Question

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

Answer 1

Answer:

hiii

your answer is here !

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.

follow me !

Answer 2

Given

Mirror is diverging

Hight of image = 2.5cm

focal length = 20cm

object distance = - 25cm

we know that

1/f = 1/v + 1/u

1/v = 1/f + 1/u

v = uf/(u+f)

v = 50/22.5= 11.11cm

we know that

m = -v/u = ho/hi

= > -11.11/25= ho/hi

=> hi = 11.11*2.5/25 = 1.11cm

nature

erect

11.11cm form behind the mirror .