Question

the image of an object formed by a mirror is real inverted and is of magnification -1 if the image is at a distance 30 from the mirror where is the object placed find the position of the image if the object is not move 20 CM towards the mirror what is the nature of the image​

Answer 1

Magnification is negative so the image is virtual and erect .

Also Magnification is 1 so, this means that the image is of the same size as the object.

This can only happen if the object is placed at the centre of curvature.

So u = v

u= v = 30cm

u = -30cm ,

Now we know that u = radius of curvature = -30cm,

So f = R/2 = -30/2 = -15cm

If object is moved 20 cm towards the mirror then the object distance ,u = -10cm

By mirror formula,

1/f = 1/u + 1/v

1/f - 1/u = 1/v

-1/15 - (-1)/10= 1/v

-1/15 + 1/10 = 1/v

-2 + 3/30 = 1/v, 1/30 = 1/v

v = 30 cm

m = -v/u = -30/-10 = 3

Since magnification is positive we can tell that the image is virtual and erect and magnification is greater than 1 so image is enlarged.

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.