An object is placed at a distance of 6 cm from a convex mirror of focal length 12 cm. Find the position and nature of the image.

Distance of the object from the mirror, $u$ = $-$6 cm

Focal length of the mirror, $f$ = 12 cm 

To find: Distance or position of the image, $v$, and the nature of the image.

Solution:

From the mirror formula, we know that-

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

Substituting the given values we get-

$\frac {1}{12}=\frac {1}{v}+\frac {1}{(-6)}$

$\frac {1}{12}=\frac {1}{v}-\frac {1}{6}$

$\frac {1}{12}+\frac {1}{6}=\frac {1}{v}$

$\frac {1}{v}=\frac {1+2}{12}$

$\frac {1}{v}=\frac {3}{12}$

$\frac {1}{v}=\frac {1}{4}$

$v=+4cm$

Thus, the distance of the image $v$ is 4 cm from the mirror, and the positive sign implies that the image forms behind the mirror (on the right).

Now, from the magnification formula, we know that-

$m=-\frac {v}{u}$

Substituting the given values we get-

$m=-\frac {4}{(-6)}$

$m=\frac {4}{6}$

$m=\frac {2}{3}$

$m=+0.6$

Thus, the magnification, $m$ is 0.6 which is less than 1, which means the image is small in size, and the positive sign implies that the image is virtual and erect.

Hence, the image is virtual, erect and small in size.