Curved Mirrors and the Mirror Formula

1Two mirrors, two jobs

At a blind bend or the exit of an underground car park, a round convex mirror on a pole lets a driver see traffic coming round the corner: the whole road fits into a mirror the size of a tray. A concave dish does the opposite job: it gathers sunlight from a wide area and sends it to one small region, hot enough to boil water. In S2 you located images in curved mirrors with ray diagrams and used the formula $\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\backslash frac\{1\}\{f\}\; =\; \backslash frac\{1\}\{u\}\; +\; \backslash frac\{1\}\{v\}$. In this lesson you prove that formula from the geometry of a ray diagram, find out when it holds, and measure a focal length from a graph.

2The terms, and why $f=\frac{r}{2}f\; =\; \backslash frac\{r\}\{2\}$

A spherical mirror is a piece cut from a hollow sphere. Its pole P is the centre of the reflecting surface; its centre of curvature C is the centre of the sphere, a distance $rr$ (the radius of curvature) from P; and the principal axis is the line through C and P. Rays parallel to the axis and close to it either converge to a point in front of a concave mirror or appear to diverge from a point behind a convex mirror. That point is the principal focus F (Figure 1.1).

Figure 1.1: Parallel rays and the principal focus. A concave mirror brings them to a real focus; a convex mirror spreads them as if from a virtual focus behind it.

Now follow one ray carefully (Figure 1.2). A ray parallel to the axis at height $hh$ strikes a concave mirror at M. Every radius of a sphere is perpendicular to its surface, so CM is the normal at M. The ray and the axis are parallel, so the angle of incidence equals the angle MCP; call it $\theta \backslash theta$. By the law of reflection the reflected ray also makes $\theta \backslash theta$ with CM. Triangle CXM therefore has equal angles $\theta \backslash theta$ at C and at M, so it is isosceles with $CX=XMCX\; =\; XM$. Dropping a perpendicular from X onto CM cuts CM (length $rr$) in half, so

For a ray close to the axis (a paraxial ray), $\theta \backslash theta$ is small and $\cos \theta \approx 1\backslash cos\backslash theta\; \backslash approx\; 1$, so $CX\approx \frac{r}{2}CX\; \backslash approx\; \backslash frac\{r\}\{2\}$: every paraxial ray crosses the axis at the same point, halfway between C and P. That point is F, and

Figure 1.2: The normal at M is the radius CM. A paraxial ray crosses the axis at F; a ray far from the axis (larger θ) crosses at X, nearer the mirror.

For a ray far from the axis, $\cos \theta \backslash cos\backslash theta$ is noticeably less than 1, so $CX>\frac{r}{2}CX\; >\; \backslash frac\{r\}\{2\}$ and the ray crosses the axis at X, between F and P. Rays at different heights then meet at different points and the image is blurred. This is spherical aberration, which you meet again in lesson 4.

3Deriving the mirror formula from a ray diagram

Figure 1.3 shows an object AB beyond C. Two rays from the top B are enough to locate its image, and both can be followed exactly:

• Ray 1 strikes the pole P. The axis is the normal at P, so the ray reflects at the same angle $ii$ on the other side of the axis.
• Ray 2 passes through C. It meets the mirror along a radius, so it strikes the mirror normally and returns along its own path.

Figure 1.3: The two rays used in the derivation. Both reflected rays pass through B′, so A′B′ is the image of AB.

Treat the part of the mirror near P as flat (the paraxial condition), so that A′ is a distance $vv$ from P. Then look for two pairs of similar triangles.

Triangles BAP and B′A′P. Both have a right angle on the axis and the angle $ii$ at P, so they are similar:

Triangles BAC and B′A′C. Both have a right angle on the axis, and the angles at C are vertically opposite, so they are similar:

Equate (1) and (2) and cross-multiply:

Divide every term by $uvruvr$:

Equation (1) also gives the linear magnification, $m=\frac{v}{u}m\; =\; \backslash frac\{v\}\{u\}$. The same steps work for a convex mirror and for objects at other positions, provided the signs below are used; that is what makes one formula serve every case.

4One sign convention for the whole unit

This unit uses the convention you met in S2 and S3, "real is positive". All distances are measured from the pole (for a lens, from the optical centre).

Some books use the Cartesian convention instead, in which distances measured in the direction the incident light travels are positive. The formulas then carry different signs, but every physical answer (where the image is, how big, real or virtual) comes out the same. Choose one convention and never mix the two in one calculation.

5Experiment 1.1: the focal length of a concave mirror from a graph

Apparatus: a concave mirror in a holder, a lamp box with a cross-wire object, a small white screen, and a metre rule (Figure 1.4).

Figure 1.4: Finding image positions. The mirror is turned very slightly so that the image falls on a screen beside the incident beam.

1. Place the object about 15 cm from the mirror. Move the screen until the image of the cross-wire is sharp. Record $uu$ and $vv$.
2. Repeat for at least five more object distances, up to about 50 cm.
3. Calculate $\frac{1}{u}\backslash frac\{1\}\{u\}$ and $\frac{1}{v}\backslash frac\{1\}\{v\}$ in ${\mathrm{m}}^{-1}\backslash mathrm\{m^\{-1\}\}$ and plot $\frac{1}{v}\backslash frac\{1\}\{v\}$ against $\frac{1}{u}\backslash frac\{1\}\{u\}$.

Why a straight line? Rearranging the mirror formula gives

which has the form $y=mx+cy\; =\; mx\; +\; c$ with gradient $-1-1$ and intercept $\frac{1}{f}\backslash frac\{1\}\{f\}$ on the $\frac{1}{v}\backslash frac\{1\}\{v\}$ axis. By symmetry the line also cuts the $\frac{1}{u}\backslash frac\{1\}\{u\}$ axis at $\frac{1}{f}\backslash frac\{1\}\{f\}$. Using both intercepts, and checking that the gradient is close to −1, tests the formula as well as measuring $ff$. A quick check: when the object sits at C, the image falls on the object itself ($u=v=ru\; =\; v\; =\; r$), so half that distance is $ff$.

Table 1.2: Illustrative results for one mirror.

Figure 1.5: The results of Table 1.2. The line has gradient −1.00, as the formula predicts.