Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2021 / Section B / 9

Use the original diagram's positive distances: $u$ from the object to the vertex, $d$ from the vertex to the image, and $R$ from the vertex toward the centre of curvature on the outgoing side. Let a paraxial light ray strike at height $Y$. Its incoming inclination is $\alpha \simeq Y/u$ above the optical axis, its outgoing inclination is $\beta \simeq Y/d$ below it, and the inward surface normal is inclined by $\theta \simeq Y/R$ below it. Thus the incidence and refraction angles relative to that normal are respectively $\alpha +\theta$ and $\theta -\beta$.

The local tangent-plane form of Snell's law, using the small-angle approximation, givesSubstitute the three height-to-distance ratios and cancel $Y$:This refraction at a spherical surface result keeps only first-order ray angles. The surface sag shifts the height's axial reference only at higher order. Distances for other object, image or curvature orientations must be signed consistently; the displayed positive-distance form follows the particular orientation in the paper.

Keep the paper's $R_1,R_2>0$ as magnitudes for the two convex faces. With the standard signed-curvature convention measured to the right, the first surface has signed radius $+R_1$ and the second $-R_2$. This explains why the requested formula contains a sum of the two reciprocal magnitudes.

Let the axial lens thickness be $\ell$, and let $d_1$ be the distance from the first vertex to the intermediate focus that its refracted rays would reach without the second surface. The first refraction at a spherical surface givesAt the second face those rays are still converging toward that intermediate focus, so its object distance is negative, $u_2=-d_2$, with $d_2=d_1-\ell$. Applying the signed spherical-surface equation again givesIn the thin lens limit, $d_2\simeq d_1$. Adding cancels the intermediate-focus terms:The focal length is the image distance for parallel incident light rays, $u\to \infty$, hence the lensmaker's equation isThe same surrounding refractive index $n_1$ is assumed on both sides. For $n_2>n_1$ this biconvex thin lens has positive focal length. With signed radii $R_a=R_1$, $R_b=-R_2$, the equivalent familiar expression is $1/f=(n_2/n_1-1)(1/R_a-1/R_b)$.

Use positive object distance $u=20\mathrm{cm}$ and focal length $f=4\mathrm{cm}$. The thin lens equation givesThe transverse magnification is $M=-v/u=-1/4$, so the signed height isThe real image lies $5.0\mathrm{cm}$ beyond the lens, is inverted, and has positive height magnitude $0.50\mathrm{cm}$.

For $u=8\mathrm{cm}$ and negative focal length $f=-8\mathrm{cm}$,The transverse magnification is $M=-v/u=+1/2$, soThus the virtual image is upright, half the object's height, and $4.0\mathrm{cm}$ from the lens on the object's side. A parallel incident light ray diverges after the lens as though it originated at the near optical focus; the central light ray continues undeviated in the ideal thin lens model. Their backward extensions meet at the image tip, $(-4,1)\mathrm{cm}$.