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.

(N/A) Object distance,$u = -20\, cm$.
Object height,$h = 5\, cm$.
Radius of curvature,$R = 30\, cm$.
Focal length,$f = R/2 = 15\, cm$.
According to the mirror formula,$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{15} - (\frac{1}{-20}) = \frac{1}{15} + \frac{1}{20} = \frac{4+3}{60} = \frac{7}{60}$.
$v = \frac{60}{7} \approx 8.57\, cm$.
The positive value of $v$ indicates that the image is formed behind the mirror.
Magnification,$m = -\frac{v}{u} = -\frac{8.57}{-20} = 0.428$.
The positive value of magnification indicates that the image is virtual and erect.
Height of the image,$h' = m \times h = 0.428 \times 5 = 2.14\, cm$.
Thus,the image is virtual,erect,and $2.14\, cm$ in size,formed $8.57\, cm$ behind the mirror.