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# 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.

Given:

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.