Reflection by spherical mirrors

Stand in front of the back of a shiny steel spoon and your face turns upside down and shrinks; flip the spoon around and it turns right-side up and grows huge as you bring it close. Nothing about your face changed — only the curve of the reflecting surface did. That single curved surface, a small cut out of a hollow sphere and silvered on one side, is a spherical mirror, and it is the opening act of the Class 12 chapter on Ray Optics and Optical Instruments. A spherical mirror bends the ordinary law of reflection — angle of incidence equals angle of reflection — around a curved surface, and the result is that parallel rays of light are gathered to a point (a concave mirror) or spread apart as if from a point (a convex mirror). Everything that follows in this unit, from the algebraic mirror formula to magnification to the sign convention that keeps all of it consistent, rests on being able to trace a handful of standard rays through a curved mirror and read off where an image forms, whether it is real or virtual, upright or inverted, bigger or smaller than the object. This article builds that ray-tracing picture from the ground up: the vocabulary of pole, centre of curvature, radius, focus and focal length; the geometric reason $R = 2f$ holds for rays close to the axis; the four standard construction rays; and — reusing and extending the two ray-diagram worked examples from the previous pass — how object position alone decides everything about the image a concave mirror produces, while a convex mirror gives the same tame answer every single time. By the end you should be able to sketch any of the six standard concave-mirror cases from memory and explain, in one line, why your car's side mirror always says "objects are closer than they appear."

Key takeaways

A spherical mirror is a small silvered cap of a sphere; silvered on the inside it is concave (converging), silvered on the outside it is convex (diverging).
Landmark points: pole $P$ (mirror centre), centre of curvature $C$ (centre of the parent sphere), radius of curvature $R = PC$ , principal axis (line through $P$ and $C$ ), focus $F$ (where axis-parallel rays converge or appear to diverge from), focal length $f = PF$ .
For rays close to the axis (paraxial rays), geometry gives the fixed relation $R = 2f$ — the focus always sits halfway between pole and centre of curvature.
Four standard rays locate any image: parallel-to-axis reflects through $F$ ; through $C$ reflects straight back; through $F$ reflects parallel to axis; onto the pole reflects symmetrically about the axis.
A concave mirror's image changes character with object distance — real, inverted and diminished for a distant object; real, inverted and magnified for an object between $F$ and $C$ ; virtual, erect and magnified for an object between $F$ and the pole.
A convex mirror gives only one outcome for any real object: virtual, erect, diminished — which is exactly why it is chosen for vehicle mirrors, giving a wide, safe field of view.
Key idea: a spherical mirror obeys the ordinary law of reflection at every point, and because R = 2f fixes the focus at the sphere's half-radius, four standard rays are enough to locate any image — a concave mirror's image type depends on where the object sits relative to F and C, while a convex mirror always gives a small upright virtual image.

The core insight

Because a spherical mirror's focus sits at half its radius of curvature ( $R = 2f$ ), four standard rays — parallel-to-axis through $F$ , through $C$ straight back, through $F$ parallel to axis, and onto the pole symmetrically — are all that is needed to trace any image, with a concave mirror's image type swinging from real-inverted-small to virtual-erect-large as the object moves inside $F$ , while a convex mirror always yields a small, upright, virtual image.

Where it fits in the syllabus

This is the opening topic of Unit IX, "Ray Optics and Optical Instruments," in CBSE Class 12 Physics, and it is squarely in the prescribed scope: reflection of light and spherical mirrors are named explicitly ahead of the mirror formula and magnification. The board treats this stage as qualitative and geometric — ray diagrams and the vocabulary of pole, focus, and centre of curvature — while the algebraic mirror equation $\dfrac1v + \dfrac1u = \dfrac1f$ and the sign convention that makes it usable belong to the very next article in this chapter. Nothing here strays into aberrations or lens topics, which the 2025-26 syllabus keeps out of this chapter's theory content.

The physical picture

A spherical mirror is not a whole sphere — it is a small bowl-shaped patch cut from one, with the reflecting silvering on either the inner surface (concave) or the outer surface (convex). Because the patch is small compared to the sphere it came from, every point on the mirror is treated as lying on the plane through the pole for the purpose of measuring distances, even though the surface is genuinely curved for the purpose of reflecting light. Five landmarks fix the geometry. The pole $P$ is the geometric centre of the mirror's reflecting surface. The centre of curvature $C$ is the centre of the sphere the mirror is sliced from, and the radius of curvature $R$ is the distance $PC$ — the sphere's own radius. The principal axis is the straight line through $P$ and $C$ , the natural "spine" of the mirror. The focus $F$ is the special point on the principal axis where rays travelling parallel to the axis converge, for a concave mirror, or from which they appear to diverge, for a convex mirror, after reflecting; the focal length is the distance $f = PF$ .

R = 2f

where: $R$ is the radius of curvature (metres), and $f$ is the focal length (metres). This relation holds for rays that travel close to the principal axis and make small angles with it — the paraxial approximation that the whole ray-diagram method quietly relies on. Physically it says the focus is always the midpoint of the segment from the pole to the centre of curvature: a mirror with a tighter curve (small $R$ ) has a short focal length and bends rays sharply, while a nearly flat mirror (large $R$ ) has a long focal length and bends rays gently, smoothly approaching a plane mirror as $R \to \infty$ . Locating an image for any object needs only ordinary reflection — angle of incidence equals angle of reflection, measured from the normal, which at any point on a spherical mirror is simply the line from that point to $C$ (a radius always meets a circle at right angles to the tangent). Tracing that law ray by ray is tedious, so ray diagrams use four rays whose reflected paths are known in advance without any calculation:

Figure

Four standard rays traced from an object arrow to a concave mirror, showing the axis parallel ray reflecting through the focus, the ray through the centre of curvature reflecting straight back, the ray through the focus reflecting parallel to the axis, and the ray to the pole reflecting symmetrically, with the real inverted image formed where the rays cross

The four rays: a ray parallel to the principal axis reflects through the focus (concave) or seems to come from the focus (convex); a ray through — or aimed at — the centre of curvature strikes the mirror along the normal and reflects straight back on itself; a ray through — or aimed at — the focus reflects parallel to the axis; and a ray striking the pole reflects symmetrically about the axis, since the axis itself acts as the local normal there. Any two of these four are enough to pin down where the image of a point forms; where the reflected rays actually cross, the image is real, and where only their backward extensions cross, it is virtual.

Derivation AWhy R = 2f for a spherical mirror

The goal is to show, using nothing but the geometry of a circle and the ordinary law of reflection, that the focus of a spherical mirror always sits exactly halfway between the pole and the centre of curvature — the relation $R = 2f$ that every ray diagram in this chapter leans on. The argument uses only a triangle's exterior-angle property and the small-angle approximation for a ray that stays close to the axis; no calculus is needed.

Step 1. Set up the ray and the normal. Consider a ray travelling parallel to the principal axis and striking the mirror at a point $M$ , close to the pole $P$ so that the ray makes a small angle with the axis. Join $M$ to the centre of curvature $C$ ; because $C$ is the centre of the sphere, $MC$ is a radius, and a radius is always perpendicular to the mirror surface at that point — so $MC$ is the normal at $M$ .

Step 2. Apply the law of reflection at $M$ . Let $\theta$ be the angle the incoming ray makes with the normal $MC$ . Since the incoming ray is parallel to the axis, the angle between the normal $MC$ and the axis, call it $\theta$ as well, equals the angle of incidence (alternate angles between the parallel ray and the axis, cut by the transversal $MC$ ). By the law of reflection, the reflected ray also leaves at angle $\theta$ to the normal, so the reflected ray crosses the axis at some point $F$ with $\angle MFP = 2\theta$ measured from the axis on the far side — this follows from the exterior-angle property of triangle $MCF$ , where the exterior angle at $C$ equals the sum of the two remote interior angles, both equal to $\theta$ .

Step 3. Bring in the paraxial approximation. Because $M$ is taken close to the pole, the height $MP$ is small compared with $R$ and $f$ , so the angles are all small and the horizontal foot of $M$ can be treated as landing essentially at $P$ itself. In this limit, triangle $MCP$ and triangle $MFP$ both become right-angled at $P$ for practical purposes, and the small-angle relations $\tan\theta \approx PM/PC$ and $\tan 2\theta \approx PM/PF$ hold.

Step 4. Combine and simplify. From Step 2, the angle at $F$ is twice the angle at $C$ for the same small height $PM$ , which for small angles means $PC \approx 2\,PF$ directly from the doubling relation between $\tan\theta$ and $\tan2\theta$ at small $\theta$ (since $\tan 2\theta \approx 2\tan\theta$ when $\theta$ is small). Writing $PC = R$ and $PF = f$ , this is exactly $R = 2f$ : the centre of curvature lies twice as far from the pole as the focus does, for any paraxial ray, independent of which parallel ray you chose to trace. That independence is precisely why all paraxial rays parallel to the axis meet at the same single point $F$ — the defining property of a focus.

Worked examples

Board level - image of an object beyond C

Question: An object is placed beyond the centre of curvature $C$ of a concave mirror of focal length $f = 15\ \text{cm}$ (so $R = 30\ \text{cm}$ ). Using ray-diagram rules only, describe the nature, position, and relative size of the image formed, and state which two standard rays you would draw.

Step 1. Mark the principal axis with pole $P$ , focus $F$ at $15\ \text{cm}$ from $P$ , and centre of curvature $C$ at $30\ \text{cm}$ from $P$ ; place the object arrow beyond $C$ , say around $40\ \text{cm}$ from $P$ .

Step 2. From the object tip, draw the axis-parallel ray; after striking the mirror it reflects through $F$ . From the same tip, draw the ray aimed at $C$ ; since it travels along the normal at the point it strikes, it reflects straight back along the identical line.

Step 3. The two reflected rays are converging as they leave the mirror and physically cross at a point between $F$ (at $15\ \text{cm}$ ) and $C$ (at $30\ \text{cm}$ ) — closer to the mirror than the object was. Because the rays genuinely intersect in front of the mirror, the image is real; because it forms nearer the pole than the object, it is diminished; and because the reflected rays cross before meeting, the image is inverted. This is the same geometry used by a reflecting telescope's objective mirror, which produces a small, real, inverted image of a very distant astronomical object. Final answer: real, inverted, diminished image located between F and C.

Competitive level - object between pole and focus, then a numeric check

Question: A candle is placed close to a concave mirror of focal length $f = 15\ \text{cm}$ , between the pole and the focus, at object distance such that later chapters would use $u$ close to $10\ \text{cm}$ . First use the ray-diagram method (no formula) to classify the image; then, given that this chapter's later mirror-formula article would compute $v = -30\ \text{cm}$ and magnification $m = +3$ for this exact placement, check that your ray-diagram classification agrees with those signed numbers.

Step 1. Place the object between $P$ and $F$ (inside the focal length). Draw the axis-parallel ray from the object tip; it reflects through $F$ on leaving the mirror.

Step 2. Draw the ray aimed at $C$ ; a ray from a point inside $F$ , extended backward through the mirror, is understood to travel toward $C$ , and its reflection retraces that same line outward, diverging away from the mirror rather than converging.

Step 3. The two reflected rays spread apart in front of the mirror and never actually meet there; extending them backward, behind the mirror, they do meet. So the image is virtual (formed by extensions, not real ray crossings), erect (the extensions meet on the same side as the object's orientation), and enlarged, sitting farther behind the mirror than the object sits in front of it — this is the everyday shaving or makeup-mirror effect.

Step 4. Cross-check against the numbers quoted from the mirror-formula article: $v = -30\ \text{cm}$ (negative $v$ with the mirror's usual sign convention flags a virtual image, since a real image in a concave mirror would carry a different sign) and $m = +3$ (a positive magnification greater than $1$ means erect and three times enlarged). Both signed results match the ray-diagram conclusion exactly: virtual, erect, magnified. Final answer: virtual, erect, enlarged image, consistent with v = -30 cm and m = +3. Timing note: in a board exam, the ray-diagram classification alone (Steps 1-3) is worth full marks and should take under three minutes with a ruler; do not attempt the signed mirror-formula arithmetic unless the question explicitly asks for $v$ and $m$ , since that belongs to the next article's toolkit, not this one.

Common traps

Believing a convex mirror can ever form a real image of a real object — for every real object position in front of a convex mirror, the image is always virtual, erect, and diminished, with no exceptions to memorise.
Mixing up the centre of curvature $C$ with the focus $F$ — remember $R = 2f$ , so $C$ sits exactly twice as far from the pole as $F$ does, never at the same point.
Assuming any two rays drawn at random will correctly locate the image — only the four standard rays (parallel to axis, through $C$ , through $F$ , onto the pole) have reflection paths that are known without recomputation; a random ray needs the full law of reflection worked out afresh at its own point of incidence.
Carrying over "image size equals object size," which is true only for a flat mirror — for spherical mirrors the image size depends strongly, and non-trivially, on exactly where the object sits relative to $F$ and $C$ .

You should be able to…

Label pole, centre of curvature, radius of curvature, principal axis, and focus correctly on a blank spherical-mirror diagram, for both concave and convex mirrors.
State and justify the relation $R = 2f$ , and explain in one sentence why it follows from paraxial-ray geometry.
Name all four standard construction rays and describe, without hesitation, what each one does after reflecting from a concave or a convex mirror.
Draw a ray diagram and correctly classify the image (real/virtual, erect/inverted, magnified/diminished/same-size) for a concave-mirror object placed beyond $C$ , at $C$ , between $C$ and $F$ , at $F$ , and between $F$ and the pole.
Explain, using ray-diagram reasoning alone, why a convex mirror gives only virtual, erect, diminished images for any real object, and connect this to its use as a vehicle mirror.
Distinguish a real image (formed where reflected rays truly cross) from a virtual image (formed only where their backward extensions appear to cross).

Ask the tutor: stuck on which two standard rays to pick for an object sitting exactly at C, or want a walkthrough of all six concave-mirror image cases back to back — ask the AI tutor to sketch them with you.

Why does a spoon flip my face upside down on one side but not the other?

The inside (bowl side) of the spoon acts like a small concave mirror, and when your face is farther from it than its focal length, ray tracing puts the crossing point of the reflected rays in front of the spoon, which always means an inverted, real image — that is the flip you see. The outside (back) of the spoon is convex, and a convex mirror only ever produces a virtual, erect image for a real object like your face, so it never flips, only shrinks. Same spoon, same law of reflection at every point, but opposite curvature gives opposite behaviour.

Do I need to memorise all six object-position cases for a concave mirror, or can I just draw the diagram each time?

Drawing it is safer and, with practice, just as fast as memorising, because the two rays through $F$ and through $C$ never change — only the object's starting position changes. What is worth memorising is the pattern: as the object moves from far away (beyond $C$ ) inward toward the mirror, the real image grows from small to large while staying real and inverted, until the object crosses $F$ , at which point the image flips to virtual, erect, and growing even larger as the object approaches the pole. Understanding that trend lets you sanity-check any diagram you draw under exam time pressure.

Why is the normal at any point on a spherical mirror just the line to the centre of curvature, and not something I have to construct separately?

A basic circle fact does the work: at any point on a circle, the radius drawn to that point is always perpendicular to the tangent to the circle there — and the mirror's surface at that point behaves exactly like that tangent for a ray hitting a very small patch of it. Since $C$ is the centre of the sphere the mirror is cut from, the line from any point on the mirror to $C$ is a radius of that sphere, so it is automatically the normal. This is exactly why the "ray through $C$ reflects straight back on itself" rule works without any extra construction — the ray is already travelling along the normal.

Is this whole chapter — ray diagrams, mirror formula, lenses — actually asked as long-answer in CBSE boards?

Yes, spherical mirrors and the broader ray optics chapter are a reliable source of both short (2-mark ray-diagram sketch, "state the nature of image") and long (5-mark, combining ray diagram plus the mirror-formula numerics from the next article) board questions, and the chapter carries real weight in most CBSE Class 12 physics papers. This particular sub-topic — pure ray-diagram reasoning without algebra — tends to appear as the diagram-and-explain part of a longer question, so being fluent with the four standard rays by hand, without a calculator, is worth the practice time.