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). 

Answer 1

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.