An object $3.0 \text{ cm}$ high is placed perpendicular to the principal axis of a concave lens of focal length $7.5 \text{ cm}$. The image is formed at a distance of $5.0 \text{ cm}$ from the lens. Calculate $(i)$ distance at which object is placed,and $(ii)$ size and nature of image formed.

(A) Given: Object height $O = 3.0 \text{ cm}$,focal length $f = -7.5 \text{ cm}$,image distance $v = -5.0 \text{ cm}$.
$(i)$ Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$:
$\frac{1}{u} = \frac{1}{v} - \frac{1}{f} = \frac{1}{-5.0} - \frac{1}{-7.5} = -0.2 + 0.1333 = -0.0667 \text{ cm}^{-1}$.
Thus,$u = -15 \text{ cm}$. The object is placed $15 \text{ cm}$ in front of the concave lens.
(ii) Magnification $m = \frac{I}{O} = \frac{v}{u}$.
$I = \frac{v}{u} \times O = \frac{-5.0}{-15.0} \times 3.0 = 1.0 \text{ cm}$.
Since the magnification is positive $(m = +0.33)$,the image is virtual,erect,and $1.0 \text{ cm}$ in size.