Q.A convex lens held some distance above a 6 cm long pencil produces its image of SOME size. On shifting the lens by a distance equal to its focal length, it again produces the image of the SAME size as earlier. Determine the image size.

[!TLDR] For the two image sizes to match after shifting the lens by exactly f, the lens must start at object distance f/2 (virtual image, m=2) and end at 3f/2 (real image, m=2) …

For a thin lens at object distance p (magnitude) from a fixed object, the magnitude of magnification works out to |m| = f/|p-f|, independent of whether the image is real or virtual. Shifting the lens AWAY from the object by exactly f changes p to p'=p+f; requiring |m(p)|=|m(p')| gives f/|p-f| = f/|p'-f| = f/|p|, so |p-f|=p, which (for 0<p<f) gives f-p=p, i.e. p=f/2. At this starting position p=f/2 (object within the focus, so a VIRTUAL image), |m| = f/|f/2-f| = f/(f/2) = 2. After shifting away by f, the new position is p'=3f/2 (object beyond 2f is not reached; between f and 2f, so a REAL image), and |m(p')| = f/(3f/2-f) = f/(f/2) = 2 as well -- confirming the same magnitude of magnification, hence the same image size, at both positions (though the FIRST image is virt …

Express the magnitude of magnification as a function of object distance p, |m|=f/|p-f|, set the value at p equal to the value at the shifted position p+f, solve for the unique starting position …

Assuming the two images must both be real (or both virtual) since they are 'the same size', missing that the printed answer's scenario actuall …