Question

An object 4cm in size is placed at 25cm in front of a concave mirror of focal length 15cm .find the position,nature and size of the image formed.

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Answer 1

Object distance from the mirror is 25 cm, height of the object is 4 cm and the focal length of mirror is 15 cm.

Using sign convention:

u = -25 cm, h = 4 cm and f = -15 cm

Using the mirror formula

1/f = 1/v + 1/u

Substitute the known values in the above formula,

1/(-15) = 1/v + 1/(-25)

-1/15 = 1/v - 1/25

1/v = 1/25 - 1/15

1/v = 1/5(1/5 - 1/3)

1/v = 1/5(-2/15)

v = -75/2

v = -37.5 cm

(Negative sign mean image formed is on the left side of the mirror.)

Therefore, the image distance from the mirror is 37.5 cm.

Using Magnification formula:

m = -v/u = h'/h

Substitute the values,

-(-37.5)/(-25) = h'/4

-1.5 = h'/4

h' = -1.5*4

h' = -6 cm

(Height of the image is bigger then that of the object height.)

Therefore, the height of the image is 6 cm.

Nature of the image:

Image formed is real, inverted and larger in size.

Answer 2

Answer

position:- between pole and f  and -2/75 or-37.5
nature:- virtual, erect  
sige of the image:-  -6cm

Explanation:

Given that f = -15 cm; u = -25 cm ; ho= 4 cm; v = ?

The mirror formula is 1/f = 1/u + 1/v = 1/-15 = 1/-25 + 1/v

1/v = 1/25 - 1/15

1/v = (3-5)/ 75 =-2/ 75

v =-37.5 cm

The screen should be placed 37.5 cm from the pole of mirror and the image is real.

Magnification, m = -v/u =hi/ho= hi / 4 = 37.5 / 25

hi = (37.5 / 25 )x 4 = -6 cm

So the image is enlarged and inverted.