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# A concave mirror produces three times enlarged virtual image of an object placed at 10 cm in front of it. Calculate the radius of curvature of the mirror.

Given:

Distance of the object from the mirror $u$ = $-$10 cm

Magnification, $m$ = 3

To find: Focal length $(f)$, and the radius of curvature $(R)$.

Solution:

From the magnification formula, we know that-

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

Substituting the given values in the magnification formula we get-

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

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

$v=10\times{3}$

$v=+30cm$

Thus, the distance of the image, $v$ is 30 cm from the mirror, and the positive sign implies that the image forms behind the mirror (on the right).

Now, 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-

$\frac{1}{f}=\frac{1}{30}+\frac{1}{(-10)}$

$\frac{1}{f}=\frac{1}{30}-\frac{1}{10}$

$\frac{1}{f}=\frac{1-3}{30}$

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

$\frac{1}{f}=\frac{-1}{15}$

$f=-15cm$

Thus, the focal length, $f$ of the mirror is **15 cm.**

We know that the radius of curvature, $R$ is equal to twice the focal length $f$ of the mirror.

$R=2f$

$R=2\times {(-15)}$

$R=-30cm$

Hence, the radius of curvature of the mirror is **30 cm.**