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Answered

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

Sagot :

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

We know

[tex]\boxed{\sf m=-\dfrac{v}{u}}[/tex]

[tex]\\ \sf\longmapsto 3=-\dfrac{-40}{u}[/tex]

[tex]\\ \sf\longmapsto 3=\dfrac{40}{u}[/tex]

[tex]\\ \sf\longmapsto u=\dfrac{40}{3}[/tex]

[tex]\\ \sf\longmapsto u=13.3cm[/tex]

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