Question

A mirror forms an erect image 40cm from the object and one third its height where must the mirror be situated ​

Answer 1

• image distance=-40cm=v
• Object distance=u=?
• Magnification=m=3

We know

$\boxed{\sf m=-\dfrac{v}{u}}$

$\\ \sf\longmapsto 3=-\dfrac{-40}{u}$

$\\ \sf\longmapsto 3=\dfrac{40}{u}$

$\\ \sf\longmapsto u=\dfrac{40}{3}$

$\\ \sf\longmapsto u=13.3cm$