Question

an object iof height 1.2m is placed before a concave mirror of focal length 20cm so that real image is formed at a distance of 60cm from it

Answer 1

heigh of object =1.2m
=120cm
focal length = -20cm
radius of curvature = 2×20=40cm

image distance = -60cm(since , real)

by magnification formula
1/v + 1/u = 1/f
1/-60 + 1/u = 1/-20
1/u = 1/-20 + 1/60

by solving this equation, you will get,
1/u = -1/30
u, object distance = -30cm , object is placed between focus and centre and image will form beyond centre of curvature.

by magnification,

m = -v/ u
m = -(-60)/-30
m , magnification = -2

therefore, size of image is larger than object. Lets check

m = height of image/ height of object
-2 = height of image/ 1.2m
height of image = 2.4m (larger than object)

Answer 2

Answer:

u = 30 cm

$\display \text{h _i = 2.4 cm}$

Explanation:

We have given :

Focal length = 20 cm

Image distance =  60 cm

We have to find object distance .

We know :

1 / f = 1 / v + 1 / u

Since image formed is real :

f = - 20 cm and v = - 60 cm

1 / u = 1 - 3 / 60

u = - 30 cm

Hence object distance is 30 cm .

Also given object height = 1.2 cm

We know :

h_i / h_o = - v / u

h_i / 1.2 = - 60 / 20

h_i = - 2.4 cm

Hence image height is 2.4 cm .