Question

Question 15
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, $u=-27cm$
Focal length, $f=-18cm$
Height of the object, $h_0=7.0cm$
According to mirror formula,
$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$
$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$
$\frac{1}{v}=-\frac{1}{18}+\frac{1}{27}=-\frac{1}{54}$
Image distance,$v=-54cm$
Screen must be placed at a distance of 54 cm from the mirror in front of it.
$\frac{h_i}{h_0}=-\frac{v}{u}$
$\frac{h_i}{7}=\frac{54}{-27}=-2$
Height of the image, $h_i=-2\times 7=-14cm$
Thus, the image is of 14 cm length and is inverted image.