An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed so that a sharp focussed image can be obtained? Find the size and nature of image.

#### Solution

Distance of the object from the mirror '*u*' = −27 cm

Height of the object '*h _{o}*' = 7 cm

Focal length of the mirror '

*f*' = −18 cm

We have to find the distance of the image '

*v*' and height of the image '

*h*'.

_{i}Using the mirror formula, we get

`1/f=1/v+1/u`

⇒ `1/-18=1/v-1/27`

⇒ `1/-18=1/v-1/27`

⇒ `1/27-1/18=1/v`

⇒ `(-1)/54=1/v`

⇒ v=-54 cm

Thus, the distance of the image '*v*' is 54 cm.

Now, using the magnification formula, we get

`m=h_i/h_o=-v/u`

`h_i/i=-(-54)/(-27)=-2/1`

`h_i=-14 cm`

Therefore, the height of the image '*h _{i}*' is 14 cm.

Again, using the magnification formula, we get

`m=-v/u=-((-54))/((-27))`

m=-2

Thus, the image is real, inverted and large in size.