Question

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

Answer 1

Ho = 1.2m.

F = -20cm. [sign convention].

V = 60.

By using mirror formula,

1/U + 1/V = 1/F.

1/U = 1/F-1/V.

1/U = 1/(-20)-(1/60).

1/U = 1/20+60.

1/U = 3+1/60 = 4/60 = 1/15.

1/U = 1/15.

U = -15.

M = Hi/Ho= -V/U [ where Hi is height of image and Ho is height of object].

Hi/Ho= -V/U.

Hi/1.2= -60/-15.

Hi = 60×1.2/15 = 4×1.2 = 4.8cm

HENCE,
POSITION OF OBJECT IS -15 AND THE HEIGHT OF IMAGE IS 4.8CM

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 .