"$A$ convex lens can form a magnified erect as well as magnified inverted image of an object placed in front of it." Draw ray diagrams to justify this statement, stating the position of the object with respect to the lens in each case. An object of height $4 \text{ cm}$ is placed at a distance of $20 \text{ cm}$ from a concave lens of focal length $10 \text{ cm}$. Use the lens formula to determine the position of the image formed.

(N/A) $1$. For a magnified erect image: The object must be placed between the optical center $(O)$ and the principal focus $(F_1)$ of the convex lens.
$2$. For a magnified inverted image: The object must be placed between the principal focus $(F_1)$ and twice the focal length $(2F_1)$ of the convex lens.
$3$. Numerical calculation for the concave lens:
Given: Focal length $f = -10 \text{ cm}$ (for concave lens), Object distance $u = -20 \text{ cm}$.
Using the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\frac{1}{v} - \frac{1}{-20} = \frac{1}{-10}$
$\frac{1}{v} + \frac{1}{20} = -\frac{1}{10}$
$\frac{1}{v} = -\frac{1}{10} - \frac{1}{20} = \frac{-2 - 1}{20} = -\frac{3}{20}$
$v = -\frac{20}{3} \approx -6.67 \text{ cm}$.
The image is formed at a distance of $6.67 \text{ cm}$ in front of the concave lens.