A small candle, 2.5cm in size is placed at 27cm in front of a concave mirror of radius of curvature 36cm. 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?

Given, 

object distance = $u=-27cm$

radius of curvature=$R=-36cm$

focal length = $f=\frac{R}{2}=-18cm$

we know that by the mirror formula distance of the image is, $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

$-\frac{1}{27}+\frac{1}{v}=-\frac{1}{18}$

$\frac{1}{v}=\frac{1}{27}-\frac{1}{18}$

$\frac{1}{v}=\frac{2-3}{54}$

$\frac{1}{v}=\frac{-1}{54}$

$v=-54cm$

Thus to get a sharp image the screen is placed 54cm away from the mirror.

Now magnification is given by, 

$$\begin{gathered} -\frac{-54}{-27}=\frac{h\text{'}}{2.5} \\ \Rightarrow h\text{'}=\frac{-54\times 2.5}{27} \end{gathered}$$

$\Rightarrow h\text{'}=-5cm$

Hence the height of the image is -5cm, negative sign implies the image is inverted and formed behind the mirror.

To obtain the image the screen should be moved away if the candle is moved closer to the mirror.