$A$ student focussed the image of a candle flame on a white screen using a convex lens. He noted down the position of the candle,screen,and the lens as under:
Position of candle $= 12.0\, cm$
Position of convex lens $= 50.0\, cm$
Position of the screen $= 88.0\, cm$
$(i)$ Where will the image be formed if he shifts the candle towards the lens at a position of $31.0\, cm$?
$(ii)$ What will be the nature of the image formed if he further shifts the candle towards the lens?
$(iii)$ Draw a ray diagram to show the formation of the image in case $(ii)$ as said above.

(N/A) First,calculate the focal length $(f)$ of the lens:
Object distance $u = 12.0 - 50.0 = -38.0\, cm$
Image distance $v = 88.0 - 50.0 = +38.0\, cm$
Using the lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\frac{1}{38} - \frac{1}{-38} = \frac{1}{f} \implies \frac{2}{38} = \frac{1}{f} \implies f = 19.0\, cm$
$(i)$ New object position $= 31.0\, cm$. New $u = 31.0 - 50.0 = -19.0\, cm$. Since the object is at the focus $(u = -f)$,the image will be formed at infinity.
$(ii)$ If the candle is shifted further towards the lens (i.e.,$u < f$),the object lies between the optical center and the focus. The image formed will be virtual,erect,and magnified.
$(iii)$ The ray diagram shows the object placed between $F$ and $O$,resulting in a virtual image formed on the same side as the object.