Question

the image of an object formed by a lens is of magnification -1 if the distance is 30cm from the mirror ,where is the object placed?find the position of image of the image if the object is now moved 20cm toword the mirror .what is the nature of the image obtained ? Justify your answer with the help of ray diagram

Answer 1

magnification (m) = -1
therefore -v/u = -1
therefore v=u
let u = x 
then 2x = 60 cm (from question)
x= 30 cm 
therefore u and v = 30cm
If object is moved 20 cm towards the lens then v will also move 20 cm but away from the lens.

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.