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 'ho' = 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 'hi'.
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 'hi' 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.
