(b)i. Distinguish between a real image and a virtual image.
Draw a ray diagram to show how a converging lens may be used to form a real diminished image of an object.
(a) Lens displacement (conjugate-foci) experiment (outline). With the object and screen a fixed distance \(D\) apart, there are two lens positions giving a sharp image; let their separation be \(L = (x_1 - x_2)\). By the displacement method the focal length satisfies
\[ f = \frac{D^2 - L^2}{4D} \;\Rightarrow\; D^2 - L^2 = 4f\,D. \]
So a graph of \((D^2 - L^2)\) against \(D\) is a straight line through the origin of slope \(s = 4f\); hence \(K = \dfrac{s}{4} = f\), the focal length of the converging lens.
Precautions: place object, lens and screen so their centres are at the same height (aligned on one axis); focus for the sharpest image and read positions at eye level to avoid parallax.
(b)(i) A real image is formed by the actual intersection of refracted rays, can be caught on a screen, and is inverted (for a single converging lens). A virtual image is formed where rays only appear to come from (their backward projections meet), cannot be caught on a screen, and is upright.
Ray diagram: with the object placed beyond \(2F\), a converging lens forms a real, inverted, diminished image between \(F\) and \(2F\) on the far side.