(N/A) $(i)$ Yes,the lens can produce an image of the complete object. Although the intensity of the image will be reduced because less light passes through the lens,the image will still be formed at the same position.
$(ii)$ The ray diagram shows the object placed beyond $2F_1$. The image is formed between $F_2$ and $2F_2$ on the other side of the lens.
$(iii)$ Given: Height of object $h = 4\, cm$,focal length $f = +20\, cm$,object distance $u = -15\, cm$.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{20} + \frac{1}{-15} = \frac{3 - 4}{60} = \frac{-1}{60}$
Therefore,$v = -60\, cm$.
The negative sign indicates that the image is formed on the same side as the object.
Nature: Virtual and erect.
Size of the image: $m = \frac{v}{u} = \frac{-60}{-15} = +4$.
$h' = m \times h = 4 \times 4\, cm = +16\, cm$. The image is magnified and erect.