Question

An object of height 1.2m is placed before a concave mirror of focal length 20cm so that a real image is formed at a distance of 60cm from it. Find the position of an object. What will be the height of the image formed? (2)

Answer 1

Hey mate..
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Given,

Height of the object, h = 1.2 m

Focal length of a concave mirror, f = -20 cm

Object Distance, u = -60 cm

Image Distance, v = ?

Height of the image , h' = ?

With the help of mirror formula we have,

$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \\ \\ = > \frac{1}{f } - \frac{1}{u} = \frac{1}{v} \\ \\ = > \frac{u - f}{fu} = \frac{1}{v} \\ \\ = > \frac{fu}{u - f} = v \\ \\ > \frac{( - 60)( - 20)}{ - 60 - ( - 20)} = v \\ \\ = > \frac{1200}{ - 40} = v \\ \\ = > v = - 30cm$

Thus,

The image distance is 30 cm on the same side of the object. The image is real.

Also,

Magnification, m = h' / h = -v / u

=> h' = -v (h) / u

= - (-30)(120) / (-60)

= 3600 / -60

= -60 cm

Or , -0.6 m

The image size is 0.6 m and it is laterally inverted and half of the size of the object.

Hope it helps !!

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 .