Question

A 4 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 24 cm. The distance of the object from the lens is 16 cm. Find the position, size and nature of the image formed, using the lens formula.

Answer 1

Height of object ho = 4 cm

Focal lenght of convex lens = 24 cm

Object distance u = -16 cm

Image distance v = ?

Image height hi = ?

Lens formula: 1/v  -  1/u  =  1/f

                         1/v  = 1/f  +  1/u

                                =  1/24  +  1/(-16)

                                = -1/48

⇒  v = - 48 cm

The image distance is 48 cm on the same side of the lens as the object.

Magnification m = v/u = (-48)/(-16) = 3 = hi/ho = hi/3

⇒   hi = 3 x 3 = 9 cm

The image is enlarged and erect

Answer 2

Answer:

Height of the object, h = +4cm

Focal length (convex lens) = +24cm

Object distance, u = -16cm

By lens formula: 1/v - 1/u = 1/f

=> 1/v = 1/f + 1/u

=> 1/v = 1/24 + 1/(-16)

=> 1/v = 1/24 - 1/16

=> 1/v = -1/48 (taking LCM)

So, V = -48 cm

Magnification, m = v/u

=> m = -48/-16

=> m = +3

Now, m = h’/h (where, h’ is the height of the image)

=> 3 = h’/4

=> h’ = +12 cm

Thus, the image is enlarged and erect