Question

A point object is placed at a distance of $10{\rm{ cm}}$ and its real image is formed at a distance of $30{\rm{ cm}}$ from a concave mirror. If the object is moved by $0.2\,{\rm{cm}}$ towards the mirror, the image will shift by about.

Answer 1

We know that the reciprocal of the focal length of a mirror is equal to the addition of the reciprocals of the distance of the object from the mirror and the distance of the image from the mirror. Let us assume the focal length of a mirror is $f$, the distance of the object from the mirror is $u$ and the distance of the image from the mirror is $v$ then the formula for the mirror becomes-
$\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}$

Complete step by step answer:
Given:
The distance between the object and the mirror $u = - 10{\rm{ cm}}$
The distance between the real image and the mirror $v = - 30{\rm{ cm}}$
Negative sign (-) indicates that the distance is measured away from the mirror.
So, from the mirror formula, we have-
$\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}$
Substituting the values into the formula we get,
$\begin{array}{l} \dfrac{1}{f} = \dfrac{1}{{\left( { - 10} \right)}} + \dfrac{1}{{\left( { - 30} \right)}}\\\ \dfrac{1}{f} = \dfrac{{ - \left( {3 + 1} \right)}}{{30}}\\\ \dfrac{1}{f} = \dfrac{{ - 4}}{{30}} \end{array}$
Solving this we get,
$f = - 7.5{\rm{ cm}}$
Negative sign (-) indicates the distance is measured away from the mirror.
Now, since the object is moved towards the mirror, the value of the distance between the object and the mirror also the distance between the image and the mirror changes, but the focal length of the mirror remains unchanged.
Given, the object is moved towards the mirror $\Delta u = 0.2{\rm{ cm}}$
So now, the new distance between the object and the mirror is,
$\begin{array}{l} u' = - 10 + 0.2\\\ u' = - 9.8{\rm{ cm}} \end{array}$
Substituting these values into the mirror formula we get,
$\begin{array}{l} \dfrac{{ - 4}}{{30}} = \dfrac{1}{{\left( { - 9.8} \right)}} + \dfrac{1}{{v'}}\\\ \dfrac{1}{{v'}} = \dfrac{{ - 4}}{{30}} + \dfrac{1}{{9.8}}\\\ \dfrac{1}{{v'}} = \dfrac{{ - 23}}{{735}} \end{array}$
Solving this we get,
$v' = - 31.956{\rm{ cm}}$
So, the change in the distance of the image = New distance between the image and the mirror – Original distance between the image and the mirror
$\Delta v = v' - v$
Substituting the values, we get,
$\begin{array}{l} \Delta v = \left( { - 31.956} \right) - \left( { - 30{\rm{ }}} \right)\\\ \Delta v = - 1.956{\rm{ cm}} \end{array}$
We can write this as,
$\Delta v = 1.956{\rm{ cm}}$(Away from the mirror)
Therefore, the image will shift about $1.956{\rm{ cm}}$ away from the mirror.

Note: The mirror formula used to solve this question is the same for a convex mirror as well as a concave mirror. The sign convention used depends upon the position of the object and the image relative to the mirror. If the object or the image is moving closer to the mirror, then the value of distance is taken positive, otherwise it is taken negative.