Question

A concave mirror forms a real image of height $2cm$ of an object $0.5cm$ high placed $10cm$ away from the mirror. Find the position of the image and the focal length of the mirror.

Answer 1

Hint: To solve this question, we need to use the formula for the magnification in the case of mirrors. There are two formulas of the magnification, one is in terms of heights of the object and the image, and the second is in terms of the object and the image distance.

Formula Used
The formula used to solve this question is
$\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$, $f$ is the focal length of a mirror, $v$ is the image distance and $u$ is the object distance.
$m = - \dfrac{v}{u}$
$m = \dfrac{{h'}}{h}$
Where, $m$ is the magnification, $v$ is the image distance, $u$ is the object distance, $h'$ is the height of the image, and $h$ is the height of the object.

Complete step-by-step solution
We know that the magnification produced by an optical instrument is given by
$m = \dfrac{{h'}}{h}$
According to the question, $h = 0.5cm$ and $h' = 2cm$
So, we get the magnification as
$m = \dfrac{2}{{0.5}}$
$m = 8$ (1)
Also, we know that the magnification formula for a mirror is given by
$m = - \dfrac{v}{u}$ (2)
According to the question, the object is placed $10cm$ away from the mirror.
So, $u = - 10cm$ (3)
Substituting (1) and (3) in (2), we get
$8 = - \dfrac{v}{{ - 10}}$
Multiplying by $10$ on both the sides
$v = 80cm$
Hence, the image is positioned $80cm$ away from the mirror, at the far side of the mirror.
Now, from the mirror equation, we have
$\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$
Substituting the values of $u$ and $v$ from above, we get
$\dfrac{1}{f} = \dfrac{1}{{80}} + \dfrac{1}{{ - 10}}$
Taking the LCM
$\dfrac{1}{f} = \dfrac{{1 - 8}}{{80}}$
$\dfrac{1}{f} = - \dfrac{7}{{80}}$
Taking reciprocal, we get
$f = - \dfrac{{80}}{7}cm$
$f = - 11.43cm$

Hence, the focal length of the concave mirror is $11.43cm$

Note: While solving the questions related to mirrors, take proper care of the Cartesian sign conventions. The signs of the image distance and the object distance are very important. Also, for a concave mirror, check that the sign of focal length must be negative, according to the sign convention. If it is not so, then there must be some error in the calculation.