To find the image- distance for varying object- distances in case of a convex lens, a student obtains on a screen a sharp image of a bright object placed very far from the lens. After that he gradually moves the object towards the lens and each time focuses its image of the screen. (a) In which direction - towards or away from the lens, does he move the screen to focus the object? (b) What happens to the size of image - does it increase or decrease? (c) What happen when he moves the object very close to the lens?

This question tests the understanding of image formation by a convex lens. A convex lens is a converging lens. When an object is at infinity, the rays are parallel and converge at the focus. As the object distance ($u$) decreases, the image distance ($v$) increases, which explains why the screen must be moved away. The magnification ($m = v/u$) also increases in magnitude, meaning the image size increases. When $u < f$, the refracted rays diverge, meaning they do not meet on the other side of the lens to form a real image; instead, they appear to meet on the same side, forming a virtual image which cannot be caught on a screen.