#### Question

An object 2 cm tall stands on the principal axis of a converging lens of focal length 8 cm. Find the position, nature and size of the image formed if the object is:

(i) 12 cm from the lens

(ii) 6 cm from the lens

State one practical application each of the use of such a lens with the object in position (i) and (ii).

#### Solution

Converging lens is a convex lens

Given:

Focal length (f) = +8 cm

Height of the object (h) = +2

(a)

(i) Object distance (u) =-2

Lens formula is given as:

`1/f=1/v-1/u`

`1/8=1/v-1/-12`

`1/8=1/v+1/12`

`1/v=1/8-1/12`

`1/v=(3-2)/24`

`1/v=1/24`

`v=24` cm

Image is at a distance of 24 cm from the convex lens; therefore, we have:

Magnification =`v/u`

Magnification (*m*) =`24/-12`

*m* =-2

Hence, the image is real and inverted.

(b) Object distance (u)=-6

According to lens formula:

`1/f=1/v-1/u`

`1/8=1/v-1/-6`

`1/8=1/v+1/6`

`1/8-1/6=1/v`

`(3-4)/24=1/v`

`-1/24=1/v`

`v=-24` cm

Image is at a distance of 24 cm in front of the lens; therefore , we have :

Magnification` (m) =v/u `

`m=(-24)/-6`

m=4

`m=h_i/h_o`

`m=h_i/2`

`h_i=2xx4`

`h_i=8` cm

Height of the image is 8 cm.

Here, height is positive; therefore, image is virtual and erect.

The practical application for case (1) is that it can be used as a corrective lens for a farsighted person and for case (2), it can be used as a magnifying lens for reading purposes.