Question

1. 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

magnification , m = -1
h1 = size of object
h2 = size of image

m = -1 => h1 / h2 = -1 => h1 = - h2
=> size of object = size of image
=> negative sign shows image is real and inverted...

Plane mirror can form image of same size but it always form virtual image so magnification produced by plane mirror is always positive and equal to 1...Convex mirror never give negative magnification as it always form virtual and erect image...

So the mirror is discussed above is a Concave mirror..and we know that Concave mirror can give magnification negative as well as positive according to object position placed infront of the mirror...

As m = -1 => this means that size of object is same as size of object...This happens only if object is placed at the Center of curvature ( C )...

Note : when an object is placed at C , then
Size of object = Size of image
Object distance = Image distance

As image distance = 30 cm then object distance = 30cm also...

For ray diagram see snap...

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.