Question

Find the size, nature and position of the image formed when an object of size $1cm$ is placed at a distance of $15cm$ from a concave mirror of focal length $10cm$ .

Answer 1

Hint: Use the mirror equation to derive the position of image. Use the magnification of a mirror formula to obtain the size of the object. Accordingly depict its nature using the sign of the image height.


Formula Used:
The mirror equation
$\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$
Where $v$ is the image distance,$u$ is the object distance and $f$ is the focal length.
Also the magnification of mirrors is given by
$m = \dfrac{{ - v}}{u} = \dfrac{{{h_i}}}{{{h_o}}}$
Where $v$ is the image distance,$u$ is the object distance and $f$ is the focal length, $m$ is the magnification, ${h_i}$ is the height of the image and ${h_o}$ is the height of object.


Complete step by step answer:
Given,
 $u = - 15cm$ {Using Sign Convention}
$f = - 10cm$ {Using sign convention since focal length of concave mirror is negative}
${h_o} = 1cm$ We can calculate $v$ by using the following equation,
$\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$
Substituting the values of $u$ and $f$ in the above equation we get,
$\dfrac{1}{v} + \dfrac{1}{{ - 15}} = \dfrac{1}{{ - 10}}$
$ \Rightarrow \dfrac{1}{v} = \dfrac{1}{{ - 10}} + \dfrac{1}{{15}}$
Taking L.C.M,
$ \Rightarrow \dfrac{1}{v} = \dfrac{{2 - 3}}{{30}}$
$ \Rightarrow \dfrac{1}{v} = \dfrac{{ - 1}}{{30}}$
Taking the reciprocal, we get
$ \Rightarrow v = - 30cm$
Thus the image is formed $30cm$ in front of the mirror.
Now using the equation,
$m = \dfrac{{ - v}}{u} = \dfrac{{{h_i}}}{{{h_o}}}$
We substitute the values here,
$m = \dfrac{{ - ( - 30)}}{{ - 15}} = \dfrac{{{h_i}}}{{1cm}}$
On solving we get,
$m = - 2 = \dfrac{{{h_i}}}{{1cm}}$
Thus the image is twice the size of object.
On solving further we get that,
${h_o} = - 2cm$
The negative sign indicates that the image is virtual and inverted. This is also because the image is formed at the same side of the mirror with the object.
Hence finally, $v = - 30cm$ , $\left| {{h_o}} \right| = 2cm$ and the image is real and inverted.

Note: Do not confuse the mirror formula with the lens formula and write the magnification formula correctly. Also do not forget to use sign convention correctly as it will cause an error in the answer. Take care of the negative sign in the image height.