Question

an arrow 2.5 cm high is placed at a distance of 25 cm from a diverging mirror of focal length 20 centimeter. find the nature, position and size of the image formed

Answer 1

h1 = +2.5cm
u = -25cm
f = +20cm

1/f = 1/v + 1/u
1/20 = 1/v -1/25
1/v = 1/20 + 1/25
1/v = 9/100
v = +11.11 cm

m = h2/h1 = -v/u
h2/2.5 = -11.11/-25
h2 = +1.11 cm

Hence the image will be formed at 11.11 cm behind the mirror, which will be virtual, erect and of 1.11cm size.

Answer 2

Answer:v=11.11cm

Height of image =1.11cm

Nature = virtual and erect

Explanation:

h1 = +2.5cm

u = -25cm

f = +20cm

1/f = 1/v + 1/u

1/20 = 1/v -1/25

1/v = 1/20 + 1/25

1/v = 9/100

v = +11.11 cm

m = h2/h1 = -v/u

h2/2.5 = -11.11/-25

h2 = +1.11 cm

Hence the image will be formed at 11.11 cm behind the mirror, which will be virtual, erect and of 1.11cm size.