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# A shop security mirror 5.0 m from certain items displayed in the shop produces on-tenth magnification.**(a)** What is the type of mirror?**(b) **What is the radius of curvature of the mirror?

**(a)** A **convex mirror** is used for shop security purposes, as it gives a wider field of view, and always forms virtual, erect, and small-sized images of a large number of objects at the same time.

**(b)** Given:

Convex Mirror

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

Magnification, $m$ = $\frac {1}{10}$

To find: Radius of curvature, $R$.

Solution:

From the magnification formula, we know that-

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

Substituting the given values we get-

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

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

$10v=5$

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

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

$v=+0.5m$

Thus, the distance of the image $v$ is 0.5 m from the mirror, and the positive sign implies that the image forms behind the mirror (on the right).

Now, using the mirror formula, we get-

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

Substituting the given values we get-

$\frac {1}{f}=-\frac {1}{5}+\frac {1}{0.5}$

$\frac {1}{f}=-\frac {1}{5}+\frac {10}{5}$

$\frac {1}{f}=\frac {-1+10}{5}$

$\frac {1}{f}=\frac {9}{5}$

$f=\frac {5}{9}m$

Thus, the focal length of the mirror $f$ is $\frac {5}{9}$.

Now,

We know that radius of curvature, $R$ is equal to twice the focal length $f$. It is given as-

$R=2f$

Putting the value of $f$, we get-

$R=2\times\frac {5}{9}$

$R=\frac {10}{9}$

$R=+1.1m$

Thus, the radius of curvature, $R$ of the mirror is **1.1 m.**