Question

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

Answer 1

ANSWERS :-


Object size , h1 = + 2.5 cm,

object distance, u = –25 cm

focal length of diverging mirror, f = +20cm

image distance, v = ?

image size , h2 = ?


using mirror formula,

1/u + 1/v = 1/f

or,

1/v = 1/f – 1/u

and putting,

u = -25cm and f = +20cm , we get


1/v = 1/20 – (1/-25)

= 5 + 4 /100

= 9/100

or ,
v = 100/9 = 11.1 cm

⚡⚡⚡since image is at the back of the mirror at 11.1 cm from the mirror, it must be virtual and erect.

if h2 age the size of image, then as

m = h2 /h1 = –v/u

h2/2.5 = –11.1/–25

or,

h2 = 2.5 * 11.1 /25

h2 = 1.11 cm

This is size of the image.

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❤BE BRAINLY ❤

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Answer 2

Answer:

Explanation:

Nature:virtual and erect

Size:10cm

Position:-100cm in the same side as the object