 1/- 20 = (- 1 + 2)/2u

⇒ - 1/20 = 1/2u

⇒ u = - 10 cm

Hence, the position of the object will be at - 10 cm.

Answer 2

Given : Focal Length of Concave Mirror = f = - 20 cm & forms an image having twice the size of object.

Need To Find : Position of an object.

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

❍ Formula for Magnification (m) is given by :

$\dag\:\:\boxed {\sf{Magnification =\bigg( \dfrac{-Size\:of\:image\:}{Size\:of\:Object} \:or\:\dfrac{-v}{u}\bigg) }}$

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

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

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

$\qquad:\implies \bf{\bigg( v = -2 u\bigg) \:\qquad \longrightarrow Eq.1}$

$\dag\:\frak {\underline { As,\:We\:know\:that\::}}$

$\dag\:\:\boxed {\sf{ \dfrac{1}{Focal\:Length} = \bigg( \dfrac{1}{Size\:of\:object \:} + \dfrac{1}{Size\:of\:image}\bigg)}}$

Or ,

$\dag\:\:\boxed {\sf{ \dfrac{1}{f} = \bigg( \dfrac{1}{u \:} + \dfrac{1}{v}\bigg)}}$

Where,

• F or Focal Length is -20 cm . ⠀⠀⠀⠀⠀[ Given that ]
• v = -2u ⠀⠀⠀⠀⠀[ Eq. 1]
• u = ?

⠀⠀⠀⠀⠀⠀$\underline {\frak{\star\:Now \: By \: Substituting \: the \: Given \: Values \::}}$

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

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

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

$:\implies \sf { \dfrac{1}{-20} = \dfrac{1}{2u \:} }$

$:\implies \sf { 2u = - 20 }$

$:\implies \sf { u = \cancel {\dfrac{-20}{2}} }$

$:\implies \bf { u = - 10 \: cm }$

Therefore,

⠀⠀⠀⠀⠀$\therefore {\underline{ \sf {  Hence\,:The\:Position \:of\:the\:object \:will\:be\:at\:\bf{-10\: cm}}}}$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀