In short

A spherical mirror is a cut out of a reflecting sphere. It has a pole P (centre of the reflecting surface), a centre of curvature C (centre of the original sphere), and a focus F midway between them. The radius of curvature R and focal length f are related by

\boxed{\;f \;=\; \tfrac{1}{2}R\;}

For paraxial rays (close to the principal axis), the mirror produces an image at distance v of an object at distance u, obeying the mirror formula (Cartesian sign convention, incident light travels along -x, all distances measured from the pole):

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

The magnification is

\boxed{\;m \;=\; \frac{h'}{h} \;=\; -\frac{v}{u}\;}

Sign convention (Cartesian):

  • Distances measured from the pole, with the principal axis as the x-axis.
  • Incident light travels along -\hat{x} (from right to left onto the mirror).
  • Distances in the direction of incident light are negative; against it are positive.
  • Heights above the principal axis are positive; below are negative.

Consequence for f: concave mirror f < 0; convex mirror f > 0.

Mirror type f sign Image formed for real object Used in
Concave negative real/inverted if $ u
Convex positive always virtual, erect, diminished vehicle side mirrors, security mirrors, crossroads

The two special rays used in every ray diagram:

  1. A ray parallel to the principal axis reflects through the focus (concave) or appearing to come from the focus (convex).
  2. A ray through the centre of curvature retraces its path (it hits the mirror along the normal).

Pick up a steel tablespoon from your kitchen and look at both sides in afternoon light. Hold the concave side (the inner bowl) close to your nose: your face is huge, right-side-up, and slightly stretched — that is a magnifying shaving mirror. Now pull the spoon slowly away from your face. At some distance, the image suddenly flips upside down, shrinks, and becomes sharp. Pull further and the upside-down image gets tinier. Flip the spoon to the convex side (the back of the bowl): your face is small, right-side-up, and no matter how close or far you hold it, the image never flips.

The spoon is two spherical mirrors glued back-to-back. Everything interesting about image formation — magnification, inversion, real versus virtual — is visible on that one spoon. This chapter explains why each image looks the way it does, derives the mirror formula from the geometry of reflection, and gives you the tools to predict any image's position, size, and orientation from three numbers: object distance, focal length, and the sign convention.

Anatomy of a spherical mirror

A spherical mirror is a piece of the inside or outside of a sphere, coated so it reflects light. Three points define its geometry:

The line through P and C is the principal axis. The distance PC is the radius of curvature R. The distance PF is the focal length f.

Anatomy of a concave mirror A curved reflecting surface on the right, principal axis drawn horizontally. Pole P at the intersection of axis and mirror surface, focus F at half the radius, centre of curvature C at the full radius from the pole. Labels for each point. $P$ (pole) $F$ (focus) $C$ (centre of curvature) $f$ $f$ $R = 2f$ principal axis
A concave mirror cut from a sphere. The pole $P$ is where the principal axis meets the surface. The focus $F$ sits exactly halfway between $P$ and the centre of curvature $C$ — so $f = R/2$.

Why f = R/2 — a geometric proof

The relation f = R/2 is not an experimental accident. It falls out of the law of reflection applied to a paraxial ray. "Paraxial" means the ray is close to and nearly parallel to the principal axis — the regime in which spherical mirrors behave like ideal focusing mirrors.

Geometric proof that f = R/2 for a spherical mirror A parallel ray hits a concave mirror at point M close to the principal axis. The radius CM is the normal at M. The reflected ray crosses the axis at F, the focus. Triangle CMF is isoceles with CF = FM, which in the paraxial limit equals CP/2. $P$ $C$ $F$ parallel ray $M$ normal $CM$ $\theta$ $\theta$
A ray parallel to the principal axis strikes the mirror at $M$. The normal at $M$ is the radius $CM$. By the law of reflection the angle on each side of $CM$ is $\theta$, which makes triangle $CMF$ isoceles: $FM = FC$. For paraxial rays ($\theta$ small), $FM \approx FP$, giving $CF = FP$ — i.e., $F$ is halfway between $C$ and $P$.

Step 1. A ray parallel to the principal axis hits the mirror at point M. The normal at M is the radius CM (any radius of a sphere meets the surface perpendicularly).

Why: the normal to a sphere at any point is the line from the centre of the sphere to that point. That line is the radius. The law of reflection measures angles from this normal.

Step 2. The law of reflection: the angle of incidence equals the angle of reflection, and both are measured from the normal. Call this angle \theta. The reflected ray crosses the principal axis at some point F.

Step 3. Triangle CMF: the angle MCF (at C) equals \theta because the incident ray is parallel to the axis and CM is a transversal — alternate angles. The angle CMF (at M) equals \theta by the law of reflection.

Why: with two angles of the triangle equal, the two sides opposite them must be equal. That is the isoceles-triangle theorem.

Step 4. So FM = FC. In the paraxial limit (\theta small, so M is very close to P), FM \approx FP. Therefore

FC \;=\; FP

F is midway between C and P. So PF = PC/2, i.e.,

\boxed{\;f \;=\; \frac{R}{2}\;}

This is a geometric fact, not an empirical one — it comes directly from the law of reflection and the fact that the mirror is a sphere.

What breaks the proof for non-paraxial rays? Step 4 used FM\approx FP, which fails when M is far from the axis. In that case different rays focus at different points along the axis — an aberration called spherical aberration. That is why the parabolic mirrors used in ISRO's DSN Byalalu antenna and the Giant Metrewave Radio Telescope (GMRT) near Pune are not spheres: a parabola focuses all parallel rays to a single point, not just paraxial ones. But for small apertures (shaving mirrors, classroom concave mirrors, car side mirrors), spherical mirrors are good enough and much cheaper to manufacture.

The sign convention — where signs go wrong

Every JEE aspirant has lost marks on mirror problems by using the wrong sign. The Cartesian sign convention fixes the confusion once and for all.

Rules.

  1. The pole P is the origin. The principal axis is the x-axis.
  2. Incident light travels from right to left — i.e., along -\hat{x}.
  3. All distances measured in the direction of the incident light (to the left of P) are negative. All distances measured against the incident light (to the right of P, behind the mirror) are positive.
  4. Heights above the principal axis are positive; below are negative.
Cartesian sign convention for mirrors A vertical mirror surface in the middle. Left side labelled "incident light" with negative x axis. Right side labelled "behind mirror" with positive x axis. Above the axis positive y, below negative y. $P$ incident light $\rightarrow$ distances here are **negative** behind mirror distances here are **positive** heights $+$ heights $-$
Cartesian sign convention: pole at origin, incident light travels leftward ($-\hat{x}$). Distances to the left of $P$ (in the direction of travel) are negative; to the right (behind the mirror) are positive. Heights above the axis are positive.

Consequences for standard quantities.

Once you internalise these rules, every mirror problem reduces to arithmetic on signed numbers.

Deriving the mirror formula

Take a concave mirror. Put an object AB of height h on the principal axis at distance |u| from the pole (with u < 0). Trace two rays from the tip B to find where the image tip B' forms. The intersection of the two reflected rays is the image tip.

Ray 1: From B parallel to the axis → reflects through F. Ray 2: From B through F → reflects parallel to the axis. (Either ray 1 or ray 2 plus a ray through C gives you the image; I use 1 and 2.)

Deriving the mirror formula for a concave mirror An object of height h on the principal axis between infinity and the centre of curvature. Two rays trace from the top of the object: one parallel to the axis reflecting through F, one through F reflecting parallel. They cross at the image between F and C. $P$ $C$ $F$ $B$ $A$ $h$ ray 1 (parallel) ray 2 $B'$ $A'$ $h'$
Object $AB$ (height $h$, at distance $u$ from the pole), concave mirror, two rays trace the image $A'B'$ (height $h'$, at distance $v$). Similar triangles give the mirror formula.

Step 1. Triangles ABP and A'B'P are similar (both right-angled at A/A'; share the angle at P).

\frac{A'B'}{AB} \;=\; \frac{A'P}{AP} \;\Rightarrow\; \frac{h'}{h} \;=\; \frac{v}{u}

Why: both triangles have their right angle on the principal axis; the third angle (at P) is shared. So the triangles are AA-similar. The ratio of corresponding sides equals the ratio of distances from the pole.

With the sign convention, h is positive (object above axis), h' is negative (image below axis because it is inverted), and u and v are both negative (both in front of the mirror). So a ratio v/u is a ratio of two negative numbers — positive. But h'/h is negative. A sign-corrected version:

\frac{h'}{h} \;=\; -\frac{v}{u}

Step 2. Triangle DPF (where D is the top of the mirror at P) and triangle A'B'F are similar (the reflected ray 1 is straight; both triangles have the same angle at F).

\frac{A'B'}{DP} \;=\; \frac{A'F}{PF}

Here DP = AB = h (the parallel ray has height h at the mirror).

\frac{h'}{h} \;=\; \frac{v - f}{f}

Why: A'F is the distance from the image to the focus, which is v - f (with signs). PF is f. In the Cartesian convention these are algebraic distances, so signs propagate automatically.

Step 3. Equate the two expressions for h'/h.

\frac{v}{u} \;=\; \frac{v - f}{f}

Why: both ratios equal h'/h (taking the unsigned form from Step 1), so they are equal to each other. From here on it is pure algebra.

Step 4. Cross-multiply and expand.

vf \;=\; u(v - f) \;=\; uv - uf

Step 5. Divide through by uvf.

\frac{vf}{uvf} \;=\; \frac{uv}{uvf} - \frac{uf}{uvf}
\frac{1}{u} \;=\; \frac{1}{f} - \frac{1}{v}

Step 6. Rearrange.

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

Why: this is the standard form of the mirror formula. It holds for both concave and convex mirrors, for real and virtual objects and images, as long as the Cartesian sign convention is applied consistently.

Magnification. From Step 1 (with the sign fix),

\boxed{\;m \;=\; \frac{h'}{h} \;=\; -\frac{v}{u}\;}

A positive m means erect image; negative means inverted. |m| > 1 means enlarged; |m| < 1 means diminished.

Explore: drag the object, watch the image form

Below is a concave mirror with focal length f = 15 cm (radius of curvature 30 cm). Drag the red object along the principal axis and watch the image track it: for |u| > |f| the image is real and inverted; for |u| < |f| it is virtual and erect. The simulation traces the two standard rays (parallel → focus, and focus → parallel) and computes the image position from the mirror formula.

Concave mirror ($f = 15$ cm). Drag the red arrow (the object) along the axis. Between $C$ and infinity the image is real, inverted, diminished or enlarged depending on the object's position. Inside $F$ the image is virtual, erect, and enlarged — this is the shaving-mirror regime.

And a convex mirror — the kind mounted on every Indian Railways shunting locomotive, every Maruti Swift's left side window, and every dark stairway corner for security:

Convex mirror ($f = +15$ cm). No matter where you put the object, the image is behind the mirror, virtual, erect, and smaller than the object. That "smaller" property is the whole point — a convex mirror gives the driver a wide field of view, showing many vehicles compressed into one small mirror.

Image formation — the six cases for a concave mirror

Using the mirror formula \frac{1}{v} + \frac{1}{u} = \frac{1}{f} with f < 0 (concave), the image depends on where the object is.

| Object position | Image position | Nature | |m| | |---|---|---|---| | At infinity | At F | real, inverted, highly diminished | \to 0 | | Beyond C (|u| > 2|f|) | Between F and C | real, inverted, diminished | < 1 | | At C (|u| = 2|f|) | At C | real, inverted, same size | = 1 | | Between F and C (|f| < |u| < 2|f|) | Beyond C | real, inverted, enlarged | > 1 | | At F (|u| = |f|) | At infinity | no image | \to \infty | | Between P and F (|u| < |f|) | Behind mirror | virtual, erect, enlarged | > 1 |

The last row is the shaving-mirror regime: you hold the mirror close enough that your face is inside its focal length (\sim 25 cm for a typical mirror), and the image becomes a virtual, enlarged, upright version of your face — perfect for shaving or applying kajal.

For a convex mirror, the image is always virtual, erect, and diminished regardless of object position. That is why vehicle side mirrors in India (and the "objects in mirror are closer than they appear" warning in cars) use convex mirrors: the compressed image lets the driver see a wide stretch of the road behind.

Worked examples

Example 1: The Indian Railways convex side mirror

A convex mirror on the side of a freight locomotive has focal length f = +50 cm. A goods wagon of height 2.0 m stands 8.0 m behind the mirror. Find the position, nature, and size of the image the driver sees.

Convex mirror on a locomotive showing a wagon behind Side view: a goods wagon (2 m tall) 8 m to the left, a convex mirror at the driver's position, and a virtual image of the wagon formed behind the mirror, smaller. wagon 2.0 m, 8 m away image (virtual) mirror
A convex mirror on the locomotive. The real wagon (left) produces a virtual, smaller, erect image behind the mirror.

Step 1. Apply the sign convention. Object is in front of the mirror (real), so u < 0.

u = -800 cm (convert 8 m = 800 cm)

f = +50 cm (convex, positive)

h = +200 cm (wagon is above the axis)

Why: consistently applying the Cartesian convention means all signs are determined by the geometry, not by guesswork. The object is on the left of the mirror — in the direction of incident light — so its distance is negative.

Step 2. Apply the mirror formula.

\frac{1}{v} + \frac{1}{u} \;=\; \frac{1}{f}
\frac{1}{v} \;=\; \frac{1}{f} - \frac{1}{u} \;=\; \frac{1}{50} - \frac{1}{-800} \;=\; \frac{1}{50} + \frac{1}{800}
\frac{1}{v} \;=\; \frac{16}{800} + \frac{1}{800} \;=\; \frac{17}{800}
v \;=\; \frac{800}{17} \;\approx\; +47.06\text{ cm}

Why: v came out positive — meaning the image is behind the mirror. That is a virtual image. For a convex mirror this is always the case.

Step 3. Compute the magnification.

m \;=\; -\frac{v}{u} \;=\; -\frac{47.06}{-800} \;=\; +0.0588

Why: m is positive, so the image is erect (same orientation as the wagon). |m| < 1, so the image is diminished.

Step 4. Compute the image height.

h' \;=\; m\cdot h \;=\; 0.0588 \times 200 \;=\; 11.8\text{ cm}

Result: The image is 47 cm behind the mirror, virtual, erect, and only 11.8 cm tall — about 1/17 the size of the wagon. The driver sees a small, upright image of a 2 m wagon that is actually 8 m behind him.

What this shows: every property of a convex side mirror in one calculation. The image is compressed (good — lets the driver see a wide angle of track) and virtual (harmless — the driver just uses it to judge distance). The "objects in mirror are closer than they appear" warning in Indian cars is literally the mirror formula: |v| = 47 cm, but the actual distance to the wagon is 800 cm. The brain interprets the small image as far, when in fact the object is close.

Example 2: A shaving mirror at the right distance

A dentist's concave mirror (focal length 10 cm) is held 5 cm from a patient's tooth. Find the image's position, magnification, and nature.

A concave mirror making a virtual magnified image of a tooth A tooth (object) between the pole and the focus of a concave mirror. The image forms behind the mirror, larger and erect. $P$ $F$ $C$ tooth image (virtual)
The tooth sits inside the focal length — between $P$ and $F$. The image is virtual (behind the mirror), erect, and enlarged.

Step 1. Sign convention.

u = -5 cm (object in front, so negative)

f = -10 cm (concave, so negative)

Step 2. Mirror formula.

\frac{1}{v} \;=\; \frac{1}{f} - \frac{1}{u} \;=\; \frac{1}{-10} - \frac{1}{-5} \;=\; -\frac{1}{10} + \frac{1}{5} \;=\; \frac{-1 + 2}{10} \;=\; \frac{1}{10}
v \;=\; +10\text{ cm}

Why: v > 0 means the image is behind the mirror — virtual. Notice the clean arithmetic: f = -10, u = -5, and the answer fell out in one step once signs were respected.

Step 3. Magnification.

m \;=\; -\frac{v}{u} \;=\; -\frac{10}{-5} \;=\; +2

Why: the dentist sees a virtual, erect, two-times-enlarged image of the tooth. Erect because m > 0; enlarged because |m| > 1. This is why dental mirrors and shaving mirrors work.

Result: Image at 10 cm behind the mirror, virtual, erect, magnification +2. The tooth appears twice as large and the right way up.

What this shows: whenever a real object is placed inside the focal length of a concave mirror (|u| < |f|), the image is virtual, erect, and enlarged. That is the shaving-mirror / magnifying-mirror regime. For a dentist looking at a small cavity or a close-up detail on a tooth, this 2× virtual image is exactly what is needed.

Example 3: A security mirror on a Chennai stairwell

A convex security mirror is mounted at the top of a Chennai apartment stairwell, focal length 60 cm. A 1.8 m tall person walks up the stairs and is 3.0 m from the mirror. Find the position, size, and nature of the image seen on the closed-circuit camera watching the mirror.

Step 1. Sign convention.

u = -300 cm, f = +60 cm, h = +180 cm.

Step 2. Mirror formula.

\frac{1}{v} \;=\; \frac{1}{60} - \frac{1}{-300} \;=\; \frac{1}{60} + \frac{1}{300}

Find a common denominator (300):

\frac{1}{v} \;=\; \frac{5}{300} + \frac{1}{300} \;=\; \frac{6}{300} \;=\; \frac{1}{50}
v \;=\; +50\text{ cm}

Why: virtual image, 50 cm behind the mirror. For any convex mirror, you will always find |v| < |f| — the virtual image never gets further behind than the focal length.

Step 3. Magnification.

m \;=\; -\frac{v}{u} \;=\; -\frac{50}{-300} \;=\; +\frac{1}{6}

Step 4. Image height.

h' \;=\; m\cdot h \;=\; \tfrac{1}{6}\times 180 \;=\; 30\text{ cm}
A convex security mirror and a walking person A convex mirror on the right, a 1.8 m person on the left, virtual small image behind the mirror. person, 1.8 m image (30 cm)
A 1.8 m person is compressed into a 30 cm image behind the security mirror — a 6× demagnification that packs a wide field of view into the camera's frame.

Result: Image at 50 cm behind the mirror, virtual, erect, height 30 cm (one sixth of the actual person).

What this shows: a convex mirror with focal length comparable to the distance of interest (|u|/|f| = 5) compresses the scene by a factor of about 6. That is exactly the property you want for a security camera — the mirror packs a large field of view into a small image for the camera to capture. The "objects in mirror are closer than they appear" warning on your car's left-side mirror is the honest statement that |v| \ll |u|, so the brain misreads the small image as "far" when the object is actually close.

Common confusions

You have the mirror formula, the sign convention, and the three regimes. What follows is for readers who want the derivation for a convex mirror, the exact connection to the thin-lens formula, and spherical aberration quantified.

Deriving the mirror formula for a convex mirror

The derivation for a concave mirror in the body of this article used the geometry of a real image between F and C. For a convex mirror the image is virtual — behind the mirror — but the same formula works, because the Cartesian sign convention automatically handles it.

Ray diagram for a convex mirror showing a virtual image Object in front of a convex mirror; parallel ray reflects so it appears to come from F behind the mirror; chief ray through centre of curvature (behind) retraces. The two reflected rays, extended backward, meet at the virtual image behind the mirror. $P$ $F$ $C$ obj virtual extension virtual image
A convex mirror. The parallel ray reflects as if it came from $F$ (behind); the reflected rays, extended backward, meet at the virtual image behind the mirror.

Setup. Object at u < 0 (real, in front); focus at f > 0 (behind, so positive). Image at v > 0 (virtual, behind mirror).

Same similar-triangle argument as before gives

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

without modification. You can verify: with u = -300, f = +60 (from Example 3), the formula gives 1/v = 1/60 + 1/300 = 1/50, so v = +50. The positive v automatically encodes "virtual image behind the mirror." You did not need a separate formula for convex mirrors — the signs did the work.

Link to the thin-lens formula

The mirror formula

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

has a near-twin in the refraction world:

\frac{1}{v} - \frac{1}{u} \;=\; \frac{1}{f}\quad\text{(thin lens)}

The sign difference (+ vs -) comes from the fact that a mirror reverses the direction of propagation (so v and u are measured on the same side of the optical element), while a lens lets light through (so v is measured on the opposite side of the element from u). Both are consequences of the same principle — Fermat's least-time principle — applied to the appropriate geometry.

Spherical aberration — how bad is it?

The paraxial derivation of f = R/2 fails for rays far from the axis. How far off does the focal point drift for a ray at finite height h from the axis?

A ray parallel to the axis at height h hits a concave mirror of radius R at an angle \theta from the normal, where \sin\theta = h/R. For paraxial rays, \sin\theta \approx \theta, but for larger h the deviation matters.

The reflected ray crosses the principal axis at

x_\text{actual} \;=\; R\left(1 - \frac{1}{2\cos\theta}\right)

Why: this comes from the law of reflection applied to the exact geometry, without the small-angle approximation. Derivation: the incident ray, the normal CM, and the reflected ray all lie in one plane; the reflected ray makes angle 2\theta with the incident direction. Working out the geometry algebraically gives the formula.

Paraxial limit: \cos\theta \to 1, so x_\text{actual}\to R/2 = f. Good.

For finite \theta: \cos\theta < 1, so 1/(2\cos\theta) > 1/2, which makes x_\text{actual} < R/2. The ray focuses closer to the mirror than the paraxial focus.

Numerical example. For a mirror with R = 30 cm and a ray at h = 6 cm (so \sin\theta = 0.2, \theta\approx 11.5°, \cos\theta\approx 0.98):

x_\text{actual} \;=\; 30\times\left(1 - \frac{1}{2\times 0.98}\right) \;=\; 30\times(1 - 0.510) \;=\; 30\times 0.490 \;=\; 14.7\text{ cm}

vs paraxial prediction f = 15 cm. A 0.3 cm difference, or 2%. For a star seen through a large telescope, that 2% is enormous and blurs the star. For your shaving mirror, it is invisible.

Why parabolas win. A parabola y^2 = 4fx has the geometric property that every ray parallel to its axis reflects through the same focus F, regardless of height. That is the Fermat-least-time condition built into the shape. The ISRO antennas at Byalalu, the Ooty Radio Telescope, and the GMRT near Pune are all parabolic precisely because they need to focus plane radio waves from distant sources to a single feed — no spherical aberration allowed.

Power of a mirror

By analogy with lenses, the power of a mirror is

P_\text{mirror} \;=\; \frac{1}{f}

measured in dioptres (D) when f is in metres. A concave mirror of focal length 20 cm has power P = 1/0.20 = 5 D, and the sign is negative by the Cartesian convention (concave f < 0). A convex mirror of focal length 50 cm has P = +2 D. Combining mirrors in sequence (as in a Cassegrain telescope) sums the powers — but the full system requires tracking the sign convention at each surface, which is why astronomical optics is its own specialty.

Where this leads next