
Optional unit · institution-dependent · Interactive lesson
Spherical Mirrors
A flat mirror only ever hands back an image the same size as the object. Bend the surface and the reflected rays begin to meet: the mirror gains a focal length, a magnification you can choose, and a sign convention you break at your peril. Draw the rays first, then do the arithmetic.
UPII-13.04 · Lesson 4 of 8 in this unit. Position is not a completion score.Optional unit · institution-dependent. Your university may teach this optics or introductory modern-physics material separately. Follow your institution’s required sequence. Optional work is not a required completion gate.
Build the model
Curve a mirror and its reflected rays acquire a focus at R/2.
The key idea
A spherical mirror is a polished cap of a sphere of radius R, and within the paraxial region near its axis it sends every ray that arrived parallel to that axis to a single focal point at a distance f = R/2 from the vertex: really through it, in front of the mirror, if the cap is concave, and only by backward extension, behind the mirror, if it is convex.
Example. A concave mirror ground to R = 40 cm has f = 20 cm: axial sunlight converges 20 cm in front of the vertex, while a ray entering 6 cm off-axis actually crosses about 2.3 mm short of that point.
The complete physical model and assumptions
A sphere is not a focusing shape, but near its axis it behaves like one. Apply the law of reflection at each point of a spherical cap, where the normal is the radius itself: a ray arriving parallel to the axis at height h turns through twice the angle θ its radius makes with the axis, and crosses the axis a distance R/(2 cos θ) from the centre of curvature. Let h shrink so cos θ → 1, and every such ray crosses at the same place, f = R/2 in front of the vertex. That single number now replaces the whole sphere. The same paraxial step — every sine and tangent replaced by its angle — turns the geometry into the mirror equation 1/s + 1/s′ = 1/f, and the one ray that strikes the vertex hands over the magnification m = −s′/s. Everything after that is sign discipline: s′ > 0 means the light really converges in front of the mirror and a card held there catches the image; s′ < 0 means only the backward extensions meet, behind the glass, and nothing is there to catch. The model stops where the paraxial step was taken. Open the aperture and rays at height h cross about h²/(4R) short of the paraxial focus, so a fast mirror images a star as a small caustic patch rather than a point — spherical aberration, the price of grinding a sphere instead of a paraboloid.
1A sphere focuses only near its axis
Reflection alone gives no focus: a flat mirror sends parallel light away still parallel. Curving the surface gives every point a different normal, and on a sphere that normal is the point's own radius. Take a ray arriving parallel to the axis at height h on a concave mirror of radius R. It strikes where the radius makes an angle θ with the axis, with sin θ = h/R, and because the incoming ray is parallel to the axis, θ is also the angle of incidence. Reflection turns the ray through 2θ, and the triangle made by the centre of curvature C, the strike point and the axis crossing is isosceles — two equal angles θ — which puts the crossing R/(2 cos θ) from C, that is, R − R/(2 cos θ) in front of the vertex. Nothing so far assumes h is small. Now assume it: cos θ → 1 and every ray crosses at R − R/2 = R/2. For R = 40 cm the focus lies 20 cm in front of the vertex, and a ray 6 cm off-axis crosses 2.3 mm short of it.
2Three angles give the mirror equation
Now let the light leave a point O on the axis, a distance s in front of the vertex, and follow one ray that meets the mirror at height y. Call α the angle the incoming ray makes with the axis, β the angle of the radius (the normal) and γ the angle of the reflected ray. The angle of incidence is β − α, the angle of reflection is γ − β, and the law of reflection sets them equal, so 2β = α + γ. Paraxially each angle equals its own tangent: α ≈ y/s, β ≈ y/R, γ ≈ y/s′. Substituting and cancelling y gives 2/R = 1/s + 1/s′. That cancellation is the whole point — y has vanished, so every paraxial ray from O arrives at the same s′, which is what makes an image rather than a smear. Writing f = R/2 puts it in working form, 1/s + 1/s′ = 1/f.
3Principal rays locate the image before any arithmetic
Three rays from the object tip are worth drawing because you already know where each goes. A ray parallel to the axis reflects through F. A ray through F reflects parallel to the axis. A ray aimed along the radius, at C, strikes the surface head-on and returns along itself. Any two of them cross at the image tip and the third is your check. A fourth ray strikes the vertex, where the axis is the normal, and reflects at an equal angle on the other side of it; the two similar triangles it makes are where m = −s′/s comes from. On a convex mirror the same three rules hold with F and C behind the surface: the parallel ray reflects as though it had come from F, the ray aimed at F returns parallel, the ray aimed at C returns along itself. Those reflections diverge, so it is their backward extensions that meet — drawn dashed, behind the mirror, giving a virtual image every time.
4Declare the signs once, then never argue with them
Measure everything from the vertex, with light arriving from the object side. Then s > 0 for a real object in front; s′ > 0 for an image in front, where rays truly cross and a card catches it; s′ < 0 for an image behind, where only the extensions meet. R and f are positive for a concave mirror, whose C lies in front, and negative for a convex one. Magnification is m = −s′/s, negative when inverted. Run a concave mirror of f = 20 cm through the whole range: at s = 60 cm, s′ = 30 cm and m = −0.50, real, inverted, reduced; at s = 40 cm, which is R, s′ = 40 cm and m = −1.00, so the image lands back on the object; at s = 30 cm, s′ = 60 cm and m = −2.00, real, inverted, magnified; at s = 20 cm the reflected rays leave parallel and there is no image; at s = 10 cm, s′ = −20 cm and m = +2.00, virtual, upright, magnified — the shaving mirror.
5A convex mirror packs the world into one focal length
A car's passenger-side mirror is convex with R = −2.0 m, so f = −1.0 m. A vehicle 8.0 m back gives 1/s′ = 1/f − 1/s = −1.000 − 0.125 = −1.125 m⁻¹, so s′ = −0.89 m and m = −s′/s = +0.11: an upright virtual image 0.89 m behind the glass at one ninth scale, so a 4.5 m car appears 0.50 m long. Send the object to infinity and s′ → f = −1.0 m; bring it against the glass and s′ → 0. Every image this mirror can form therefore lies inside the 1.0 m of virtual space behind it, upright and reduced, which is exactly why it is fitted — a wide field of view squeezed into a small pane. It is also why the pane carries a warning: the eye reads distance partly from angular size, and this mirror shrinks everything by a factor that itself shrinks with range.
6Open the aperture and the focus smears
Every step above sent cos θ to 1. Put it back. A ray at height h crosses the axis R − R/(2 cos θ) from the vertex, short of R/2 by (R/2)(sec θ − 1) ≈ h²/(4R). Rays at different heights therefore cross at different points, and the envelope of those crossings is a caustic surface rather than a point. This is spherical aberration, and it is a property of the sphere itself, not a fault in the grinding. It grows fast: the longitudinal miss goes as h² and the transverse blur radius in the paraxial focal plane as about h³/(2R²), so halving the aperture sharpens the image roughly eightfold. Two cures exist. Stop the mirror down and pay in light. Or change the shape — a paraboloid brings all rays parallel to its axis to one exact point, which is why telescope primaries are parabolic, at the cost of blurring everything off-axis.
Key equations
Read the units, assumptions and limits before substituting.
Paraxial focal length
f = R/2
One number replaces the whole sphere: the focus sits halfway from the vertex to the centre of curvature.
R is the radius of curvature in m, positive for a concave mirror and negative for a convex one; f takes the same sign.The mirror equation
1/s + 1/s′ = 1/f
Fix f and s and the image has no freedom left: s′ is determined, and its sign says real or virtual.
s object distance, s′ image distance, f focal length, all in m and all measured from the vertex.Transverse magnification
m = h′/h = −s′/s
Its sign reports orientation — negative means inverted — and its size reports the stretch across the axis.
h object height and h′ image height in the same unit; m is a pure number.Magnification from the focal length
m = f/(f − s)
The design form: you are usually told the magnification wanted, not the image distance.
f and s in the same unit; obtained by eliminating s′ between the mirror equation and m = −s′/s.Axial stretch of the image
ds′/ds = −(s′/s)² = −m²
Depth is magnified as m², not m, so a 4× image is 16 times as sensitive to a slip in focus.
Dimensionless; differentiate 1/s + 1/s′ = 1/f at fixed f. Negative: the image retreats as the object advances.Longitudinal spherical aberration
Δ ≈ h²/(4R)
Sets the usable aperture: the miss grows as h², so stopping a mirror down sharpens it quickly.
h is the ray's height above the axis in m; Δ in m is how far short of the paraxial focus it crosses.Interactive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Slide the object in towards R and watch the image slide out past C and grow; at s = R exactly, object and image stand together on C at the same size. Then change R and watch F move to half of it.
Live results
- FOCAL LENGTH f = R/2
- 20 cm
- IMAGE DISTANCE s′
- 30.0 cm
- MAGNIFICATION m
- -0.50
- IMAGE HEIGHT h′
- -6.0 cm
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Radius of curvature R: 40 cm; Object distance s: 60 cm
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
Covering half a concave mirror does not cut the image in half. The expectation is that the top half of the mirror makes the top half of the image, because that is how a photograph behaves when it is torn. Imaging does not work that way: every part of the mirror that the object can see collects light from every point of the object, and the mirror equation sends all of it to the same image point. Mask the lower half and the entire image survives — it simply loses about half its light and dims, while its aberrations change because the surviving rays now strike one side of the axis only. The neighbouring slip is to say the image is on the mirror. A real image hangs in the air at s′, where a card intercepts it; a virtual image has nothing at its location at all, and the eye is merely receiving a cone of rays that diverge as though from a point behind the glass.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
A concave shaving mirror has a radius of curvature of 50 cm. A face sits 15 cm in front of it. Find the focal length, the image distance and the magnification, and say whether a card could catch the image.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- f = R/2 = 25 cm, positive because the mirror is concave.
- Rearrange the mirror equation: 1/s′ = 1/f − 1/s = 1/25 − 1/15. Over a common denominator of 75: 3/75 − 5/75 = −2/75 cm⁻¹.
- So s′ = −75/2 = −37.5 cm. The negative sign places the image 37.5 cm behind the mirror, so it is virtual.
- m = −s′/s = −(−37.5)/15 = +2.50: upright and two and a half times life size, which is the point of the mirror.
- Nothing crosses in front of the mirror — the reflected rays there are still diverging — so no card can catch it. Only an eye, receiving that diverging cone, sees the image.
Answer. f = 25 cm, s′ = −37.5 cm (37.5 cm behind the mirror) and m = +2.50: a virtual, upright image 2.50 times life size, and no card can catch it.
MediumExample 2
A concave mirror throws a real image of a 2.0 cm tall lamp filament onto a screen 1.80 m in front of the mirror, and the image is 4.0 times the filament's height. Find where the filament sits, the focal length and radius the mirror needs, the image height, and how far the image moves if the filament is nudged 1.0 mm further away.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- A real image caught on a screen is inverted, so m = −4.0. With m = −s′/s and s′ = 1.80 m, that gives s = s′/4.0 = 0.45 m.
- Mirror equation: 1/f = 1/s + 1/s′ = 1/0.45 + 1/1.80 = 2.2222 + 0.5556 = 2.7778 m⁻¹, so f = 0.36 m and R = 2f = 0.72 m.
- Check the regime: f = 0.36 m < s = 0.45 m < R = 0.72 m, and the band between F and C is exactly the one that gives a real, inverted, magnified image.
- Image height: h′ = m h = −4.0 × 2.0 cm = −8.0 cm, the minus sign meaning inverted.
- Focus sensitivity: ds′/ds = −m² = −16, so pushing the filament 1.0 mm further from the mirror pulls the image 16 mm towards it. Solving exactly gives s′ = 1.784 m, a shift of 15.8 mm — the screen end is sixteen times fussier than the lamp end.
Answer. s = 0.45 m, f = 0.36 m and R = 0.72 m; the image is 8.0 cm tall and inverted, and a 1.0 mm nudge of the filament moves it about 16 mm.
HardExample 3
A concave mirror of radius 1.20 m collects light arriving parallel to its axis from a star. Find where a ray striking 15 cm above the axis crosses the axis, how far that is from the paraxial focus, and the radius of the blur it makes in the paraxial focal plane. Then say what stopping the aperture down to 7.5 cm does.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 6 steps revealed.
- Paraxial focus first: f = R/2 = 0.600 m in front of the vertex.
- The normal at the strike point is the radius, so the angle of incidence obeys sin θ = h/R = 0.15/1.20 = 0.1250, giving θ = 7.181° and cos θ = 0.99216.
- The isosceles triangle puts the crossing R/(2 cos θ) = 0.6047 m from the centre of curvature, so it lies R − 0.6047 = 1.200 − 0.6047 = 0.5953 m in front of the vertex.
- Longitudinal miss: Δ = 0.600 − 0.5953 = 4.7 mm, which the small-angle estimate h²/(4R) = 0.15²/(4 × 1.20) = 4.7 mm reproduces.
- That ray leaves the mirror at 2θ = 14.36° to the axis, so by the paraxial focal plane it has climbed Δ tan 2θ = 4.743 mm × 0.2560 = 1.2 mm off the axis: the star is a disc about 2.4 mm across, not a point.
- The blur radius goes as h³/(2R²), so halving h to 7.5 cm divides it by about eight, to 0.15 mm — paid for with three quarters of the light, since the collecting area falls fourfold.
Answer. The ray crosses 0.5953 m from the vertex, 4.7 mm short of the paraxial focus, and blurs the star into a disc about 2.4 mm across; stopping down to 7.5 cm cuts the blur radius to about 0.15 mm.
Quick check
Select an answer and explain it before checking.
Review & continue
A spherical mirror is a polished cap of a sphere of radius R, and within the paraxial region near its axis it sends every ray that arrived parallel to that axis to a single focal point at a distance f = R/2 from the vertex: really through it, in front of the mirror, if the cap is concave, and only by backward extension, behind the mirror, if it is convex.
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.
Optional supporting lessons
These chapter resources can help with earlier ideas; they are not additional required Physics II completion tasks.