Thin Lenses and the Lens Formula

1One lens, three jobs

The same kind of converging lens can throw a large, bright picture onto a classroom wall (a projector), shrink a whole landscape onto a small sensor (a camera), or make small print look bigger (a magnifying glass). What decides which job it does is only where the object sits compared with the lens's focal length. In S3 you drew ray diagrams for lenses and used the lens equation. This lesson explains why a lens has a focal length at all, derives the lens formula from a ray diagram, and gives you three graphical ways to measure $ff$.

2Types and characteristics of lenses

A lens is a piece of transparent material with two refracting surfaces, at least one of them curved. A lens that is thicker at the centre bends parallel light towards the axis and is converging; one that is thinner at the centre spreads it out and is diverging (Figure 3.1). In air, this holds whenever the lens material has the higher refractive index.

Figure 3.1: The six common lens shapes.

Each lens has an optical centre O, through which rays pass without deviation (the two faces there are nearly parallel, and a thin lens produces no noticeable sideways shift), and a principal axis through O at right angles to the lens. Rays parallel to the axis converge to, or appear to diverge from, a principal focus. Light can pass either way, so a lens has two principal foci, F and F′, one on each side and equally far from O. The plane through F at right angles to the axis is the focal plane: parallel rays arriving at an angle to the axis are focused at one point in that plane.

Why does a lens have a focal length? Each face refracts the light. For a thin lens of refractive index $nn$ in air with faces of radii $R_{1}R\_1$ and $R_{2}R\_2$, applying Snell's law at both faces to paraxial rays gives the lens-maker's formula

where, in the real-is-positive convention, a radius is positive for a face that bulges outwards (convex) and negative for a hollow (concave) face, and a plane face has $\frac{1}{R}=0\backslash frac\{1\}\{R\}\; =\; 0$. Two things follow at once: a more strongly curved lens (smaller $RR$) has a shorter focal length, and because $nn$ is slightly different for each colour, so is $ff$ (lesson 4).

3Ray diagrams and the lens formula

Three rays from the top B of an object are easy to draw (Figure 3.2): a ray parallel to the axis refracts through F′; a ray through the optical centre goes straight on; a ray through F leaves parallel to the axis. Any two locate the image; the third is a check. In a thin-lens diagram we draw all the bending at the central line of the lens.

Figure 3.2: Ray diagram for an object between F and 2F. The image is real, inverted and magnified.

Derivation. Let $OA=uOA\; =\; u$, $O{A}^{\mathrm{\prime }}=vOA\text{'}\; =\; v$ and $O{F}^{\mathrm{\prime }}=fOF\text{'}\; =\; f$. The ray parallel to the axis meets the lens at C, with $OC=ABOC\; =\; AB$.

Triangles ABO and A′B′O are similar (right angles on the axis, vertically opposite angles at O):

Triangles OCF′ and A′B′F′ are similar (right angles, vertically opposite angles at F′), and $OC=ABOC\; =\; AB$:

Equating (1) and (2): $\frac{v}{u}=\frac{v-f}{f}\backslash frac\{v\}\{u\}\; =\; \backslash frac\{v\; -\; f\}\{f\}$, so $vf=uv-ufvf\; =\; uv\; -\; uf$. Dividing every term by $uvfuvf$:

It is the same formula as for a mirror, with the same sign convention: real objects and images positive, virtual ones negative, $ff$ positive for converging and negative for diverging lenses.

A diverging lens always forms a virtual, upright, diminished image of a real object, between F and the lens (Figure 3.3b).

Figure 3.3: Virtual images. Dashed lines show where the refracted rays appear to come from.

4Experiment 3.1: graphical determination of focal length

Apparatus: a converging lens in a holder, an illuminated object (a cross-wire or a lit letter of known height $h_{o}h\_o$), a white screen with a millimetre scale, and a metre rule.

1. Set the object at a distance $uu$ a little more than $ff$ from the lens. Move the screen until the image is sharp. Record $vv$ and the image height $h_{i}h\_i$.
2. Repeat for at least five more object distances. Calculate $m=\frac{h_i}{h_o}m\; =\; \backslash frac\{h\_i\}\{h\_o\}$.
3. Plot $mm$ against $vv$.

Why a straight line? Multiply the lens formula by $vv$: $\frac{v}{u}+1=\frac{v}{f}\backslash frac\{v\}\{u\}\; +\; 1\; =\; \backslash frac\{v\}\{f\}$, so

which is a straight line of gradient $\frac{1}{f}\backslash frac\{1\}\{f\}$ that cuts the $mm$ axis at −1 and the $vv$ axis at $v=fv\; =\; f$. Other straight-line graphs work too: $\frac{1}{v}\backslash frac\{1\}\{v\}$ against $\frac{1}{u}\backslash frac\{1\}\{u\}$ (as for the mirror in lesson 1), or $uvuv$ against $(u+v)(u\; +\; v)$, whose gradient is $ff$ because $uv=f(u+v)uv\; =\; f(u\; +\; v)$.

Table 3.1: Illustrative results.

Figure 3.4: Magnification against image distance for the results in Table 3.1.