An object 4 cm high is placed at a distance of 6 cm. In front of a concave mirror of a focal length of 12 cm. Find the height of the image formed.

An object

4 cm

high is placed at a distance of

6 cm

. In front of a concave mirror of a focal length of

12 cm

. The height of the image formed is

8 cm

.
From sign conventions:

focal length f = - 12 cm,

[Negative: as the mirror is concave]

object distance  u = - 6 cm,

Height of the object,

h_{o} = 4 cm

From the mirror formula,

\frac{1}{v} + \frac{1}{u} = \frac{1}{f},

\Rightarrow \frac{1}{v} = - \frac{1}{u} + \frac{1}{f}

\Rightarrow \frac{1}{v} = - \frac{1}{- 6} + \frac{1}{- 12}

\Rightarrow \frac{1}{v} = \frac{+ 2 - 1}{12}

\Rightarrow v = 12 cm

Now, from the magnification formula for mirrors,

m = - \frac{v}{u}

\Rightarrow m = - \frac{12}{- 6}

\Rightarrow m = + 2

Also,

m = - \frac{height of the  image}{height of the object}

2 = \frac{height of the image}{4}

height of the image = 8 cm

An object $4 cm$ high is placed at a distance of $6 cm$ . In front of a concave mirror of a focal length of $12 cm$ . Find the height of the image formed.