An object $5 \,cm$ in length is held $25 \,cm$ away from a converging lens of focal length $10 \,cm$. Find the position,size,and the nature of the image formed.

(N/A) Given:
Object height,$h_o = 5 \,cm$
Object distance,$u = -25 \,cm$
Focal length of converging lens,$f = +10 \,cm$
Using the lens formula:
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{10} - \frac{1}{25} = \frac{5 - 2}{50} = \frac{3}{50}$
$v = \frac{50}{3} \approx 16.67 \,cm$
The positive value of $v$ indicates that the image is real and formed on the other side of the lens at a distance of $16.67 \,cm$.
Magnification,$m = \frac{v}{u} = \frac{16.67}{-25} \approx -0.667$
Also,$m = \frac{h_i}{h_o}$
$h_i = m \times h_o = -0.667 \times 5 \approx -3.33 \,cm$
The negative sign of $h_i$ indicates that the image is inverted. Thus,the image is real,inverted,and diminished in size.