Answer –

We are given that,

Height of the candle is h =2.5 cm

Let the image size be h’

Object distance is u = -27 cm

Radius of the concave mirror, is R = -36 cm

Focal length of the concave mirror is,

$f=\frac{R}{2}=-18cm$

v represents image distance

$\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$

$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$

$=\frac{1}{-18}-\frac{1}{-27}=\frac{-3+2}{54}$$=-\frac{1}{54}$

Therefore, image distance is v = -54cm

As a result, the distance between the screen and the mirror should be 54cm for a sharp image. Image magnification is:

$m=\frac{h’}{h}=-\frac{v}{u}$

$∴ h’=\frac{-v}{u}\times h$

$=-(\frac{-54}{-27})\times2.5$

= – 5cm

The image of the candle has a height of 5cm. The image is inverted and virtual since there is a negative sign. If the candle is moved closer to the mirror, the picture must be obtained by moving the screen away from the mirror.