$(a)$ Explain the following terms related to spherical lenses:
$(i)$ Optical centre
$(ii)$ Centres of curvature
$(iii)$ Principal axis
$(iv)$ Aperture
$(v)$ Principal focus
$(vi)$ Focal length
$(b)$ $A$ converging lens has a focal length of $12 \, cm$. Calculate at what distance should the object be placed from the lens so that it forms an image at $48 \, cm$ on the other side of the lens.

(N/A) Definitions:
$(i)$ Optical centre: The central point of a lens through which a ray of light passes without any deviation.
$(ii)$ Centres of curvature: The centres of the two imaginary spheres of which the lens surfaces are parts.
$(iii)$ Principal axis: An imaginary line passing through the optical centre and the centres of curvature of the lens.
$(iv)$ Aperture: The effective diameter of the circular outline of a spherical lens.
$(v)$ Principal focus: $A$ point on the principal axis where light rays parallel to the principal axis converge (in convex lens) or appear to diverge (in concave lens) after refraction.
$(vi)$ Focal length: The distance between the optical centre and the principal focus of the lens.
$(b)$ Given: Focal length $f = +12 \, cm$ (for a converging lens),Image distance $v = +48 \, cm$ (real image on the other side).
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
$\frac{1}{12} = \frac{1}{48} - \frac{1}{u}$
$\frac{1}{u} = \frac{1}{48} - \frac{1}{12}$
$\frac{1}{u} = \frac{1 - 4}{48} = \frac{-3}{48} = -\frac{1}{16}$
Therefore,$u = -16 \, cm$. The object should be placed at a distance of $16 \, cm$ in front of the lens.