Question

A shop security mirror $5.0\,{\text{m}}$ from certain items displayed in the shop produces one-tenth magnification.
A. What is the type of mirror?
B. What is the radius of curvature of the mirror?

Answer 1

Check which types of mirrors are used in security shops. Use the formula for the magnification of mirror and the mirror formula. These formulae give the relation between the magnification of the mirror, object distance, image distance and focal length. Then use the relation between the radius of curvature and focal length of the mirror.

Formula used: The magnification $m$ of a lens is given by
$m = - \dfrac{v}{u}$ …… (1)

Here, $v$ is the image distance and $u$ is the object distance.

The mirror equation is
$\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}$ …… (2)

Here, $f$ is the focal length, $v$ is the image distance and $u$ is the object distance.

Complete step by step answer:
A. The convex mirrors increase the field of vision and also the image size using these mirrors is less than the object size but using these mirrors a wide area can be seen.

Hence, the mirrors which are used in a security shop are convex mirrors.

B. The items are placed at a distance of $5.0\,{\text{m}}$ from the mirror and the magnification of the mirror is $\dfrac{1}{{10}}$.

Determine the image distance for the given mirror.

Rearrange equation (1) for image distance $v$.
$v = - mu$

Substitute $\dfrac{1}{{10}}$ for $m$ and $- 5.0\,{\text{m}}$ for $u$ in the above equation.
$v = - \left( {\dfrac{1}{{10}}} \right)\left( { - 5.0\,{\text{m}}} \right)$
$\Rightarrow v = 0.5\,{\text{m}}$

Hence, the image distance is $0.5\,{\text{m}}$.

Determine the focal length of the mirror.

Substitute $- 5.0\,{\text{m}}$ for $u$and $0.5\,{\text{m}}$ for $v$in equation (2).
$\dfrac{1}{f} = \dfrac{1}{{ - 5.0\,{\text{m}}}} + \dfrac{1}{{0.5\,{\text{m}}}}$
$\Rightarrow \dfrac{1}{f} = - 0.2 + 2$
$\Rightarrow \dfrac{1}{f} = 1.8$
$\Rightarrow f = \dfrac{1}{{1.8}}$
$\Rightarrow f = 0.55\,{\text{m}}$

Hence, the focal length of the mirror is $0.55\,{\text{m}}$.

The radius $R$ of curvature of the mirror is twice as that of the focal length.
$R = 2f$

Substitute $0.55\,{\text{m}}$ for $f$ in the above equation.
$R = 2\left( {0.55\,{\text{m}}} \right)$
$\Rightarrow R = 1.1\,{\text{m}}$

Hence, the radius of curvature of the mirror is $1.1\,{\text{m}}$.

Note:
The object distance is taken negative because the object is placed in front of the mirror and distances are taken negative in front of the mirror and positive behind the mirror.