Question

An object 2cm tall is kept 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 at 12cm from the lens. Also find the magnification
produced by the lens.

Answer 1

Answer:

m = -2 , I = -4cm

The image will be formed beyond C (2f) , It will be magnified.

         v = 24cm

Explanation:

 Given u = -12cm , O = 2cm , f = 8cm

              1/f = 1/v - 1/u

               1/8 = 1/v + 1/12

                  1/8 - 1/12 = 1/v

                          v = 24cm

                     m = v/u = -24/12 = -2

                       also, m = I/O = -2

                                       I = -2 x 2

                                          I = -4cm

                           

Answer 2

Answer:

Given : Height of the object (h) = 2 cm

Focal length (f) = +8 cm [ Converging lens = Convex lens]

Object distance (u) = - 12 cm

To find : Height of the image (h') = ?

Image distance (v) = ?

Solution : By lens formula,

$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\\ \\ \frac{1}{8} + \frac{1}{ - 12} = \frac{1}{v} \\ \\ \frac{3 - 2}{24} = \frac{1}{v} \\ \\ v = 24cm$Magnification,

m = h'/h = v/u

=> h'/2 = 24/-12

h' = -4cm.

Nature : Real and inverted (Because v is positive )

Size : Enlarged