An object is kept at a distance of 5 cm in front of a convex mirror of focal length 10 cm. Calculate the position and magnification of the image and state its nature.

Convex Mirror

Distance of the object, $u$ = $-$5 cm

Focal length of the mirror, $f$ = 10 cm

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

Solution:

From the mirror formula, we know that-

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

Substituting the given values we get-

$\frac {1}{10}=\frac {1}{v}+\frac {1}{(-5)}$

$\frac {1}{10}=\frac {1}{v}-\frac {1}{5}$

$\frac {1}{10}+\frac {1}{5}=\frac {1}{v}$

$\frac {1}{v}=\frac {1+2}{10}$

$\frac {1}{v}=\frac {3}{10}$

$v=\frac {10}{3}$

$v=+3.3cm$

Thus, the distance of the image $v$ is 3.3 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-

$m=-\frac {3.3}{(-5)}$

$m=-\frac {3.3}{(-5)}$

$m=\frac {3.3}{5}$

$m=+0.66$

Thus, the magnification is $0.66$ 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.