Question

An arrow $2.5\,cm$ high is placed at a distance of $25\,cm$ from a diverging mirror of focal length of $20\,cm$ . Find the nature, position and size of the image formed.

Answer 1

Hint: In this question, we have to find the height of the image, position and the magnification of the image. The given mirror is a diverging mirror, so it means that it is a convex mirror. So to solve this question, we will use the mirror formula i.e. $\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$. And after this we will use the magnification formula. By substituting the values in the formula, we will get the result.

Complete step by step answer:
In this question, we have the height of the object, let it be ${h_1}$ , i.e.
${h_1} = + 2.5\,cm$
Distance of the object from the mirror be
$u = - 25\,cm$
We have the focal length of the convex mirror,
$f = 20\,cm$
Now we have to find the position of the image i.e. $v$, height of the image which is ${h_2}$ and its magnification i.e. $m$.Now we will use the mirror formula i.e.
$\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$
By substituting the values back in the equation we have:
$\dfrac{1}{{20}} = \dfrac{1}{v} + \dfrac{1}{{( - 25)}}$
By arranging the similar terms together we have:
$\dfrac{1}{v} = \dfrac{1}{{20}} - \left( { - \dfrac{1}{{25}}} \right)$

It can be written as :
$\dfrac{1}{v} = \dfrac{1}{{20}} + \dfrac{1}{{25}}$
By taking the LCM we can write,
$\dfrac{1}{v} = \dfrac{{5 + 4}}{{200}}$
$\Rightarrow \dfrac{1}{v} = \dfrac{9}{{100}}$
By cross multiplication, it gives us
$v = \dfrac{{100}}{9} = 11.1cm$
So it gives us the distance of the image i.e. $v = 11.1cm$ , behind the mirror.
Now we will use the magnification formula :
$m = \dfrac{{{h_2}}}{{{h_1}}} = - \dfrac{v}{u}$
By equating the other two parts we have:
$\dfrac{{{h_2}}}{{{h_1}}} = - \dfrac{v}{u}$

By substituting the values, we can write:
$\dfrac{{{h_2}}}{{2.5}} = \dfrac{{ - 11.1}}{{ - 25}}$
BY cross multiplication we have,
${h_2} = \dfrac{{11.1 \times 2.5}}{{25}} = 1.11$ .
So the height of the image is $1.1cm$
We can calculate m , by the same formula:
$m = - \dfrac{v}{u} = \dfrac{{ - 11.1}}{{ - 25}}$
It gives $m = 0.44$.

Hence in convex lenses, the image is virtual, erect and smaller in size.

Note: We should note that to identify the nature of the object, like magnification, we use magnification equation, which states
$m = \dfrac{{height\,of\,image}}{{height\,of\,object}} \\
\Rightarrow m = - \dfrac{{distance\,of\,image}}{{distance\,of\,object}}$
If we get the value of $m$ in positive, then the image is magnified and if we get the value of m in negative, then the image is diminished.