(D) $(i)$ For power $P = +2 \text{ D}$,the lens is a convex lens. Focal length $f = \frac{1}{P} = \frac{1}{+2} = +0.5 \text{ m} = +50 \text{ cm}$.
$(ii)$ For power $P = -4 \text{ D}$,the lens is a concave lens. Focal length $f = \frac{1}{P} = \frac{1}{-4} = -0.25 \text{ m} = -25 \text{ cm}$.
$(b)$ Given: Power $P = -4 \text{ D}$,object distance $u = -100 \text{ cm}$.
Focal length $f = \frac{1}{P} = -25 \text{ cm}$.
Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$:
$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{-25} + \frac{1}{-100} = \frac{-4 - 1}{100} = \frac{-5}{100} = -\frac{1}{20}$.
Therefore,$v = -20 \text{ cm}$. The image is formed at a distance of $20 \text{ cm}$ on the same side as the object.