Question

An object of height 1.2 m height is placed 30cm before a concave mirror of focal length 20cm so that a real image is formed distance 60 cm from it find the position of an object what will be the height of the image formed

Answer 1

Object is always place to the left side of a mirror therefore the object distance (u) is always negative.

If the image is formed in front of the mirror(real image) on the left side then the image distance (v) will be negative.

The focus of concave mirror is in front of the mirror on the left side so the focal length of a concave mirror is considered negative.

Given:

Focal length (f)= -20 cm ( concave mirror)

Image distance (v)= -60cm (real image)

Height of the object (h1)= 1.2cm

Mirror Formula

1/v + 1/u = 1/f

(-1/60)+(1/u) = (-1/20)

1/u= -1/20 + 1/60

1/u = (-3+1)/60= -2/60= -1/30

1/u= -1/30

u = -30

Object distance(u) = -30 cm

Magnification (m) = -v/u

m = - (-60) /-30= 60/-30= -2

m = h2/h1

-2= h2/1.2

-2×1.2= h2

h2= -2.4 cm

Height of real image (h2)= -2.4 cm

Real image is formed below the axis. So the height of a real image is negative.

Hence, the position of an object is -30 cm(to the left of mirror) & the height of the real image formed is -2.4 cm

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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 .