# 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.

Question:

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?

Solution:

Size of the candle, h= 2.5 cm

Image size = h

Object distance, u= −27 cm

Radius of curvature of the concave mirror, R= −36 cm

Focal length of the concave mirror, $f=\frac{R}{2}=-18 \mathrm{~cm}$

Image distance = v

The image distance can be obtained using the mirror formula:

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

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

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

$\therefore v=-54 \mathrm{~cm}$

Therefore, the screen should be placed 54 cm away from the mirror to obtain a sharp image.

The magnification of the image is given as:

$m=\frac{h^{\prime}}{h}=-\frac{v}{u}$

$\therefore h^{\prime}=-\frac{v}{u} \times h$

$=-\left(\frac{-54}{-27}\right) \times 2.5=-5 \mathrm{~cm}$

The height of the candle’s image is 5 cm. The negative sign indicates that the image is inverted and real.

If the candle is moved closer to the mirror, then the screen will have to be moved away from the mirror in order to obtain the image.