= Solution
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> gives
$$
\frac{n_1}{u}+\frac{n_2}{d_1}=\frac{n_2-n_1}{R_1}.
$$
At 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 gives
$$
-\frac{n_2}{d_2}+\frac{n_1}{v}=\frac{n_1-n_2}{-R_2}=\frac{n_2-n_1}{R_2}.
$$
In the <thin lens> limit, $d_2\simeq d_1$. Adding cancels the intermediate-focus terms:
$$
\boxed{\frac1u+\frac1v=\left(\frac{n_2}{n_1}-1\right)\left(\frac1{R_1}+\frac1{R_2}\right).}
$$
The <focal length> is the image distance for parallel incident <light rays>, $u\to\infty$, hence the <lensmaker's equation> is
$$
\boxed{f=\left[\left(\frac{n_2}{n_1}-1\right)\left(\frac1{R_1}+\frac1{R_2}\right)\right]^{-1}.}
$$
The 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 $1/f=(n_2/n_1-1)(1/R_a-1/R_b)$.
Back to article page
