A small candle, $2.5\text{cm}$ in size is placed at $27\text{cm}$ in front of a concave mirror of radius of curvature $36\text{cm}$. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved?

In the direction of incident ray, distance is considered positive.
From mirror formula:
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
Here,
$u=-27\text{cm}$
$f=\frac{R}{2}=-18\text{cm}$
Height of object $\left(h\right)=2.5\text{cm}$
Putting values, we get
$\frac{1}{v}+\left(\frac{1}{-27}\right)=\left(\frac{1}{-18}\right)$
The size of image is $5\text{cm}$.
If candle move closer to mirror
As $u\to f,v\to \infty );$
For u < f, image is virtual.
Final Answer: $v=-54\text{cm}$. The image is real, inverted and magnified. The size of the image is $5.0\text{cm}$. As $u\to f,v\to \infty$; for u < f, image is virtual.