Question

An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.

Answer 1

Object distance, $u = −20\ cm$
Object height, $h = 5\ cm$
Radius of curvature, $R = 30\ cm$
Radius of curvature $= 2 × \text {Focal length}$
$R = 2f$
$f = 15\ cm$
According to the mirror formula,
$\frac 1v+\frac 1u=\frac 1f$

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

$\frac 1v=\frac {1}{15}+\frac {1}{20}$

$\frac 1v=\frac {4+3}{60}$

$\frac 1v=\frac {7}{60}$
$v= 8.57$
The positive value of v indicates that the image is formed behind the mirror.
Magnfication, $m=-\frac {\text {Image\ distance}}{\text{Object\ distance}}$
$m =-\frac {8.57}{-20}$
$m=0.428$
The positive value of magnification indicates that the image is formed is virtual.
Magnfication, $m=-\frac {\text {Height \ of the \ Image}}{\text{Height \ of the \ Object}}$
$m = \frac {h'}{h}$
$h'=m \times h$
$h' =0.428 \times 5$
$h'=2.14\ cm$
The positive value of image height indicates that the image formed is erect.

Therefore, the image formed is virtual, erect, and smaller in size.