Question

An object of height 2·5 cm is placed 12 cm in front of a lens. An image is produced 4 cm behind the lens. Calculate:


a) the focal length of the lens


b) the magnification of the lens


c) the height of the image.

Answer 1

Given :

Size of the object, h = 2.5 cm

Object distance, u = - 12 cm

Image distance, v = 4 cm

Solution :

a. Focal length of the lens = f

By using the formula, $\sf \dfrac{1}{f} = \sf \dfrac {1}{v} - \sf \dfrac{1}{u}$

• $\sf \dfrac{1}{f} = \sf \dfrac {1}{4} - \sf \dfrac{1}{(-12)}$

• $\sf \dfrac{1}{f} = \sf \dfrac{3+1}{12}$

• $\sf \dfrac{1}{f} = \sf \dfrac{4}{12}$

• $\sf \dfrac{1}{f} = \sf \dfrac{1}{3}$

• $\sf f = 3 \; cm$

b. Magnification of the lens = m

• $\sf m = \dfrac{v}{u}$

• $\sf m = \dfrac{4}{(-12)}$

• $\sf m = \dfrac{1}{(-3)}$

• $\sf m = -0.33$

c. Height of the image = h'

• $\sf \dfrac {h'}{h} = \dfrac{v}{u}$

• $\sf h' = \dfrac{v × h}{u}$

• $\sf h' = \dfrac{4 × 2.5}{(-12)}$

• $\sf h' = \dfrac{10}{(-12)}$

• $\sf h' = -0.83 \; cm$