Question

A real image, ${\left( {\dfrac{1}{5}} \right)^{{\text{th}}}}$ of the size of the object, is formed at a distance of $18{\text{cm}}$ from a mirror. Determine the nature of the mirror and calculate its focal length.

Answer 1

Hint: Two types of spherical mirrors exist – concave mirror and convex mirror. The convex mirror is a diverging mirror whereas the concave mirror is converging. The real image formed depends on the converging nature of the mirror. The focal length of the mirror can be obtained using the magnification of the mirror and the mirror equation.

Formulas used:
The focal length of a mirror is given by, $\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$ where $u$ is the object distance and $v$ is the image distance.
The magnification of the mirror is given by, $m = \dfrac{{ - v}}{u}$ where $u$ is the object distance and $v$ is the image distance.

Complete step by step answer:
Step 1: Based on the fact that the concave mirror is converging and the convex mirror is diverging in nature determine the nature of the given mirror.
As a concave mirror is converging, the rays incident on it will converge at a point upon reflection. The images formed by the concave mirror are thus real (or virtual but only if the object is placed very close to the mirror).
So the nature of the given mirror is concave.
Step 2: Using the relation for magnification of the mirror, obtain the object distance.
Here it is given that the image formed is real and the magnification is thus $m = - \dfrac{1}{5}$ .
By sign convention, the distance of the real image from the mirror is given to be $v = - 18{\text{cm}}$. Let $u$ be the object distance.
 The magnification of the mirror can be expressed as $m = \dfrac{{ - v}}{u}$
$ \Rightarrow u = \dfrac{{ - v}}{m}$ ------ (1)
Substituting for $m = - \dfrac{1}{5}$ and $v = - 18{\text{cm}}$ in equation (1) we get,
$\Rightarrow u = \dfrac{{ - \left( { - 18} \right)}}{{\left( {\dfrac{{ - 1}}{5}} \right)}} = - 90{\text{cm}}$
Thus the object distance is $u = - 90{\text{cm}}$ .
Step 3: Express the relation for the focal length of the mirror.
The focal length of the given mirror can be expressed as $\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$ ------- (2)
Substituting for $u = - 90{\text{cm}}$ and $v = - 18{\text{cm}}$ in equation (2) we get,
$\Rightarrow \dfrac{1}{f} = \dfrac{{ - 1}}{{18}} + \dfrac{{ - 1}}{{90}} = \dfrac{{ - \left( {5 + 1} \right)}}{{90}} = \dfrac{{ - 1}}{{15}}$
$ \Rightarrow f = - 15{\text{cm}}$

Therefore, we obtain the focal length of the concave mirror as $f = - 15{\text{cm}}$.

Note:
A convex mirror only forms virtual images. For a real image, magnification is negative so we have $m = - \dfrac{1}{5}$. The given image is said to be real. This suggests that the image is formed in front of the mirror. The object of a concave mirror is always placed on its left. By sign convention, all the distances to the left (in front) of the mirror are taken to be negative. So we have $u = - 90{\text{cm}}$ and $v = - 18{\text{cm}}$. The focal length of the concave mirror is negative and this matches our calculated result.