The thin lens equation $\frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f}$ takes the focal length as given. But a lens maker starts with a piece of glass and two curved surfaces to grind. How do the shape and the material set $f$? Each curved surface refracts the light a little, and the two effects add.
Key idea
A thin lens of index $n$ in air, with surface radii $R_1$ (the surface the light meets first) and $R_2$, has $$\frac{1}{f} = (n - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right).$$ Sign rule: $R > 0$ when the surface's center of curvature is on the outgoing side (where the light goes), $R < 0$ when it is on the incoming side, and $R = \infty$ for a flat surface. In a medium of index $n_m$, replace $n$ by $n/n_m$.
Why it works
For one spherical surface between indices $n_1$ and $n_2$, paraxial rays obey $\frac{n_1}{s} + \frac{n_2}{s'} = \frac{n_2 - n_1}{R}$ (this comes from Snell's law with small angles). Apply it twice. At the first surface, from air into glass: $$\frac{1}{d_o} + \frac{n}{s_1'} = \frac{n - 1}{R_1}.$$ The image it makes is the object for the second surface. The lens is thin, so that object sits at distance $-s_1'$ (on the outgoing side, a virtual object). From glass into air: $$\frac{n}{-s_1'} + \frac{1}{d_i} = \frac{1 - n}{R_2}.$$ Add the two equations: the $\frac{n}{s_1'}$ terms cancel, leaving $\frac{1}{d_o} + \frac{1}{d_i} = (n - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$. That is the thin lens equation, with the right side as $\frac{1}{f}$.
Worked examples
Worked example: A double-convex lens
A glass lens ($n = 1.50$) bulges outward on both sides, with radii 20 cm (front) and 30 cm (back). Find its focal length.
- Front surface: its center is beyond the lens, on the outgoing side, so $R_1 = +20$ cm.
- Back surface: it bulges toward the outgoing side, so its center is on the incoming side: $R_2 = -30$ cm.
- $\frac{1}{f} = 0.50\left(\frac{1}{20} - \frac{1}{-30}\right) = 0.50\left(\frac{1}{20} + \frac{1}{30}\right) = 0.50 \times 0.0833 = 0.0417\ \text{cm}^{-1}$.
Answer: $f = +24.0$ cm (converging)
Worked example: Flat on one side
A plano-convex lens ($n = 1.50$) has one flat face and one face of radius 15 cm. Find $f$ with the flat face toward the light, and again with the curved face toward the light.
- Flat face first: $R_1 = \infty$, $R_2 = -15$ cm. $\frac{1}{f} = 0.50\left(0 + \frac{1}{15}\right)$, so $f = 30$ cm.
- Curved face first: $R_1 = +15$ cm, $R_2 = \infty$. $\frac{1}{f} = 0.50\left(\frac{1}{15} - 0\right)$, so again $f = 30$ cm.
Answer: $f = 30$ cm either way (a thin lens has the same focal length in both directions)
Worked example: The same lens under water
The double-convex lens from the first example ($f = 24.0$ cm in air) is put in water ($n_m = 1.33$). What is its focal length now?
- Replace $n - 1 = 0.50$ by $\frac{n}{n_m} - 1 = \frac{1.50}{1.33} - 1 = 0.128$.
- The radii are unchanged, so $f$ grows by the factor $\frac{0.50}{0.128} = 3.91$.
- $f = 24.0 \times 3.91 = 93.9$ cm.
Answer: $f \approx 94$ cm: the lens is much weaker in water, which is why your eyes blur when you open them underwater
Common mistake
Giving the back surface the wrong sign. For a lens that bulges on both sides, the two centers of curvature are on opposite sides, so $R_1$ and $R_2$ have opposite signs, and the two terms add in $\frac{1}{R_1} - \frac{1}{R_2}$. If you give both radii the same sign you get the difference instead, and a converging lens can come out far too weak or even diverging. Check: a lens thicker in the middle than at the edges must have $f > 0$ in air.