Question

A converging mirror has a focal length of 20 cm. Where should an object be placed so that its
virtual image will be twice as tall as the object?

Answer 1

GiveN :

• Focal length (F) = - 20 cm
• Concave Mirror
• Magnification (m) = 2

To FinD :

• Object Position

SolutioN :

As we know that formula of magnification is :

$\implies \sf{m = \dfrac{-v}{u}} \\ \\ \implies \rm{2 = \dfrac{- v}{u}} \\ \\ \implies \rm{-v = 2u} \\ \\ \implies \sf{v = -2u \: \: \: \: \: \: ...(1)}$

___________________

Now, use mirror formula :

$\implies \sf{\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}} \\ \\ \\ \implies \rm{\dfrac{1}{f} = \dfrac{1}{-2u} + \dfrac{1}{u}} \\ \\ \\ \implies \rm{\dfrac{1}{-20} = \dfrac{-1 + 2}{2u}} \\ \\ \\ \implies \rm{\dfrac{-1}{20} = \dfrac{1}{2u}} \\ \\ \\ \implies \rm{u = \dfrac{-20}{2}} \\ \\ \\ \implies \rm{u = -10} \\ \\ \\ {\Large{\implies{\boxed{ \sf{u = -10 cm}}}}}$

Answer 2

f = +20 cm

let (height of object) = h

so, m = 2h ( given that 2times magnified

real image )

m = 2h/h = 2

Again,

m = v/u

2 = v/u

》 v =2u

Now, 1/f = 1/v - 1/u

1/20 = 1/2u - 1/u

1/20 = 1-2/2u

1/20 = -1/2u

》 u = -10cm ( object distance )

$\rule{200}{2}$

Refer attachment for diagram