Question

ans pls fast with step by step explaination​

Answer 1

Given :-

Height of Object = h = 4 cm

Object distance = u = -16 cm (Sign Convention)

Focal Length = f = 24 cm

Let height of Image = h'

From Mirror Formula,

1/v + 1/u = 1/f

1/v - 1/16 = 1/24

1/v = 1/24 + 1/16

1/v = 5/48

v = 48/5 cm.

v = 9.6 cm

m = h/h' = -v/u

4/h' = 9.6/16

h' = 64/9.6

h' = 6.6 cm

Hence,

Image distance = v = 9.6 cm

Height of Image = h' = 6.6 cm

Nature of Image = Virtual, Erect and larger than object.