Question

A $2cm$ high object is placed at a distance of $32cm$ from a concave mirror. The image is real, inverted and $3cm$ in size. Find the focal length of the mirror and the position where the image is formed.

Answer 1

This problem can be solved by using the formula for the magnification produced by the mirror in terms of the height of the image and the object. From the value of the magnification, the image distance can be found which when input in the mirror formula, the focal length can be found.
Formula used:
$m=\dfrac{{{h}_{i}}}{{{h}_{o}}}$
$m=-\dfrac{v}{u}$
$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$

Complete step by step answer:
We will find out the magnification and then from the magnification find out the image distance. By using the image distance in the mirror formula, we will find out the focal length of the mirror. Hence, let us proceed to do that.
The magnification $m$ produced by a mirror is given by
$m=\dfrac{{{h}_{i}}}{{{h}_{o}}}$ --(1)
Where ${{h}_{i}},{{h}_{o}}$ are the image and object sizes respectively.
The magnification $m$ produced by a mirror can also be written as
$m=-\dfrac{v}{u}$ ---(2)
Where $v,u$ are the image distance and object distance respectively.
The mirror formula relates the object distance $u$, image distance $v$ and focal length $f$ of a mirror by
$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$ --(3)
Now, let us analyze the question.
The height of the object is ${{h}_{o}}=2cm$.
The height of the image is ${{h}_{i}}=-3cm$ (Since the image is real and inverted, according to sign convention, the height is negative).
The object distance is $u=-32cm$ (According to sign convention, the object distance is negative).
Let the magnification produced by the mirror be $m$.
Let the image distance be $v$ and the focal length of the mirror be $f$.
Therefore, using (1), we get the magnification as
$m=\dfrac{-3}{2}=-1.5$ --(4)
Also, using (2) the magnification can be written as
$m=-\dfrac{v}{-32}$
Putting (4) in the above equation, we get
$-1.5=-\dfrac{v}{-32}$
$\therefore -1.5\times 32=v$
$\therefore v=-48cm$ --(5)
Now, according to the mirror formula, that is, (3), we get
$\dfrac{1}{v}+\dfrac{1}{-32}=\dfrac{1}{f}$
Putting (5) in the above equation, we get
$\dfrac{1}{-48}+\dfrac{1}{-32}=\dfrac{1}{f}$
$\therefore \dfrac{-2-3}{96}=\dfrac{1}{f}$
$\therefore \dfrac{1}{f}=\dfrac{-5}{96}$
$\therefore f=\dfrac{-96}{5}=-19.2cm$
Therefore, the image distance is $v=-48cm$ and the focal length is $f=-19.2cm$.
The image is formed at $48cm$ away from the mirror on the same side as that of the object.

Note:
Students must be careful and properly apply the sign conventions when solving optics problems because the signs of the values hold a lot of significance. For example, the negative sign of the image distance tells us that the image is real, inverted and is formed on the same side of the mirror as that of the object. The negative value of the magnification obtained is also indicative of the fact that the image is inverted with respect to the object. If we had not applied the sign conventions properly in the calculations, we would have arrived at a completely wrong answer.