An object $4 \ cm$ high is placed at a distance of $27 \ cm$ in front of a convex lens of focal length $18 \ cm$. Find the position,nature,and size of the image formed.

(D) Given: Object height $O = 4 \ cm$,object distance $u = -27 \ cm$,focal length $f = +18 \ cm$.
Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$,we have:
$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{18} + \frac{1}{-27} = \frac{3-2}{54} = \frac{1}{54}$.
Therefore,the image position $v = +54 \ cm$.
Using magnification formula $m = \frac{I}{O} = \frac{v}{u}$,we have:
$\frac{I}{4} = \frac{54}{-27} = -2$.
Therefore,the image size $I = 4 \times (-2) = -8 \ cm$.
The positive value of $v$ indicates that the image is formed on the other side of the lens,and the negative value of $I$ indicates that the image is real and inverted.