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

(D) Given: Size of the candle,$h = 2.5\; cm$. Object distance,$u = -27\; cm$. Radius of curvature,$R = -36\; cm$.
Focal length,$f = R/2 = -18\; cm$.
Using the mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-18} - \frac{1}{-27} = \frac{-3 + 2}{54} = -\frac{1}{54}$.
Thus,$v = -54\; cm$. The screen should be placed $54\; cm$ in front of the mirror.
Magnification $m = \frac{h'}{h} = -\frac{v}{u}$.
$h' = -\frac{v}{u} \times h = -\left(\frac{-54}{-27}\right) \times 2.5 = -5\; cm$.
The image is real,inverted,and $5\; cm$ in size.
If the candle is moved closer to the mirror (i.e.,$u$ decreases),the image distance $v$ increases,so the screen must be moved away from the mirror.