A rod of length $10{{ cm}}$ lies along the principal axis of concave mirror of focal length $10{{ cm}}$ in such a way that its end closer to the pole is $20{{ cm}}$ away from the mirror. The length of the image is then:
(A) $15{{ cm}}$
(B) $2.5{{ cm}}$
(C) $5{{ cm}}$
(D) $10{{ cm}}$

Hint: Figure out the distances of the ends of the rod with respect to the given concave mirror. Using the mirror formula, this distance can be used to evaluate the height of the image.
Mirror formula is : $\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}$............equation $(1)$

Complete step by step solution:
In the above equation, f is the focal length of the given mirror, v is the image distance and u is the object distance.
Of the two ends of the object (the rod in this case), first one is at $20$ cm away from the mirror and the second one is at $10$ cm away.
For the first end, $\dfrac{1}{f} = \dfrac{1}{{{v_1}}} + \dfrac{1}{{{u_1}}}$............equation $(1)$
Substitute the values,
$\Rightarrow -\dfrac{1}{{10}} = - \dfrac{1}{{20}} + \dfrac{1}{{{v_1}}}$
$ \Rightarrow \dfrac{1}{{{v_1}}} = \dfrac{1}{{20}} - \dfrac{1}{{10}} = \dfrac{{1 - 2}}{{20}} = - \dfrac{1}{{20}}$
$\therefore {v_1} = - 20{{ cm}}$.............equation $(2)$
Similarly, for the second end, the mirror formula can be written as
$\Rightarrow \dfrac{1}{f} = \dfrac{1}{{{v_2}}} + \dfrac{1}{{{u_2}}}$
Substituting the values,
$\Rightarrow - \dfrac{1}{{10}} = - \dfrac{1}{{30}} + \dfrac{1}{{{v_2}}}$
$\Rightarrow \dfrac{1}{{{v_2}}} = \dfrac{1}{{30}} - \dfrac{1}{{10}} = \dfrac{{1 - 3}}{{30}} = -\dfrac{1}{15}$
$\therefore {v_2} = - 15$ cm……………...equation $(3)$
Using $(1)\& (2)$ it can be concluded that the height of image will be the difference in heights of two ends
i.e. Image height $ = {v_2} - {v_1} = ( - 15) - ( - 20) = 5{{ cm}}$

Therefore, option (C) is correct.

Note: For concave mirrors, all distances measured in front of the mirror are taken as negative. As both the ${v_1}\& {v_2}$ are negative, it indicates that the images formed are in front of the mirror, and they are real and inverted.