An object is situated at $8 \, cm$ from a convex lens of focal length $10 \, cm$. Find the position and nature of the image. Draw a ray diagram to illustrate the formation of the image (not to scale).

(N/A) Given: Focal length $f = +10 \, cm$ and object distance $u = -8 \, cm$.
According to the lens formula:
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
Rearranging for $v$:
$\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$
Substituting the values:
$\frac{1}{v} = \frac{1}{10} + \frac{1}{-8} = \frac{1}{10} - \frac{1}{8} = \frac{4 - 5}{40} = -\frac{1}{40}$
Therefore,$v = -40 \, cm$.
Hence,we conclude that:
$(a)$ The image is formed at $40 \, cm$ from the lens on the same side as the object.
$(b)$ The image is virtual and erect.
$(c)$ The image is magnified,i.e.,larger in size than the object.