Past exam of the physics 2 paper of the University of Oxford / 2019 resit / 7 / Solution

TODO: the source glass-rod diagram is unavailable. The recovered prose gives a surface curvature and object distance, but the diagram's orientation, labels and any further optical surfaces cannot be verified. The following solves the surviving request for a single convex entrance surface; it is conditional on that interpretation and does not reconstruct the missing diagram or supply a rod length.

For refraction at a spherical surface, take incident light travelling rightwards from refractive index $n_0$ to $n_g$, object distance $u>0$ to the left of the vertex, image distance $v$ positive to the right, and signed curvature $R_s$ positive when the centre is inside the rod. In the paraxial approximation, applying Snell's law to a light ray at height $y$ givesAfter cancelling $y$, the image position and transverse magnification areThe magnification follows by considering a paraxial light ray through the vertex: Snell's law there gives $n_0(-y_o/u)=n_g(y_i/v)$.

For a convex entrance, $R_s=+20\mathrm{mm}$ and $u=60\mathrm{mm}$. In air, the denominator $(n_g-n_0)/R_s-n_0/u$ is $1/75\mathrm{m}{\mathrm{m}}^{-1}$, givingThis is a real, inverted image within a sufficiently long glass rod, before any subsequent surface.

For water with refractive index $1.33$, the denominator is $-13/1500\mathrm{m}{\mathrm{m}}^{-1}$, andThe emerging light rays diverge as though from an upright virtual image on the object side. The change of sign is consistent with $v$ diverging when $n_0=n_{g}u/(u+R_s)=1.20$. A reversed curvature or a second exit surface changes these answers; the unavailable diagram prevents a definitive answer for the full illustrated arrangement.