$A$ $10 \, cm$ tall object is placed perpendicular to the principal axis of a convex lens of focal length $30 \, cm$. The distance of the object from the lens is $20 \, cm$. Find the $(i)$ position,$(ii)$ nature,and $(iii)$ size of the image formed.

(D) Given: Object height $O = 10 \, cm$,object distance $u = -20 \, cm$,focal length $f = +30 \, cm$.
Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$,we have:
$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{30} + \frac{1}{-20} = \frac{2 - 3}{60} = -\frac{1}{60}$.
Therefore,the position of the image is $v = -60 \, cm$ (the image is formed on the same side as the object).
Using the magnification formula $m = \frac{I}{O} = \frac{v}{u}$,we have:
$\frac{I}{10} = \frac{-60}{-20} = 3$.
Therefore,the size of the image is $I = 30 \, cm$.
Since $v$ is negative,the image is virtual and erect. Since the height $I$ is positive and greater than $O$,the image is magnified.