Question 14
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.

Radius of curvature (R) = 30 cm
$f=\frac{R}{2}=\frac{30}{2}=15cm$
Object distance, $u=-20cm$
Height of the object, $h=5cm$
According to mirror formula,
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$
$\frac{1}{v}=\frac{1}{15}+\frac{1}{20}=\frac{7}{60}$
Image distance,$v=\frac{60}{7}=8.6cm$
Image is virtual and erect and formed behind the mirror.
$\frac{h_i}{h_0}=-\frac{v}{u}$
$\frac{h_i}{5}=\frac{-8.6}{-20}$
$h_i=2.15cm$
Size of image is 2.15 cm.