An object $5\, cm$ high is kept $25\, cm$ away from a converging lens (convex lens) of focal length $10\, cm$. What is the nature,position,and size of the image?

(D) Given: Object height $h = 5\, cm$,object distance $u = -25\, cm$,focal length $f = 10\, cm$.
Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$,we have $\frac{1}{v} = \frac{1}{u} + \frac{1}{f} = \frac{1}{-25} + \frac{1}{10} = \frac{-2 + 5}{50} = \frac{3}{50}$.
Thus,$v = \frac{50}{3}\, cm \approx 16.67\, cm$.
The positive sign indicates that the image is formed on the other side of the lens (real image).
Magnification $m = \frac{v}{u} = \frac{h'}{h}$.
$h' = h \times \frac{v}{u} = 5 \times \frac{16.67}{-25} = -3.33\, cm$.
The negative sign of $h'$ indicates that the image is inverted.
Therefore,the image is real,inverted,diminished,and formed at a distance of $16.67\, cm$ from the lens.