9.3 Refraction at Spherical Surfaces and by Lenses

An optometrist hands you a prescription reading "$-1.75$ D". The number is the power of the lens you need, in dioptres, and the minus sign says it must spread light out. The lens grinder who makes it chooses the glass and the curvatures of its two faces using a single formula — the lens maker's formula. The camera in your phone contains five or six tiny lenses designed with the same rules, and a modern intraocular lens, implanted in millions of cataract operations every year, is chosen to a quarter of a dioptre by the same arithmetic.

In Section 9.2 we refracted light at flat surfaces. Here we curve the surface. Snell's law, applied to paraxial rays at one spherical surface, gives a formula relating object and image distances; applying it at the two surfaces of a thin lens gives the lens maker's formula and the thin-lens formula. We define the power of a lens and show that, for thin lenses in contact, powers simply add — which is how spectacle lenses and camera lenses are combined.

Refraction at a Spherical Surface

Let a spherical surface of radius $R$ separate a medium of refractive index $n_1$ (where the light comes from) from one of index $n_2$. A point object $O$ on the axis sends a ray to a point $N$ of the surface at height $h$ above the pole $M$; it refracts and crosses the axis at the image $I$. Let $C$ be the centre of curvature. For paraxial rays all the angles are small, so each equals its tangent: $\angle NOM\approx\dfrac{h}{OM}$, $\ \angle NCM\approx\dfrac{h}{MC}$ and $\ \angle NIM\approx\dfrac{h}{MI}$. The normal at $N$ is $CN$. In triangle $NOC$ the exterior angle is the angle of incidence, $i=\angle NOM+\angle NCM$; in triangle $NIC$, $\angle NCM=r+\angle NIM$, so $r=\angle NCM-\angle NIM$. Snell's law for small angles, $n_1i=n_2r$, gives $$n_1\left(\frac{h}{OM}+\frac{h}{MC}\right)=n_2\left(\frac{h}{MC}-\frac{h}{MI}\right),$$ that is, $$\frac{n_1}{OM}+\frac{n_2}{MI}=\frac{n_2-n_1}{MC}.$$ Now apply the Cartesian sign convention, with distances from the pole $M$: $OM=-u$, $MI=+v$, $MC=+R$. The height $h$ has cancelled, so every paraxial ray from $O$ passes through the same $I$ — an image is formed.

Thin Lenses: the Lens Maker's Formula and the Thin-Lens Formula

Take a thin lens of refractive index $n_2$ in a medium of index $n_1$, with faces of radii $R_1$ (the face the light meets first) and $R_2$. At the first face the object $O$ forms an image $I_1$ at $v_1$: $$\frac{n_2}{v_1}-\frac{n_1}{u}=\frac{n_2-n_1}{R_1}.$$ Before that image can form, the light meets the second face. For a thin lens $I_1$ is at the same distance $v_1$ from the second face, and acts as its object, with the light now going from $n_2$ back into $n_1$: $$\frac{n_1}{v}-\frac{n_2}{v_1}=\frac{n_1-n_2}{R_2}.$$ Adding, $v_1$ cancels: $$\frac{n_1}{v}-\frac{n_1}{u}=(n_2-n_1)\left(\frac1{R_1}-\frac1{R_2}\right).$$ For an object at infinity the image is at the focus, $v=f$. Dividing by $n_1$ and writing $n_{21}=n_2/n_1$ gives the two results below.

The magnification follows from the ray through the optical centre, which goes straight on: the object and image tips lie on one line through $O$, so $\dfrac{h'}{h}=\dfrac vu$ (no minus sign, unlike the mirror). A converging lens behaves for objects beyond $F_1$ like a concave mirror (real, inverted images) and for objects inside $F_1$ gives a virtual, erect, enlarged image — the magnifying glass. A diverging lens always gives a virtual, erect, diminished image of a real object, between the lens and $F_1$.

Power of a Lens and Lenses in Contact

A lens with a short focal length bends rays sharply: a ray parallel to the axis at height $h$ is turned through an angle of about $h/f$ to reach the focus. That turning ability is what opticians quote.

Place two thin lenses of focal lengths $f_1$ and $f_2$ in contact. An object at $u$ gives, through the first lens, an image at $v_1$ with $\dfrac1{v_1}-\dfrac1u=\dfrac1{f_1}$. This image is the object for the second lens — at the same place, since the lenses touch — and the final image is at $v$ with $\dfrac1v-\dfrac1{v_1}=\dfrac1{f_2}$. Adding the two equations, $v_1$ cancels: $$\frac1v-\frac1u=\frac1{f_1}+\frac1{f_2}.$$ The pair acts as a single thin lens whose focal length $f$ satisfies $\dfrac1f=\dfrac1{f_1}+\dfrac1{f_2}$.

This is why opticians work with powers: a $+2.00$ D lens and a $-0.50$ D lens together make a $+1.50$ D lens. Camera designers combine converging and diverging lenses of different glasses so that the colour errors of one cancel those of the other while the powers still add to the value they want.

Exercises