Past exam of the electromagnetism and optics course of the University of Oxford / 2016 / 5 / Solution

Use the thin lens sign convention in which a real object on the incident side has positive object distance, a real image on the outgoing side has positive image distance, and a converging lens has positive focal length. The first image distance isFor the image distance for two separated thin lenses, the second lens has object distance $u_2=L-v_1$, which can be negative when its intermediate object is virtual. Applying the thin lens equation again givesThe image is $d_2$ beyond the second lens when $d_2>0$; $d_2<0$ denotes a virtual image on its incident side. A vanishing denominator with nonzero numerator denotes an image at infinity. The rational expression also supplies appropriate limits when the first image is at infinity; for example, $d_1=f_1$ gives $d_2=f_2$ for finite nonzero focal lengths.

As $L\to 0$, the effective focal length is $F=f_1{f}_2/(f_1+f_2)$ andFor a real incident object, $d_1>0$, and positive $F$, the real-image condition is exactly $d_1>F$, yielding the requested inequality. This includes two converging lenses. The positivity assumption matters: with $f_1=f_2=-1\mathrm{m}$ and $d_1=1\mathrm{m}$, one has $F=-0.5\mathrm{m}$ and $d_1>F$, but $d_2=-1/3\mathrm{m}$ is virtual. If $f_1+f_2=0$, the touching pair has zero net optical power and produces no real image of this real object in the ideal thin lens model.