Question

An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.

Answer 1

Object distance, $u = −25\ cm$
Object height, $h_o = 5\ cm$
Focal length, $f = +10\ cm$
According to the lens formula,
$\frac 1v-\frac1u=\frac 1f$

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

$\frac 1v=\frac {1}{10} -\frac {1}{25}$

$\frac 1v=\frac {15}{250}$

$v=\frac {250}{15}$
$v=16.66\ cm$
The positive value of v shows that the image is formed at the other side of the lens.
Magnification, $m=-\frac {\text {Image\ distance}}{\text {Object\ distance}}$
$m =-\frac vu$

$m =-\frac {16.66}{25}$
$m =-0.66$
The negative sign shows that the image is real and formed behind the lens.
Magnification, $m=-\frac {\text {Image\ height}}{\text {Object\ height}}$
$m=\frac {H_I}{H_o}$

$m=\frac {HI}{5}$
$HI=5 \times m$
$HI =5 \times (-0.66)$
$HI=-3.3\ cm$
The negative value of image height indicates that the image formed is inverted.
The position, size, and nature of image are shown in the following ray diagram.