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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.