Spherical mirrors

Look into the back of a shiny spoon and your face turns upside down; look into its front and it shrinks, cheerful and tiny. Both images come from the same simple object — a spherical mirror, a small curved patch cut from a hollow sphere and silvered on one side — obeying one compact rule. A concave mirror (silvered so it reflects from its caved-in inner surface) converges parallel rays to a real focus and can throw a real, inverted image onto a screen; a convex mirror (reflecting from the bulging outer surface) always spreads rays apart, so it only ever forms a smaller, upright, virtual image. This article opens the ISC Class 12 ray-optics unit: the Cartesian sign convention that keeps every calculation honest, the mirror formula linking object and image distance to focal length, and the magnification that tells you how the image compares to the object — the toolkit you will reuse for every mirror and lens problem this year.

The core insight

For any spherical mirror, object distance $u$ , image distance $v$ and focal length $f$ (all measured from the pole with a consistent sign convention) are tied together by $\dfrac1v+\dfrac1u=\dfrac1f$ , with $R=2f$ and magnification $m=-v/u$ .

The core idea

A spherical mirror has a pole $P$ (the centre of the mirror's surface), a centre of curvature $C$ (the centre of the sphere it was cut from) and a principal focus $F$ , the point where rays parallel to the axis converge (concave) or appear to diverge from (convex). The distance $PC$ is the radius of curvature $R$ , and $PF$ is the focal length $f$ .

Cartesian sign convention. All distances are measured from the pole. Distances measured in the direction the incident light travels (object always placed to the left, so this is "into" the mirror for a real object) are negative; distances measured against that direction are positive. Heights above the axis are positive, below are negative. Consequences worth memorising: object distance $u$ is always negative for a real object; $f$ (and $R$ ) is negative for a concave mirror, positive for a convex mirror, because $C$ and $F$ sit on opposite sides of $P$ for the two mirror types.

Radius and focal length are related for paraxial rays (rays close to and nearly parallel to the principal axis, so the small-angle approximation $\tan\theta\approx\theta$ holds):

R = 2f

where: $R$ is the radius of curvature and $f$ is the focal length, both carrying the sign convention above — so this is a relation between magnitudes as well as signs, never $R=f$ .

Mirror formula, linking object distance $u$ , image distance $v$ and focal length $f$ :

\frac{1}{v}+\frac{1}{u}=\frac{1}{f}

where: $v$ is measured from the pole to the image, $u$ from the pole to the object, and $f$ from the pole to the focus — all under the same sign convention. Given any two of $u$ , $v$ , $f$ you can solve for the third.

Linear magnification compares image size to object size:

m=\frac{h_2}{h_1}=-\frac{v}{u}

where: $h_1$ is object height and $h_2$ is image height. A negative $m$ means an inverted image (usually real); a positive $m$ means an erect image (usually virtual). $|m|>1$ means magnified, $|m|<1$ means diminished.

Figure

Ray diagram for a concave mirror forming a real inverted image of an object placed beyond the centre of curvature

Derivation

Derivation ADeriving the mirror formula from a ray diagram

The goal is to show, using only two rays and the property that similar triangles have proportional sides, that for a concave mirror the object distance $u$ , image distance $v$ and focal length $f$ obey $\dfrac1v+\dfrac1u=\dfrac1f$ — the same algebraic relation you use for every numerical, derived once so you can reproduce it under exam conditions.

Step 1. Set up the ray diagram. Take a concave mirror with pole $P$ , centre of curvature $C$ , and focus $F$ (with $PF=f$ , $PC=R=2f$ ). Place a point object $O$ on the principal axis. Draw ray $OA$ parallel to the axis, striking the mirror at $A$ near the pole; by definition of the focus, this ray reflects through $F$ . Draw a second ray from $O$ through $C$ , which — because it meets the mirror along the normal — reflects straight back along its own path. The two reflected rays meet at $I$ , the image point, where $MI$ (the perpendicular from $I$ to the axis) marks the image height.

Step 2. Identify the similar triangles. The ray $OA$ reflects at $A$ and crosses the axis at $F$ before reaching the image. In the small triangles formed near the axis, triangle $A P F$ and triangle $I M F$ are similar (both right-angled, sharing the vertical angle at $F$ ), so $\dfrac{PF}{FM} = \dfrac{AP}{MI}$ . Since $A$ is close to $P$ for a paraxial ray, $AP \approx PM'$ where $M'$ is the foot of the object's height — in practice $AP$ equals the object height $h_1$ and $MI$ equals the image height $h_2$ , so this ratio reproduces the magnification relation geometrically.

Step 3. Use the second pair of similar triangles, formed by the object, the pole, and the image, together with the axis. Triangles $OPA$ (or equivalently the object-side triangle) and the image-side triangle sharing vertex at the mirror give $\dfrac{OP}{PI} = \dfrac{AP}{MI}$ , i.e. the ratio of object distance to image distance equals the ratio of object height to image height. Combining this with the $F$ -based ratio from Step 2 and eliminating the height ratio between the two proportions yields a single distance relation between $u=OP$ , $v=PI$ and $f=PF$ .

Step 4. Apply the sign convention to the raw distances (all of $OP$ , $PF$ , $PI$ were treated as positive lengths in the geometry above) and simplify algebraically. Substituting $u\to-u$ , $v\to-v$ , $f\to-f$ as appropriate for a concave mirror and clearing the resulting fractions gives exactly

\frac{1}{v}+\frac{1}{u}=\frac{1}{f}.

The same relation, and the same steps, apply unchanged to a convex mirror once the signs of $u$ , $v$ , $f$ are substituted according to the convention — which is exactly why one formula serves both mirror types.

Worked example

A concave mirror has a radius of curvature of 30 cm. An object 2 cm tall is placed 20 cm in front of the mirror. Find the position, nature and size of the image, and repeat for a convex car mirror.

Step 1. Fix the signs. $R=-30\ \text{cm}$ (concave), so $f=R/2=-15\ \text{cm}$ . Object distance $u=-20\ \text{cm}$ (real object, always negative).

Step 2. Apply the mirror formula. $\dfrac1v=\dfrac1f-\dfrac1u=\dfrac{1}{-15}-\dfrac{1}{-20}=-\dfrac{1}{15}+\dfrac{1}{20}$ . Taking the LCM of 15 and 20 (which is 60): $-\dfrac{4}{60}+\dfrac{3}{60}=-\dfrac{1}{60}$ , so $v=-60\ \text{cm}$ .

Step 3. Interpret $v$ and find magnification. Negative $v$ places the image 60 cm in front of the mirror, on the same side as the object — a real image. Magnification $m=-\dfrac{v}{u}=-\dfrac{-60}{-20}=-3$ , so image height $h_2=m\times h_1=-3\times2=-6\ \text{cm}$ .

Answer: the image is real, inverted, 60 cm in front of the mirror, and 6 cm tall — magnified three times.

Competitive stretch

A convex car side-mirror has focal length $+20\ \text{cm}$ ; a scooter sits $u=-500\ \text{cm}$ behind the driver (5 m, real object so negative). Then $\dfrac1v=\dfrac1{20}-\dfrac1{-500}=\dfrac{25}{500}+\dfrac{1}{500}=\dfrac{26}{500}$ , giving $v=\dfrac{500}{26}\approx19.2\ \text{cm}$ — a virtual image 19.2 cm behind the mirror. Magnification $m=-\dfrac{19.2}{-500}\approx+0.038$ , an erect image about $\tfrac1{26}$ the scooter's size — why side mirrors warn "objects are closer than they appear."

Common traps

Sign convention slips. Forgetting $u$ is always negative for a real object, or that $f$ is negative for concave and positive for convex — fix the signs before touching numbers.
Confusing $R$ and $f$ . $R=2f$ , not $R=f$ ; the radius of curvature is always twice the focal length, with the same sign.
Assuming concave images are always real. An object between the pole and focus of a concave mirror gives a virtual, erect, magnified image (the shaving-mirror case) — check where $u$ lies relative to $f$ .
Misreading the sign of $m$ . Positive $m$ means erect (usually virtual); negative means inverted (usually real) — report the sign, not just the magnitude.

Ask the tutor: unsure how to place the object relative to F and C in a ray diagram, or want to work through a mixed concave-convex numerical step by step — ask the AI tutor.

Why is u always negative even though the object is clearly real and in front of the mirror?

The Cartesian convention measures every distance from the pole, taking the direction of incident light as negative. Since a real object always sits in front of the mirror, in the path the incident light is travelling from, its distance is measured against the outgoing convention and comes out negative by definition — it is a bookkeeping rule, not a statement about whether the object is real. Once you accept $u$ is negative by default, plugging numbers into the mirror formula becomes mechanical.

How can I tell if an image is real or virtual just from the sign of v?

For mirrors under this convention, a negative $v$ places the image on the same side as the object, which is where reflected rays actually converge — a real image, one you could catch on a screen. A positive $v$ places it behind the mirror, where rays only appear to diverge from, giving a virtual image you cannot project. So check the sign of $v$ after solving the mirror formula, then match it to "same side as object equals real, behind the mirror equals virtual."

Why does a convex mirror always give a smaller, upright image no matter where the object is?

A convex mirror's focus and centre of curvature both sit behind the reflecting surface (virtual), so the mirror formula with $f$ positive always returns a positive $v$ smaller in magnitude than the mirror's own focal length for any real object distance — meaning the image is always virtual, upright and diminished, regardless of how close or far the object is. This is exactly why convex mirrors are chosen for vehicle mirrors and shop-security mirrors: they always show a wide, complete (if shrunk) field of view.