Ray Optics and Optical InstrumentsReflection of Light by Spherical Mirrors
You already know the laws of reflection. The angle of reflection, between the reflected ray and the normal to the surface, equals the angle of incidence, between the incident ray and the normal. And the incident ray, the reflected ray and the normal at the point of incidence all lie in one plane (Fig. 9.1). These laws hold at every point of any reflecting surface, flat or curved. In this chapter we stick to curved surfaces that are parts of spheres.
The concave mirror below is part of a sphere whose centre is the centre of curvature C. Its geometric centre is the pole P, and the line through P and C is the principal axis.
A ray parallel to the principal axis strikes the mirror at M. Along which line is the normal to the mirror at M?
The normal at M is perpendicular to the tangent there, so it runs along the radius MC. That is how we apply the laws of reflection at a curved mirror: measure the angles from the radius through the point of incidence. The pole of a spherical mirror is its geometric centre; the geometric centre of a spherical lens is called its optical centre. For a lens, the principal axis is the line through the optical centre and its principal focus.
Sign convention
To derive formulae for mirrors and lenses we need a rule for the signs of distances. This book uses the Cartesian sign convention (Fig. 9.2):
- all distances are measured from the pole of the mirror or the optical centre of the lens, along the principal axis;
- distances measured in the same direction as the incident light are positive, and those measured against it are negative;
- heights measured upwards from the principal axis are positive, and heights measured downwards are negative.
We always draw the incident light travelling from left to right, so "along the light" means "to the right". With this one convention a single formula covers every spherical mirror, and another every spherical lens. Give each quantity its sign, then check.
Notice that a real object always has a negative u, that f is negative for a concave mirror and positive for a convex one, and that the sign of a height tells you at once whether the image is erect or inverted.
Focal length of spherical mirrors
When a parallel beam falls on a concave mirror, the reflected rays converge at a point F on the principal axis; on a convex mirror they spread out as if they came from a point F behind it (Fig. 9.3). F is the principal focus, and its distance from the pole is the focal length f. If the parallel beam is tilted a little, the rays converge in a plane through F at right angles to the axis, the focal plane.
All this is true only for paraxial rays: rays that strike the mirror close to the pole and make small angles with the principal axis. The mirror below has a radius of curvature R = 20 cm.
Send in parallel rays at different heights and see where they cross the axis.
height of the ray above the axis: h =
As the blue curve shows, the crossing point creeps towards the mirror for high rays, but for paraxial rays it settles on one point, F. Here is why that point is halfway to C.
A ray parallel to the axis strikes the mirror at M (Fig. 9.4). The radius CM is the normal at M, so the angle of incidence θ equals ∠MCP, and the reflected ray makes ∠MFP = 2θ with the axis. Drop the perpendicular MD on the axis. Then
For paraxial rays θ is small, so tan θ ≈ θ and tan 2θ ≈ 2θ, and Eq. (9.1) gives
For small θ the point D is very close to P, so FD = f and CD = R:
The mirror equation
If rays from a point actually meet at another point after reflection or refraction, that point is a real image of the first. If they do not meet but only appear to diverge from a point when extended backwards, the image is virtual. An image is a point-to-point match between object and image set up by the mirror or lens.
Every point of an object sends rays in all directions, and each of them passes through the image point. To find that point we need only two rays, and these four are the easiest:
- a ray parallel to the principal axis, which is reflected through the focus;
- a ray through the centre of curvature of a concave mirror (or heading for it, for a convex mirror), which retraces its path;
- a ray through the focus of a concave mirror (or heading for it, for a convex mirror), which is reflected parallel to the axis;
- a ray striking the pole at any angle, which reflects according to the laws of reflection.
The figure shows an object AB beyond C and its real image A′B′ (Fig. 9.5). We now find the relation between the object distance u, the image distance v and the focal length f.
For paraxial rays MP can be treated as a straight line at right angles to the axis, so PM = AB. The triangle A′B′F is similar to the triangle MPF, and since ∠APB = ∠A′PB′, the triangle A′B′P is similar to the triangle ABP:
These are lengths, with no signs yet. Light travels from the object towards the mirror, so that direction is positive. To reach B′, F and B from P we travel against the incident light, so B′P =
Substituting in Eq. (9.6):
Dividing by v gives the mirror equation:
The linear magnification m is the ratio of the height of the image h′ to the height of the object h, each with its sign:
From the triangles A′B′P and ABP, B′A′/BA = B′P/BP. With signs this is −h′/h = −v/−u, so
Note
We derived Eqs. (9.7) and (9.9) for a real, inverted image formed by a concave mirror. Used with the sign convention, they hold for every spherical mirror, concave or convex, and for real and virtual images alike (Fig. 9.6). The next step lets you check that.
One Equation, Every Image
Both mirrors below have a focal length of 12 cm: f = −12 cm for the concave one, +12 cm for the convex one. The readout gives u, v and m with their signs.
Drag the object, the blue arrow, along the axis.
The blue curve is v plotted against u. For the concave mirror it has two branches on either side of the dashed lines u = f and v = f: an object beyond F gives a real image (v negative), an object inside F a virtual one (v positive). The convex mirror's curve stays between 0 and f: its image is always virtual, erect and diminished, wherever the object is.
Finding f in the Lab
In the laboratory the focal length of a concave mirror is found by the u–v method: a candle stands in front of the mirror, and a screen is moved until a sharp image of the flame falls on it. The mirror below has f = 20 cm, but pretend you do not know that. Move the candle and record four readings.
distance of the candle from the mirror =
For a real image u, v and f are all negative, so the mirror equation in magnitudes reads 1/|v| + 1/|u| = 1/|f|. That is a straight line of slope −1. The line through your readings cuts both axes at 1/|f| = 0.05 cm⁻¹, so f = 1/0.05 = 20 cm, and the mirror is concave: f = −20 cm.
Tip
In the practical exam, draw the best straight line through all your points, read both intercepts, and average the two values of f you get. Points that lie far off the line usually mean the image on the screen was not quite sharp.
Example 9.1: The lower half of the reflecting surface of a concave mirror is covered with an opaque, non-reflecting material. What happens to the image of an object placed in front of the mirror?
Below, seven rays from the tip of the object reach seven parts of the mirror, and all of them meet again at the tip of the image. Cover the lower half of the mirror and watch the image.
Solution: you might expect the image to show only half of the object. But the laws of reflection hold at every point of the uncovered part, so it still forms an image of the whole object. Since the reflecting area is halved, the image is less intense: about half as bright here.
Slide the phone along the axis and complete all three missions.
nearer end of the phone:
Solution: each slice of the phone is a small object at its own distance u, so each has its own image distance v and its own magnification −v/u. The end that sits in the plane through C (at right angles to the axis) is imaged on the same plane, at the same size: B′C = BC. Slices nearer F are imaged farther away and larger, so the image is stretched unevenly. Yes, the distortion depends on the position: it is worst when the phone is close to F, and mild when the phone is well beyond C.
Example 9.3: An object is placed at (i) 10 cm and (ii) 5 cm in front of a concave mirror of radius of curvature 15 cm. Find the position, nature and magnification of the image in each case.
Solution: the focal length is f = −15/2 cm = −7.5 cm.
(i) u = −10 cm. From Eq. (9.7), , so v =
(ii) u = −5 cm. Now , so v =
Type negative answers with a minus sign: the sign is part of the answer, because it tells you which side of the mirror the image is on.
Example 9.4: Sitting in a parked car, you notice a jogger approaching in the side-view mirror, a convex mirror with R = 2 m. The jogger runs at 5 m/s. How fast does the image appear to move when the jogger is (a) 39 m, (b) 29 m, (c) 19 m and (d) 9 m away?
Solution: from the mirror equation, . For a convex mirror with R = 2 m, f = +1 m. For u = −39 m, m. One second later the jogger is at u = −34 m, and the image is at 34/35 m. So in that second the image moves 39/40 − 34/35 = 1/280 m.
Move the jogger and read the image speed at each distance.
jogger’s distance:
The average speeds of the image are 1/280, 1/150, 1/60 and 1/10 m/s. The jogger keeps a constant speed, yet the blue curve shoots up as the distance shrinks: the image seems to rush towards you. Anyone sitting in a parked car or bus can notice this, and so can a driver when the vehicle behind closes in at a steady speed.
Speed Round: Mirror Equation
and , with every distance measured from the pole. Get five right to move on.