Past exam of the physics 2 paper of the University of Oxford / 2017 resit / 9 / c / Solution

Substituting the stated refractive index and thickness coefficient, $\alpha =2{\mathrm{m}}^{-1}$, givesThe thin lens equation then givesThe positive image distance places a real image on the outgoing side of the optical lens, at $x=+1.75\mathrm{m}$. The object is beyond the focal point, so outgoing light rays actually converge there rather than only their backward extensions meeting.