Question

An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed, so that a sharp focused image can be obtained? Find the size and the nature of the image.

Answer 1

Object distance, $u = −27\ cm$
Object height, $h = 7\ cm$
Focal length, $f = −18\ cm$
According to the mirror formula,
$\frac 1v+\frac 1u=\frac1f$

$\frac1v=\frac 1f-\frac 1u$

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

$\frac 1v=-\frac {1}{54}$
$v=-54\ cm$
The screen should be placed at a distance of $54\ cm$ in front of the given mirror.
Magnfication, $m=-\frac {\text {Image\ distance}}{\text {Object\ distance}}$
$m =-\frac {54}{27}$
$m=-2$
The negative value of magnification indicates that the image formed is real.
Magnfication, $m=\frac {\text{Height\ of\ the\ image}}{\text {Height\ of\ the\ Object}}$
$m=\frac {h'}{h}$
$h'=m\times h$
$h'=7 \times (-2)$
$h'=-14\ cm$
The negative value of image height indicates that the image formed is inverted.