Question

the image of an object formed by a mirror is real, inverted and is of magnification -1. If the image is at the distance of 30 cm from the mirror, where is the object placed? find the position of the image if the object is now moved 20 cm towards the mirror. what is the nature of the image obtained? justify your answer with the help of Ray diagram.

Answer 1

Given :

m= -1

That means the object is placed at C and image is formed at C

Nature of image :

Real Inverted and same size.

a) If V=30cm

then u=?

m= -v/u

u=-v/m

= -(-30)/ -1 = -30 cm due to sign conventions

b)Object is now moved 20cm towards the mirror then u=10 cm [ object between p and F]

nature of image

Virtual

erect

magnified

Answer 2

Answer:

+ 30 cm

Explanation:

Given :

Magnification = - 1

Image distance = 30 cm

We know

m = - v / u

- 1 = 30 / u

u = - 30 cm

Now we have ,

1 / f = 1 / v + 1 / u

1 /  f = 1 / - 30 - 1 / 30

f = - 15 cm

When object moved 20 cm

u' = - 30 + 20 = - 10 cm

We have f = - 15 cm

We have to find new image distance v'

Again ,

1 / f = 1 / v' + 1 / u'

- 1 / 15 = 1 / v' - 1 / 10

1 / v' = 1 / 30

v' = 30.

Nature of image as :

Virtual

Erect

Magnified.