Question

the image of an object formed by a mirror is real inverted and is of magnification -1 if the images is the distance of 30 cm from the mirror where is the object placed find the position of the image of the object is no more 20cm towards the mirror what is the nature of image obtained justify your answer with the help of ray diagram

Answer 1

magnification =-1

image distance= -30

m=-v/u

-1=30/u

=>u= -30cm

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.