Past exam of the physics 2 paper of the University of Oxford / 2015 resit / 11 / b / Solution

An ideal focusing lens maps propagation angle to position in its back focal plane, with $z\simeq f\theta$ for paraxial light rays. The two-slit relative phase remains $2\pi s\sin \theta /\lambda$, so the focal-plane interference pattern has the same form with propagation distance replaced by focal length:The new peak intensity $I_{\max ,f}$ accounts for focusing and need not equal the earlier screen peak. The lens forms the angular Fraunhofer pattern at a finite focal-plane distance, so a long free propagation distance is no longer needed to obtain that angular pattern. The displayed position formula assumes the same narrow-slit and paraxial approximations.