An object of size$7cm$is placed at $27cm$ in front of a concave mirror of focal length $18cm$. At what distance from the mirror should a screen be placed, so that a sharp focused image can be obtained? Find the size and nature of the image?

Hint: A concave mirror is curved inwards. It can form real as well as virtual images for different positions of an object. Substituting the corresponding values in the mirror’s formula, which gives us the relationship between object distance, image distance and focal length, we can find the missing value. Magnification is the ratio of image height to object height.

Formulas used:
$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$

Complete answer:
A concave mirror is curved inwards. It forms real as well as virtual images for different positions of the object. By convention, its object distance and focal length are negative.
The mirror formula is given by-
$\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$ - (1)
Here,$f$is the focal length
$v$is the image distance from the mirror
$u$is the object distance from the mirror
Given,$f=-18cm$, $u=-27cm$
Substituting values in eq (1), we get,
$\begin{align}
  & \dfrac{-1}{18}=\dfrac{1}{v}+\dfrac{1}{-27} \\
 & \Rightarrow \dfrac{-1}{18}+\dfrac{1}{27}=\dfrac{1}{v} \\
 & \dfrac{1}{v}=\dfrac{-1}{54} \\
\end{align}$
$\therefore v=-54cm$
The screen should be placed at a distance of $54cm$ in front of the mirror, on the same side as the object. The object formed is real and inverted.
We know that magnification is the ratio of image height to object height. It tells us about the size of the image relative to the object. The formula for magnification is-
$M=-\dfrac{v}{u}$
Substituting values in the above equation, we get,
$\begin{align}
  & M=-\dfrac{-54}{-27} \\
 & \therefore M=-2 \\
\end{align}$
Therefore, the image is magnified and the negative sign indicates it is inverted. Also, we know that,
$\begin{align}
  & M=\dfrac{{{h}_{i}}}{{{h}_{o}}} \\
 & \Rightarrow -2=\dfrac{{{h}_{i}}}{7} \\
 & \therefore {{h}_{i}}=-14cm \\
\end{align}$
The height of the image is $14cm$ below the axis.
Therefore, the image is formed at $54cm$ in front of the mirror. It is real inverted and magnified by $2\times $ and the height of the image is $14cm$ below the axis.

Note:
The image is formed beyond the radius of curvature. Only the images formed between the centre of mirror and focus are virtual images, formed at the back of the mirror. Unlike concave mirrors, convex mirrors always form virtual images for all positions of the object.