A man holds a spherical shaving mirror of radius of curvature 60 cm, and focal length 30 cm, at a distance of 15 cm, from his nose. Find the position of image, and calculate the magnification.

Distance of the object from the mirror $u$ = $-$15 cm

Focal length of the mirror, $f$ = $-$30 cm

To find: Distance of the image, $(v)$ from the mirror.

Solution:

From the mirror formula, we know that-

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

Substituting the given values in the mirror formula we get-

$\frac{1}{(-30)}=\frac{1}{v}+\frac{1}{(-15)}$

$-\frac{1}{30}=\frac{1}{v}-\frac{1}{15}$

$\frac{1}{15}-\frac{1}{30}=\frac{1}{v}$

$\frac{1}{v}=\frac{2-1}{30}$

$\frac{1}{v}=\frac{1}{30}$

$v=30cm$

Thus, the position or distance of the image is 30 cm from the mirror, and the positive sign implies that the image is formed behind the mirror (on the right side).

Now, from the magnification formula, we know that-

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

Substituting the given values in the magnification formula we get-

$m=-\frac{30}{(-15)}$

$m=\frac{30}{15}$

$m=2$

Thus, the magnification $m$ of the mirror is 2 cm, and the positive sign implies that the image is virtual and erect.

Hence, the position of the image is 30 cm behind the mirror, the image is 2 times magnified than the object.