Focal length of spherical mirrors

Stand a shaving mirror or a torch reflector on the table, aim a thin pencil of sunlight at it parallel to its axis, and something almost magical happens: every ray you send in, no matter how far from the centre, bends and crosses the axis at the same single point. Move a sheet of paper along the axis and at one exact distance the scattered spot of light collapses into a brilliant pinprick — hot enough, if the mirror is large and the sun strong, to singe the paper. That special point is the principal focus, and the distance from the mirror's pole to it is the focal length, $f$ . It is one of the two or three numbers that completely describe how a spherical mirror behaves. The remarkable thing is how simply $f$ connects to the mirror's shape. A spherical mirror is just a cap sliced off a sphere, and that sphere has one geometric fingerprint: its radius of curvature, $R$ , the distance from the mirror's centre of curvature to any point on its surface. You might expect the relation between $f$ and $R$ to need calculus, trigonometric tables, or at least a page of algebra. It needs neither. For rays that stay close to the axis, plain geometry — the law of reflection plus one small-angle approximation — collapses everything to $f = R/2$ : the focus sits exactly halfway between the pole and the centre of curvature, for every spherical mirror, concave or convex, without exception in form. This section picks up the thread right after sign convention and pushes it to the one formula every ray-optics numerical leans on before the mirror equation takes over. You will see why the paraxial approximation is what makes $f = R/2$ true rather than approximate garbage, watch the geometry play out in a labelled ray diagram, work through a board-style numerical and a sharper competitive one, and collect the exact places students lose marks — mixing up the sign of $R$ with the sign of $f$ , or believing "shorter focal length" means "weaker mirror" when it means the reverse.

Key takeaways

Every spherical mirror has one focus. Rays parallel to the principal axis converge at (concave) or appear to diverge from (convex) a single point $F$ , called the principal focus, at distance $f$ from the pole $P$ .
The defining relation is $f = R/2$ . The focal length is exactly half the radius of curvature — proved by the law of reflection plus the small-angle (paraxial) approximation, valid for both concave and convex mirrors in the same algebraic form.
Signs must match the Cartesian convention. For a concave mirror both $f$ and $R$ are negative; for a convex mirror both are positive — the magnitude relation $|f| = |R|/2$ never changes, only the sign attached to it.
$f = R/2$ is a paraxial result, not an exact one. It holds strictly only for rays close to and nearly parallel to the principal axis; rays striking near the rim of a large-aperture mirror miss the paraxial focus slightly (spherical aberration), a qualification the syllabus mentions but does not ask you to quantify.
Smaller radius means shorter focal length means a more powerful mirror. A mirror cut from a small sphere is sharply curved, bends light through a larger angle, and brings rays to a focus close to the pole; a nearly flat mirror (large $R$ ) has a very large $f$ and behaves almost like a plane mirror.
The relation is a bridge formula. It converts between the mirror's physical shape ( $R$ , which you can sometimes measure or is given directly) and the optical quantity ( $f$ ) that the mirror formula $1/v + 1/u = 1/f$ actually needs.
Key idea: for a spherical mirror, paraxial reflection geometry forces the focal length to be exactly half the radius of curvature, f = R/2, with matching signs for concave and convex mirrors alike.

The core insight

A spherical mirror's principal focus sits at the midpoint of the line from the pole to the centre of curvature because paraxial reflection geometry forces $f = R/2$ , with $f$ and $R$ carrying the same sign for a given mirror.

Where it fits in the syllabus

This sits inside CBSE Class 12 Unit IX, "Ray Optics and Optical Instruments," directly between sign convention for mirrors and the mirror formula $1/v + 1/u = 1/f$ — both of which are live, examinable topics in the 2025-26 prescription. The syllabus lists spherical mirrors and the mirror formula explicitly, and $f = R/2$ is the standard bridge every NCERT-based numerical uses to convert a given radius of curvature into the focal length the mirror formula demands. Nothing here strays beyond the prescribed scope: the derivation uses only the law of reflection and small-angle geometry, no calculus, and large-aperture effects (spherical aberration) are mentioned only as the one qualitative caveat the syllabus itself expects, never as a derivation target.

The physical picture

Picture a concave mirror — a small cap sliced from a sphere of radius $R$ , silvered on its inner (concave) face — with pole $P$ at the centre of the cap, centre of curvature $C$ marking the sphere's own centre, and the principal axis running through both. A ray travelling parallel to this axis strikes the mirror at some point $M$ close to $P$ and reflects. Because $C$ is the centre of the sphere, the line $CM$ is a radius of the mirror at that point and therefore the normal to the mirror's surface there — this single geometric fact is the entire key to the derivation below. The law of reflection then fixes exactly how the ray bends, and simple triangle geometry shows every such parallel ray, regardless of where it strikes (as long as $M$ stays close to the axis), crosses the axis at the same point $F$ .

f = PF, \qquad R = PC, \qquad f = \frac{R}{2}

where: $f$ is the focal length (pole to focus), $R$ is the radius of curvature (pole to centre of curvature), and the relation says the focus is the midpoint of segment $PC$ . In the Cartesian sign convention used throughout this chapter, distances measured against the direction of incident light are negative; for a concave mirror the pole is taken as origin with light travelling in the positive direction, so both $C$ and $F$ lie on the "wrong" side and $f = -|f|$ , $R = -|R|$ together — the ratio $f/R = 1/2$ stays positive and unaffected by which sign convention you adopt, which is exactly why the formula looks identical for concave and convex mirrors.

Figure

Ray diagram of a concave mirror showing a ray parallel to the principal axis striking the mirror and reflecting through the principal focus, with pole, focus, and centre of curvature marked and the focal length shown as half the radius of curvature

The picture makes two things visible at once. First, the focus $F$ sits between $P$ and $C$ , never beyond $C$ — it is the midpoint. Second, the whole construction only holds together because $M$ was drawn close to the axis; drag $M$ far up toward the rim of a wide mirror and the reflected ray would cross the axis noticeably closer to the mirror than $F$ , a real effect called spherical aberration that blurs the sharp point-focus of a wide-aperture mirror. Because a smaller $R$ means $C$ (and hence $F$ , at half that distance) sits closer to $P$ , a more sharply curved mirror bends the same parallel ray through a larger angle and brings it to a focus nearer the pole — exactly the intuition that "small radius, short focal length, strong mirror" captures.

Derivation AWhy the focus sits exactly halfway between the pole and the centre of curvature

The goal of this derivation is to show, using nothing beyond the law of reflection and the small-angle approximation for rays close to the principal axis, that a ray travelling parallel to the axis and striking a spherical mirror at a point $M$ near the pole always reflects through a single fixed point $F$ located exactly midway between the pole $P$ and the centre of curvature $C$ , so that the focal length obeys $f = PF = \tfrac12\,PC = R/2$ regardless of where on the mirror (within the paraxial region) the ray happens to strike.

Step 1. Set up the triangle. Let a ray parallel to the principal axis strike the concave mirror at point $M$ , close enough to the pole $P$ that $M$ can be treated as lying almost on the axis for distance purposes (the paraxial approximation). Since $C$ is the centre of the sphere the mirror is cut from, $CM$ is a radius of that sphere, and a radius is always perpendicular to the sphere's surface at the point it touches — so $CM$ is the normal to the mirror at $M$ . Let $\theta$ be the angle the incident ray makes with this normal $CM$ .

Step 2. Apply the law of reflection. The reflected ray leaves $M$ making the same angle $\theta$ with the normal $CM$ , on the other side of it. Because the incident ray is parallel to the principal axis, the angle the incident ray makes with the normal, $\theta$ , equals the angle $MCP$ that the normal itself makes with the axis (they are alternate angles formed by a transversal, $CM$ , cutting two parallel lines — the incident ray and the principal axis). So angle $MCF = \theta$ as well, where $F$ is the point where the reflected ray crosses the axis.

Step 3. Read off the triangle $MCF$ . In triangle $MCF$ , the angle at $C$ is $\theta$ (from Step 2), and the reflected ray makes angle $\theta$ with the normal $CM$ at $M$ , which means the angle $CMF$ inside the triangle is also $\theta$ . A triangle with two equal angles has the two sides opposite those angles equal, so triangle $MCF$ is isosceles with $FM = FC$ .

Step 4. Apply the paraxial approximation and finish. Because $M$ is close to the axis, $M$ lies almost exactly above $P$ , so $FM \approx FP$ to the accuracy the paraxial approximation allows. Combining this with $FM = FC$ from Step 3 gives $FP \approx FC$ — the point $F$ is (to paraxial accuracy) equidistant from $P$ and from $C$ , which are themselves both on the axis with $C$ farther out. The only point on segment $PC$ equidistant from both ends is its midpoint, so $F$ is the midpoint of $PC$ : $PF = \tfrac12 PC$ , i.e. $f = R/2$ . Because the argument never used which point $M$ was chosen (only that it stayed paraxial), every parallel ray converges at this same $F$ , confirming it is a single, well-defined focus — and the identical argument, run with the mirror convex instead of concave, gives the same relation $f = R/2$ with the focus now virtual, behind the mirror.

Worked examples

Board level - radius of curvature from a required focal length

Question: A concave mirror is to be selected so that parallel rays converge at a point 12 cm in front of it. What should be the radius of curvature of the mirror, and where does the centre of curvature lie?

Step 1. Identify the given quantity in words first. "Parallel rays converge at a point 12 cm in front of it" is exactly the definition of the principal focus, so the focal length has magnitude $|f| = 12$ cm.

Step 2. Attach the correct sign. The mirror is concave, and in the Cartesian sign convention (pole as origin, incident light taken as travelling in the negative direction for a mirror facing the object) a concave mirror's real focus in front of it is assigned a negative value: $f = -12$ cm.

Step 3. Apply $f = R/2$ with signs carried through consistently: $R = 2f = 2 \times (-12\ \text{cm}) = -24\ \text{cm}$ .

Step 4. State the physical answer. The magnitude tells you the mirror must be cut from a sphere of radius 24 cm; the negative sign, consistent with $f$ 's negative sign, simply confirms the centre of curvature $C$ lies in front of the mirror (same side as the focus and the object), 24 cm from the pole, twice as far out as $F$ . The required radius of curvature is 24 cm (concave), with $C$ 24 cm in front of the pole.

Competitive level - ranking three mirrors by converging power, including a convex one

Question: Three mirrors are available: mirror A, concave, $R = 20$ cm; mirror B, concave, $R = 40$ cm; mirror C, convex, $R = 60$ cm. Rank them by how strongly they converge (or diverge) parallel light, and state the distance and nature (real or virtual) of the focus for each.

Step 1. Convert every radius to a focal length using $f = R/2$ , working in magnitudes first and restoring signs after: $|f_A| = 20/2 = 10$ cm, $|f_B| = 40/2 = 20$ cm, $|f_C| = 60/2 = 30$ cm.

Step 2. Restore signs from mirror type. Concave mirrors A and B take negative $f$ (real focus in front): $f_A = -10$ cm, $f_B = -20$ cm. Convex mirror C takes positive $f$ (virtual focus behind): $f_C = +30$ cm.

Step 3. Rank by converging strength. Converging (and diverging) power scales inversely with $|f|$ — a shorter focal length means a more sharply curved mirror bending light through a larger angle. Ordering by $|f|$ smallest to largest: A (10 cm) is the strongest converger, then B (20 cm), then C (30 cm) is the weakest — and C diverges rather than converges, since it is convex.

Step 4. State the focus for each. Mirror A brings parallel rays to a real focus 10 cm in front of it — a screen placed there would show a sharp bright point. Mirror B brings them to a real focus 20 cm in front. Mirror C never actually focuses the rays at all; it spreads them out as if they came from a virtual point 30 cm behind the mirror, which is why a convex mirror is used as a wide-view vehicle mirror rather than to burn paper. Ranking (strongest to weakest converger): A ( $f=10$ cm, real) $>$ B ( $f=20$ cm, real) $>$ C ( $f=30$ cm, virtual, diverging). Timing note: this should take under ninety seconds in an exam. The shortcut is to never restore the sign until the very last step — compute all three $|f|$ values from $|R|/2$ in one line, rank them purely by size (smallest magnitude = strongest mirror, full stop, regardless of concave or convex), and only then attach + for convex and − for concave to state the final positions. Trying to juggle signs while ranking is where marks get lost under time pressure.

Common traps

Writing $f = R/2$ correctly but then forgetting to give $f$ and $R$ the same sign — a concave mirror needs both negative, a convex mirror needs both positive, in the Cartesian convention used everywhere else in this chapter; mixing a positive $R$ with a negative $f$ (or vice versa) for the same mirror is an internal contradiction.
Treating $f = R/2$ as exact for every ray that hits the mirror — it is strictly a paraxial result. Rays striking far from the pole, near the rim of a wide mirror, converge slightly closer to the mirror than $F$ (spherical aberration); the syllabus expects you to know this is a limitation in words, not to compute the shift.
Reading "shorter focal length" as "weaker mirror." It is the opposite: a shorter $f$ means a more sharply curved, more powerful mirror that bends light through a larger angle and focuses it closer to the pole.
Assuming convex mirrors need a different formula from concave ones. The relation $f = R/2$ is identical in form for both; only the sign (and whether the focus is real or virtual) differs, never the underlying geometric relation.
Quoting $R$ or $f$ without units, or mixing centimetres and metres mid-calculation — a common silent error when a question gives $R$ in metres for a large mirror (like a telescope blank) and $f$ is expected in centimetres, or vice versa.

You should be able to…

State the relation $f = R/2$ and correctly attach signs for both concave and convex mirrors using the Cartesian convention.
Reproduce, in your own words, the geometric argument (normal along a radius, law of reflection, isosceles triangle, paraxial approximation) for why the focus is the midpoint of $P$ and $C$ .
Convert between a mirror's radius of curvature and its focal length in either direction, including when only magnitudes are given.
Compare two or more mirrors' converging or diverging strength directly from their radii of curvature, without first computing exact focal lengths if only a ranking is asked.
Distinguish a real focus (concave mirror, rays actually converge) from a virtual focus (convex mirror, rays only appear to diverge from a point) in both diagram and calculation.
State, in one sentence, why $f = R/2$ is a paraxial approximation and what physically goes wrong for rays far from the axis.

Ask the tutor: Want to work through the paraxial derivation with an actual protractor-style angle diagram, or practise ranking a mixed set of concave and convex mirrors by converging power under time pressure — ask the AI tutor to run either with you.

Why does f = R/2 have a factor of 2 in it — where does that number physically come from?

It comes straight out of the isosceles triangle in the derivation, not from any arbitrary rule. The reflected ray and the mirror's normal at the strike point form a triangle where two angles are forced equal by the law of reflection, which makes two of its sides equal — and that equality, combined with the paraxial approximation that the strike point sits almost on the axis, pins the focus to the exact midpoint of the segment from the pole to the centre of curvature. "Midpoint" is precisely what the factor of $\tfrac12$ in $f = R/2$ encodes; there is no deeper constant hiding in it, just triangle geometry.

If I only remember one version of the formula, should it be f = R/2 or R = 2f?

Either — they are the same statement rearranged, and an exam will sometimes hand you $R$ and ask for $f$ , sometimes the reverse. What actually costs marks is not which form you memorise but forgetting to carry the sign through the rearrangement: if a question gives $R = -40$ cm you must compute $f = -20$ cm, not $+20$ cm. Get comfortable rearranging in either direction with the sign attached from the start, rather than solving in magnitudes and guessing the sign at the end.

Does a convex mirror actually have a focus if it never brings rays to a real point?

Yes — it is called a virtual focus, and it is just as real a feature of the mirror's geometry, only the rays never actually pass through it. Extend the diverging reflected rays backward, behind the mirror, and they all appear to meet at a single point, exactly $R/2$ behind the pole. You could not catch a bright spot there on a screen (nothing physically converges), but your eye, or a camera, looking at the mirror perceives the reflected rays as if they came from that point — which is exactly how a virtual image works, and it is why the same formula $f = R/2$ applies with a positive sign rather than a different rule altogether.

Is spherical aberration something I need to calculate for the board exam?

No. The CBSE 2025-26 prescription treats aberrations as, at most, a one-line qualitative mention, not a topic with its own formulas or numericals. What you do need is the awareness that $f = R/2$ is a paraxial result — true only for rays close to the axis — so that if a question ever asks "why do wide-aperture mirrors give a blurred rather than a point focus," you can answer in one sentence (rays far from the axis do not obey the small-angle approximation used in the derivation, so they converge slightly off from $F$ ) without being asked to derive or quantify the blur itself.