Question

A lamp is placed 6.0m from a wall. On putting a lens between the lamp and the wall at a
distance of 1.2m from the lamp, a real image of the lamp is formed on the wall. The
magnification of the image is?​

Answer 1

★ Solution :-

• Distance of the lamp from the wall = 6 m

Lens is placed between the lamp and the wall

• Distance of the lens from the lamp = object distance = - 1.2 m

Object distance is negative as the distance is measured from the optical centre of the lens

• A real and inverted image of the lamp is formed on the wall

The distance between the image and the lens

= 6 - 1.2

= 4 .8 m

• The distance between the image and the lens = image distance = 4.8 m

Let the image distance be v and the object distance be u

Magnification of a lens is given by the formula ,

$\dashrightarrow\mathtt{m = \dfrac{v}{u}}$

$\dashrightarrow\mathtt{m = \dfrac{4.8}{ - 1.2}}$

$\dashrightarrow\mathtt{m = - 4}$

As the image is real and inverted image will be formed below the principal axis ,hence the magnification value is negative

The magnification of the image is 4

Answer 2

GiveN :-

• The distance of lamp from the wall = 6 m

★ From the Question it clearly states that The lens is placed between the lamp and the wall

• The distance where lens is kept = 1.2m

★ A real image of lamp is formed on tge walls

Since the image formed is real object distance is also - ve.

To FinD :-

• The Magnification produced.

SoluTioN :-

Distance between image and lens

• 6 - 1.2 = 4.8 m

★ Since image distance = 4.8m

★ Object distance = 1.2m

★ Magnification produced

$\sf{m = \frac{v}{u} }{} \\ \\ \sf{m = \frac{4.8}{ - 1.2} }{} \\ \\ \bf{ \red{★Magnification = - 4}{} }{}$

★ Since Image formed is Real which is negative the Magnification is also negative due to this reason.