Chapter 9 · Ray Optics and Optical Instruments

9.1 Reflection of Light by Spherical Mirrors

A curved mirror turns a bundle of rays into an image whose position, size and orientation follow from one equation, $\dfrac1v+\dfrac1u=\dfrac1f$ — provided every distance carries the sign that the Cartesian convention gives it.

The wing mirror on the passenger side of many cars carries a warning: objects in mirror are closer than they appear. The mirror is convex — it bulges towards you — so it shows a wide stretch of road, but every car in it looks small and therefore far away. A dentist's mouth mirror and a make-up mirror do the opposite: they are concave, and a face held close to them looks larger. The reflector behind a car headlamp, the dish of a solar cooker and the 6.5-metre gold-coated main mirror of the James Webb Space Telescope are concave too, and they work because a concave mirror sends rays parallel to its axis through a single point, the focus.

In Class 10 you drew ray diagrams for these mirrors. In this section we make them quantitative. We fix a sign convention so that one formula covers every case, show why the focal length is half the radius of curvature, derive the mirror equation and the magnification, and use them to predict where an image forms, how large it is and whether it is real or virtual.

Spherical Mirrors and the Cartesian Sign Convention

Light travels in straight lines in a uniform medium; a narrow beam is drawn as a ray, an arrow along the direction the light's energy travels. When a ray meets a smooth reflecting surface, two rules — known since antiquity — fix the reflected ray.

📐 Law 9.1 — The laws of reflection

At any point of a smooth reflecting surface, (i) the incident ray, the reflected ray and the normal to the surface at the point of incidence lie in one plane, and (ii) the angle of reflection equals the angle of incidence, both measured from the normal. For a spherical mirror the normal at any point is along the radius through that point — the line joining the point to the centre of curvature.

A spherical mirror is a piece of a hollow sphere silvered on one side. If the reflecting surface faces the centre of the sphere the mirror is concave; if it faces away from the centre the mirror is convex.

📘 Definition 9.1 — Parts of a spherical mirror

The pole $P$ is the centre of the reflecting surface. The centre of curvature $C$ is the centre of the sphere of which the mirror is a part, and the radius of curvature $R$ is the radius of that sphere. The principal axis is the line through $P$ and $C$. The width of the mirror is its aperture. Rays that travel close to the principal axis and make small angles with it are called paraxial rays.

After reflection, paraxial rays parallel to the principal axis pass through (concave mirror) or appear to diverge from (convex mirror) a point $F$ on the axis, the principal focus. The distance $PF$ is the focal length $f$.

A position on the axis is described by its distance from the pole. To let one formula serve for concave and convex mirrors, real and virtual images, we give every distance a sign.

📘 Definition 9.2 — The Cartesian sign convention

(i) All distances along the principal axis are measured from the pole (from the optical centre, for a lens). (ii) Distances measured in the same direction as the incident light are positive; distances measured against it are negative. (iii) Heights measured upwards, perpendicular to the axis, are positive; heights measured downwards are negative.

We always draw the incident light travelling from left to right, so the pole is the origin of an $x$-axis pointing to the right. The object distance is $u$, the image distance $v$, the focal length $f$ and the radius of curvature $R$. A real object in front of a mirror has $u\lt0$. A concave mirror has $R\lt0$ and $f\lt0$ (its centre and focus are in front of it); a convex mirror has $R\gt0$ and $f\gt0$.

Why is the focus half-way between the pole and the centre? Take a concave mirror and a ray parallel to the axis at height $h$. It strikes the mirror at $M$, where the normal $CM$ makes an angle $\theta$ with the axis, with $\sin\theta=h/R$. The angle of incidence is $\theta$, so the angle of reflection is $\theta$ as well, and the reflected ray crosses the axis at a point $F'$. The angle $\angle MCF'=\theta$ (alternate angles with the incident ray) and $\angle CMF'=\theta$ (the angle of reflection), so the triangle $CMF'$ is isosceles with $CF'=MF'$. Dropping a perpendicular from $F'$ to $CM$ bisects $CM$, so $$CF'=\frac{R}{2\cos\theta}.$$ For paraxial rays $\theta$ is small, $\cos\theta\approx1$, and every such ray crosses at the same point, $R/2$ from $C$ — half-way to the pole. For wider rays $\cos\theta\lt1$, $CF'\gt R/2$, and the ray crosses closer to the mirror. That blurring of the focus by wide rays is called spherical aberration; it is why large telescope mirrors are ground to a paraboloid rather than a sphere.

📐 Result 9.2 — Focal length of a spherical mirror

For paraxial rays, the focal length of a spherical mirror is half its radius of curvature, $$f=\frac R2,$$ with the signs of the Cartesian convention: $f\lt0$ for a concave mirror and $f\gt0$ for a convex mirror. The focal length depends only on the shape of the mirror — not on the colour (wavelength) of the light or on the medium in front of it.

✏️ Example 1 — Focal lengths, and a ray that misses the focus
(a) Write down, with signs, the focal lengths of a concave mirror of radius of curvature $30$ cm and a convex mirror of radius of curvature $40$ cm. (b) A ray parallel to the axis strikes the concave mirror at height $h=R/2=15$ cm. How far from the pole does the reflected ray cross the axis, and how far is that from the paraxial focus?
Signs Cartesian convention, light travelling left to right. The concave mirror's centre is in front of it: $R=-30$ cm. The convex mirror's centre is behind it: $R=+40$ cm.
(a) f = R/2 Concave: $f=-15$ cm. Convex: $f=+20$ cm.
(b) Wide ray $\sin\theta=h/R=0.5$, so $\theta=30^\circ$ and $CF'=\dfrac{R}{2\cos30^\circ}=\dfrac{30}{1.732}\approx17.3$ cm. The pole is $30$ cm from $C$, so the ray crosses $30-17.3\approx12.7$ cm from the pole — about $2.3$ cm nearer the mirror than the paraxial focus at $15$ cm.
Check In Interactive 1, set the highlighted ray to $h/R=0.50$: it reads $PF'/R=0.423$, i.e. $0.423\times30\approx12.7$ cm. ✓
Interactive 1 — Parallel Rays and the Focus of a Concave Mirror

Rays parallel to the principal axis strike a concave spherical mirror and are reflected exactly by the law of reflection (the normal at each point is the radius to $C$). Paraxial rays all cross the axis at $F$, half-way between $C$ and $P$; wide rays cross closer to the mirror. Press Play to sweep the highlighted ray from the axis to the rim, or set its height with the slider; switch the aperture to see the blur of a wide mirror.

0.50
angle $\theta$ at the mirror—
crossing point $PF'/R$—
paraxial focus $PF/R$—
shift from $F$—

The Mirror Equation and Magnification

Place a small object $AB$ upright on the axis in front of a concave mirror, with its foot $B$ on the axis, and follow two rays from its top $A$. A ray parallel to the axis reflects through $F$; a ray to the pole reflects symmetrically about the axis. They meet at $A'$, the top of the image $A'B'$. Let $M$ be the point where the first ray meets the mirror; for paraxial rays the mirror is so nearly flat that $M$ lies above $P$ and the image of $B$ is the foot $B'$ of the perpendicular from $A'$.

From the ray through $F$: triangles $A'B'F$ and $MPF$ are similar, and $MP=AB$, so $\dfrac{B'A'}{BA}=\dfrac{B'F}{FP}$.

From the ray to the pole: the angles of incidence and reflection are equal, so triangles $A'B'P$ and $ABP$ are similar: $\dfrac{B'A'}{BA}=\dfrac{B'P}{BP}$.

Equating, and writing $B'F=B'P-FP$: $\dfrac{B'P-FP}{FP}=\dfrac{B'P}{BP}$. These are lengths. Now put in the signs: $B'P=-v$, $FP=-f$, $BP=-u$ (object, image and focus are all in front of the mirror). Then $$\frac{-v+f}{-f}=\frac{-v}{-u}\quad\Rightarrow\quad\frac vf-1=\frac vu,$$ and dividing by $v$, $$\frac1v+\frac1u=\frac1f.$$ The same equation follows, with the same signs, for a convex mirror and for virtual images — that is the point of the sign convention.

📐 Result 9.3 — The mirror equation

For a spherical mirror and paraxial rays, the object distance $u$, the image distance $v$ and the focal length $f$, all measured from the pole with the Cartesian sign convention, satisfy $$\frac1v+\frac1u=\frac1f=\frac2R.$$ A negative $v$ means the image is in front of the mirror — a real image, through which the reflected rays actually pass and which can be caught on a screen. A positive $v$ means the image is behind the mirror — a virtual image, from which the reflected rays only appear to come.

📐 Result 9.4 — Linear magnification

The linear magnification of a mirror is the ratio of the height $h'$ of the image to the height $h$ of the object: $$m=\frac{h'}{h}=-\frac vu.$$ $m\lt0$: the image is inverted (and, for a real object, real). $m\gt0$: the image is erect (and virtual). $|m|\gt1$: enlarged; $|m|\lt1$: diminished. Eliminating $v$ with the mirror equation gives the useful forms $m=\dfrac{f}{f-u}=\dfrac{f-v}{f}$.

The minus sign comes from the pole ray: triangles $A'B'P$ and $ABP$ give $\dfrac{|h'|}{|h|}=\dfrac{|v|}{|u|}$, and for the real image the image points downwards ($h'\lt0$) while $u$ and $v$ have the same sign. For a concave mirror and a real object the whole story is summarised below; a convex mirror always gives a virtual, erect, diminished image between $P$ and $F$, whatever the position of the object.

Object (concave mirror)Image positionNature
at infinityat $F$real, inverted, point-sized
beyond $C$between $F$ and $C$real, inverted, diminished
at $C$at $C$real, inverted, same size ($m=-1$)
between $C$ and $F$beyond $C$real, inverted, enlarged
at $F$at infinity— (rays leave parallel)
between $F$ and $P$behind the mirrorvirtual, erect, enlarged
✏️ Example 2 — A real, enlarged image
An object $4.0$ cm tall stands on the axis $30$ cm in front of a concave mirror of radius of curvature $40$ cm. Find the position, size and nature of the image.
Signs Cartesian convention: $u=-30$ cm, $R=-40$ cm, so $f=-20$ cm; $h=+4.0$ cm.
Image distance $\dfrac1v=\dfrac1f-\dfrac1u=-\dfrac1{20}+\dfrac1{30}=\dfrac{-3+2}{60}=-\dfrac1{60}$, so $v=-60$ cm.
Size $m=-\dfrac vu=-\dfrac{-60}{-30}=-2.0$, so $h'=mh=-8.0$ cm.
Answer The image is $60$ cm in front of the mirror, real (it can be caught on a screen), inverted and twice the size: $8.0$ cm tall. The object is between $F$ and $C$, as the table predicts.
✏️ Example 3 — The same mirror, object inside the focus
The object of Example 2 is moved to $10$ cm from the same mirror. Find the image.
Signs $u=-10$ cm, $f=-20$ cm (Cartesian convention).
Image distance $\dfrac1v=-\dfrac1{20}+\dfrac1{10}=+\dfrac1{20}$, so $v=+20$ cm.
Size $m=-\dfrac{20}{-10}=+2.0$, $h'=+8.0$ cm.
Answer A virtual, erect image $8.0$ cm tall, $20$ cm behind the mirror — the magnified face you see in a make-up mirror. It cannot be caught on a screen.

Using the Mirror Equation

🧭 Problem-solving strategy — spherical mirrors
  1. Draw the light travelling left to right with the mirror at the origin. Write down $u$, $f$ (or $R$) and $h$ with their signs before doing any arithmetic: a real object has $u\lt0$; concave $f\lt0$; convex $f\gt0$.
  2. Use $\dfrac1v=\dfrac1f-\dfrac1u$ and keep the signs inside the fractions.
  3. Read the sign of $v$: negative — real image in front; positive — virtual image behind.
  4. Find $m=-v/u$ and $h'=mh$; read the sign of $m$ (inverted or erect) and its size.
  5. Check against a rough ray diagram: does the image fall where the table (or the convex-mirror rule) says it should?
✏️ Example 4 — Closer than they appear
The convex wing mirror of a car has a radius of curvature of $2.0$ m. A car $1.5$ m tall is $9.0$ m behind the mirror. Find the position and height of its image, and explain the warning printed on the mirror.
Signs Cartesian convention: $u=-9.0$ m, $R=+2.0$ m, so $f=+1.0$ m.
Image distance $\dfrac1v=\dfrac11-\dfrac1{-9.0}=1+\dfrac19=\dfrac{10}{9}$, so $v=+0.90$ m: virtual, $0.90$ m behind the mirror.
Height $m=-\dfrac{0.90}{-9.0}=+0.10$, so $h'=0.10\times1.5=0.15$ m, erect.
Why the warning The image is one-tenth the size. Your brain judges distance partly from apparent size, and a car looking ten times smaller than it is seems much farther away than $9$ m. The convex mirror gains a wide field of view at the cost of misleading sizes.
✏️ Example 5 — Choosing the object distance for a given magnification
A concave mirror has focal length $15$ cm. Where must an object be placed to obtain (a) a real image three times the size of the object, and (b) a virtual image three times the size? Find the image distance in each case.
Signs $f=-15$ cm. A real image of a real object is inverted, so $m=-3$; a virtual image is erect, so $m=+3$.
(a) m = −3 $m=-v/u=-3\Rightarrow v=3u$. Then $\dfrac1{3u}+\dfrac1u=\dfrac{4}{3u}=-\dfrac1{15}$, so $u=-20$ cm and $v=-60$ cm: object $20$ cm in front, image $60$ cm in front.
(b) m = +3 $v=-3u$: $-\dfrac1{3u}+\dfrac1u=\dfrac{2}{3u}=-\dfrac1{15}$, so $u=-10$ cm and $v=+30$ cm: object $10$ cm in front, image $30$ cm behind.
Check With $m=\dfrac{f}{f-u}$: (a) $\dfrac{-15}{-15+20}=-3$ ✓; (b) $\dfrac{-15}{-15+10}=+3$ ✓. Interactive 2 (concave, object at $20$ cm) reads $v=-60.0$ cm, $m=-3.00$.
✏️ Example 6 — Designing a shaving mirror
A man wants a mirror that shows his face, held $25$ cm away, erect and $1.5$ times its natural size. What kind of mirror is needed, and what must its radius of curvature be?
Signs $u=-25$ cm; erect and enlarged means $m=+1.5$ (Cartesian convention).
Image $v=-mu$ $=-(1.5)(-25)$ $=+37.5$ cm (virtual, behind the mirror).
Focal length $\dfrac1f=\dfrac1{37.5}+\dfrac1{-25}=\dfrac{2-3}{75}=-\dfrac1{75}$, so $f=-75$ cm.
Answer $f\lt0$: a concave mirror with $R=2f=-150$ cm, i.e. a radius of curvature of $1.5$ m. The face is inside the focus ($25\lt75$ cm), as it must be for an erect, enlarged image.
Interactive 2 — Mirror Ray-Diagram Lab

Drag the object arrow along the axis (or press Play). Three principal rays leave its tip: one parallel to the axis (reflected through, or as if from, $F$), one to the pole (reflected symmetrically) and one along the line through $C$ (reflected back on itself). Where they meet is the image — solid rays for a real image, dashed extensions behind the mirror for a virtual one. The readouts come from $\tfrac1v+\tfrac1u=\tfrac1f$ with $|f|=15$ cm; the rays are drawn to the plane of the pole (the paraxial approximation).

object distance $u$—
image distance $v$—
magnification $m$—
nature of image—
⚠️ Common mistakes with spherical mirrors

1. Dropping the signs. Putting $u=30$ cm and $f=20$ cm into the mirror equation for a concave mirror gives $v=-60$ cm — the right number by luck but with the wrong meaning in other cases. Always write $u=-30$, $f=-20$ first.

2. Using the lens formula. Mirrors: $\tfrac1v+\tfrac1u=\tfrac1f$ and $m=-v/u$. Lenses (Section 9.3): $\tfrac1v-\tfrac1u=\tfrac1f$ and $m=+v/u$.

3. "Virtual means behind the screen." A virtual image is one the rays do not pass through. For a mirror it lies behind the mirror ($v\gt0$); it is still seen by the eye, but no screen can catch it.

4. Covering half the mirror. Every part of a mirror forms the whole image. Covering half the mirror halves the brightness; it does not remove half the image.

5. Forgetting the paraxial condition. $f=R/2$ and the mirror equation hold only for rays close to the axis. Wide rays focus closer to the mirror (Interactive 1).

🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — longitudinal magnification and the speed of an image

Not required by the CBSE syllabus. Spherical mirrors and magnification are listed by JEE Main, NEET and JEE Advanced; longitudinal magnification and the speed of an image are standard applications (they use differentiation) that no syllabus names.

Differentiate the mirror equation, keeping $f$ fixed: $-\dfrac{dv}{v^2}-\dfrac{du}{u^2}=0$, so $$\frac{dv}{du}=-\frac{v^2}{u^2}=-m^2.$$ A short object lying along the axis, of length $\Delta u$, therefore has an image of length $|\Delta v|\approx m^2|\Delta u|$ — the longitudinal magnification is $m^2$, not $m$. Dividing by $dt$, an object moving along the axis with velocity $v_O$ has an image moving with velocity $v_I=-m^2v_O$: the image always moves the opposite way along the axis. For example, an object $30$ cm in front of a concave mirror of focal length $20$ cm ($m=-2$) moving towards the mirror at $2.0$ cm/s ($du/dt=+2.0$ cm/s) has an image moving at $dv/dt=-8.0$ cm/s — away from the mirror, four times as fast. These results are used in JEE problems; they are not required for the board examination.

Exercises

I
Sign convention and focal length. Use the Cartesian sign convention throughout.
  • 1 A concave mirror has a radius of curvature of $24$ cm. Write its focal length with the correct sign.
  • 2 A convex mirror has a focal length of magnitude $15$ cm. Write $R$ and $f$ with signs.
  • 3 An object is $20$ cm in front of a mirror and its image is $10$ cm behind it. Write $u$ and $v$.
  • 4 A ray parallel to the axis strikes a concave mirror at height $h=0.60R$. At what fraction of $R$ from the pole does the reflected ray cross the axis?
  • 5 A concave mirror has $f=-10$ cm. Where is its centre of curvature?
  • 6 Treat a plane mirror as a spherical mirror with $R\to\infty$. What does the mirror equation give for $v$ and $m$?
  • 7 A concave mirror forms a real image $2.0$ cm long of an upright object. What is the sign of $h'$, and why?
1 $R=-24$ cm (centre in front), $f=R/2=-12$ cm.
2 $f=+15$ cm, $R=2f=+30$ cm (both behind the mirror).
3 $u=-20$ cm (against the incident light), $v=+10$ cm (along it).
4 $\sin\theta=0.60$, $\cos\theta=0.80$; $CF'=\dfrac{R}{2(0.80)}=0.625R$, so $PF'=R-0.625R=0.375R$ — compared with $0.5R$ for paraxial rays.
5 $R=2f=-20$ cm: $20$ cm in front of the pole, on the principal axis.
6 $\tfrac1f\to0$, so $\tfrac1v=-\tfrac1u$, $v=-u$: the image is as far behind as the object is in front, and $m=-v/u=+1$ (erect, same size).
7 Negative: heights below the axis are negative, and a real image formed by a single mirror is inverted, so $h'=-2.0$ cm.
II
The mirror equation. For each, find $v$ and $m$ and describe the image.
  • 1 Concave mirror, $f=15$ cm; object $45$ cm in front.
  • 2 Concave mirror, $f=15$ cm; object $10$ cm in front.
  • 3 Concave mirror, $f=15$ cm; object $30$ cm in front.
  • 4 Convex mirror, $f=20$ cm; object $20$ cm in front.
  • 5 Concave mirror, $f=12$ cm; an object $2.5$ cm tall is $18$ cm in front. Find also the image height.
  • 6 Concave mirror; object exactly at the focus.
  • 7 A concave mirror forms a real image, four times the size of the object, on a screen $2.0$ m from the mirror. Find the object distance and the focal length.
  • 8 A convex mirror forms an image one-quarter the size of an object placed $30$ cm in front of it. Find its focal length.
1 $f=-15$, $u=-45$: $\tfrac1v=-\tfrac1{15}+\tfrac1{45}=-\tfrac2{45}$, $v=-22.5$ cm; $m=-0.50$. Real, inverted, half-size, $22.5$ cm in front.
2 $u=-10$: $\tfrac1v=-\tfrac1{15}+\tfrac1{10}=\tfrac1{30}$, $v=+30$ cm; $m=+3.0$. Virtual, erect, three times enlarged, $30$ cm behind.
3 $u=-30$ (at $C$): $\tfrac1v=-\tfrac1{15}+\tfrac1{30}=-\tfrac1{30}$, $v=-30$ cm; $m=-1$. Real, inverted, same size, at $C$.
4 $f=+20$, $u=-20$: $\tfrac1v=\tfrac1{20}+\tfrac1{20}=\tfrac1{10}$, $v=+10$ cm; $m=+0.50$. Virtual, erect, half-size, $10$ cm behind.
5 $f=-12$, $u=-18$: $\tfrac1v=-\tfrac1{12}+\tfrac1{18}=-\tfrac1{36}$, $v=-36$ cm; $m=-2.0$, $h'=-5.0$ cm. Real, inverted, $5.0$ cm tall.
6 $u=f$: $\tfrac1v=0$, the image is at infinity — the reflected rays leave parallel (the headlamp arrangement).
7 Real and inverted: $m=-4$, $v=-200$ cm, so $u=v/4=-50$ cm. $\tfrac1f=-\tfrac1{200}-\tfrac1{50}=-\tfrac5{200}$, $f=-40$ cm: object $50$ cm in front, focal length $40$ cm.
8 Virtual and erect: $m=+\tfrac14$, $v=-mu=+7.5$ cm. $\tfrac1f=\tfrac1{7.5}-\tfrac1{30}=\tfrac{4-1}{30}=\tfrac1{10}$: $f=+10$ cm.
III
Reasoning with images.
  • 1 Use the mirror equation to show that a convex mirror always forms a virtual, erect, diminished image of a real object, between the pole and the focus.
  • 2 Where must a real object be placed in front of a concave mirror to obtain a virtual image? Show that the image is then always enlarged.
  • 3 Where must an object be placed in front of a concave mirror to obtain a real image of the same size?
  • 4 An object in front of a concave mirror of focal length $20$ cm is moved from $60$ cm to $25$ cm from the mirror. How do the image position and magnification change?
  • 5 Give two reasons why convex mirrors are used at blind corners and as rear-view mirrors.
  • 6 Why does a dentist use a concave mirror, held close to a tooth?
  • 7 The lower half of a concave mirror is covered with black paper. What happens to the image of an object?
1 $f\gt0$ and $u\lt0$, so $\tfrac1v=\tfrac1f-\tfrac1u=\tfrac1f+\tfrac1{|u|}\gt\tfrac1f\gt0$: $v$ is positive (virtual) and $v\lt f$ (between $P$ and $F$). Then $m=-v/u=v/|u|\gt0$ (erect), and $m=\dfrac{f}{f-u}=\dfrac{f}{f+|u|}\lt1$ (diminished).
2 $v\gt0$ requires $\tfrac1f-\tfrac1u\gt0$; with $f=-|f|$, $u=-|u|$: $\tfrac1{|u|}\gt\tfrac1{|f|}$, i.e. $|u|\lt|f|$ — between $P$ and $F$. Then $m=\dfrac{f}{f-u}=\dfrac{|f|}{|f|-|u|}\gt1$: always enlarged.
3 $m=-1$ gives $v=u$, so $\tfrac2u=\tfrac1f$, $u=2f=R$: at the centre of curvature.
4 $u=-60$: $v=-30$ cm, $m=-0.50$. $u=-25$: $\tfrac1v=-\tfrac1{20}+\tfrac1{25}=-\tfrac1{100}$, $v=-100$ cm, $m=-4.0$. The real image moves away from the mirror (from $30$ cm to $100$ cm) and grows from half-size to four times the size.
5 The image is always erect, and the mirror gives a much wider field of view than a plane mirror of the same size (the image of a wide scene is squeezed into a small region behind the mirror).
6 With the tooth inside the focus, a concave mirror gives an erect, enlarged virtual image; the mirror also concentrates light onto the tooth.
7 The full image is still formed — every part of the mirror receives rays from every point of the object — but with less light, so it is dimmer.
IV
Applications and longer problems.
  • 1 Modelling. A solar cooker uses a concave mirror of focal length $0.50$ m. The Sun subtends an angle of $0.53^\circ$ at the Earth. Find the diameter of the image of the Sun formed by the mirror.
  • 2 The reflector of a car headlamp is a concave mirror of radius of curvature $20$ cm. Where should the bulb be placed to produce a parallel beam?
  • 3 A make-up mirror is to show a face $20$ cm away erect and twice its size. Find the radius of curvature of the mirror required.
  • 4 A shop's security mirror is convex with a radius of curvature of $1.2$ m. A customer $1.7$ m tall stands $5.4$ m from it. Find the position and height of the image.
  • 5 A spherical mirror forms a virtual image one-third the size of an object placed $24$ cm in front of it. Is the mirror concave or convex? Find its radius of curvature.
  • 6 A concave mirror of focal length $10$ cm forms an image of a $1.0$ cm tall object placed $15$ cm in front of it. Find the image. If the object is moved $1.0$ cm closer to the mirror, how far does the image move, and which way?
1 The Sun is effectively at infinity, so its image is at the focus. Its diameter is $f\alpha$ with $\alpha=0.53^\circ=9.25\times10^{-3}$ rad: $d=0.50\times9.25\times10^{-3}\approx4.6\times10^{-3}$ m $=4.6$ mm.
2 At the focus: $|f|=R/2=10$ cm in front of the pole.
3 $u=-20$ cm, $m=+2$, $v=+40$ cm; $\tfrac1f=\tfrac1{40}-\tfrac1{20}=-\tfrac1{40}$, $f=-40$ cm, $R=-80$ cm: a concave mirror of radius of curvature $80$ cm.
4 $f=+0.60$ m, $u=-5.4$ m: $\tfrac1v=\tfrac1{0.60}+\tfrac1{5.4}=\tfrac{9+1}{5.4}$, $v=+0.54$ m (virtual, behind the mirror); $m=+0.10$, image height $0.17$ m, erect.
5 A virtual diminished image of a real object can only come from a convex mirror. $m=+\tfrac13$, $v=+8.0$ cm; $\tfrac1f=\tfrac18-\tfrac1{24}=\tfrac2{24}$, $f=+12$ cm, $R=+24$ cm.
6 $f=-10$, $u=-15$: $\tfrac1v=-\tfrac1{10}+\tfrac1{15}=-\tfrac1{30}$, $v=-30$ cm, $m=-2$, $h'=-2.0$ cm (real, inverted). At $u=-14$: $\tfrac1v=-\tfrac1{10}+\tfrac1{14}=-\tfrac{4}{140}$, $v=-35$ cm. The image moves $5.0$ cm farther from the mirror — five times as far as the object moved, in the opposite direction.
🎯 Exam corner — spherical mirrors BoardJEE MainJEE AdvancedNEET

E1. Board · 3 marks With the help of a ray diagram, derive the mirror equation for a concave mirror forming a real image. Using it, show that a convex mirror always produces a virtual image of a real object.

Diagram Object $AB$ on the axis, a ray from $A$ parallel to the axis reflected through $F$, a ray from $A$ to the pole $P$ reflected symmetrically; they meet at $A'$; $A'B'$ is the real, inverted image. $M$ is where the first ray meets the (nearly flat, paraxial) mirror.
Similar triangles $\triangle A'B'F\sim\triangle MPF$: $\dfrac{B'A'}{BA}=\dfrac{B'F}{FP}$. $\triangle A'B'P\sim\triangle ABP$ (equal angles of incidence and reflection): $\dfrac{B'A'}{BA}=\dfrac{B'P}{BP}$. So $\dfrac{B'P-FP}{FP}=\dfrac{B'P}{BP}$.
Signs Cartesian convention: $B'P=-v$, $FP=-f$, $BP=-u$. Then $\dfrac{f-v}{-f}=\dfrac vu$, which rearranges to $\dfrac1v+\dfrac1u=\dfrac1f$.
Convex mirror $f\gt0$, $u\lt0$: $\dfrac1v=\dfrac1f-\dfrac1u\gt0$, so $v\gt0$ — the image is always behind the mirror, i.e. virtual. Exam tip: state the sign convention in one line before substituting; it carries a mark.

E2. Board · 1 mark Assertion (A): A concave mirror brings red and violet light parallel to its axis to the same focus. Reason (R): The focal length of a spherical mirror, $f=R/2$, depends only on its radius of curvature. (A) Both A and R are true and R is the correct explanation of A. (B) Both are true but R is not the correct explanation of A. (C) A is true, R is false. (D) A is false, R is true.

Concept Reflection obeys $i=r$ for every wavelength, so the geometry — and hence $f=R/2$ — is the same for all colours. A mirror has no colour fringes (no chromatic aberration), unlike a lens.
Answer (A). Exam tip: this is one reason large telescopes use mirrors (Section 9.4).

E3. JEE Main A concave mirror forms a real image three times the size of an object for one object position, and a virtual image three times the size for another. The two object positions are $10$ cm apart. The focal length of the mirror is (A) $10$ cm (B) $15$ cm (C) $20$ cm (D) $30$ cm.

Setup From $m=\dfrac{f}{f-u}$: $u=f\left(1-\dfrac1m\right)$. Cartesian convention, $f=-F$.
Positions $m=-3$: $u=-F\cdot\tfrac43=-\tfrac43F$. $m=+3$: $u=-F\cdot\tfrac23=-\tfrac23F$. Separation $\tfrac23F=10$ cm.
Answer (B) $F=15$ cm. Check: Example 5 — positions $20$ cm and $10$ cm. ✓

E4. NEET An object is placed $10$ cm in front of a concave mirror of focal length $15$ cm. The image is (A) real, $30$ cm in front of the mirror (B) virtual, $30$ cm behind the mirror, three times enlarged (C) virtual, $6.0$ cm behind the mirror (D) real, $6.0$ cm in front of the mirror.

Signs $u=-10$ cm, $f=-15$ cm.
Equation $\dfrac1v=-\dfrac1{15}+\dfrac1{10}=\dfrac1{30}$: $v=+30$ cm, $m=-\dfrac{30}{-10}=+3$.
Answer (B). Exam tip: (C) and (D) come from adding $\tfrac1{10}+\tfrac1{15}$ — the result of forgetting the signs.

E5. JEE Advanced (One or more correct.) A real object is placed in front of a concave mirror of focal length of magnitude $F$. Then (A) the image is the same size as the object when the object is at distance $2F$; (B) whenever the image is virtual it is enlarged; (C) a real image is always at least a distance $F$ from the mirror; (D) a real image can be erect.

Formula With $f=-F$, $u=-U$: $m=\dfrac{f}{f-u}=\dfrac{F}{F-U}$ and $v=\dfrac{uf}{u-f}=-\dfrac{UF}{U-F}$.
Options (A) $U=2F$: $m=-1$, same size — true. (B) Virtual needs $U\lt F$, then $m=\dfrac{F}{F-U}\gt1$ — true. (C) Real needs $U\gt F$; then $|v|=\dfrac{UF}{U-F}\gt F$ because $\dfrac{U}{U-F}\gt1$ — true. (D) Real images of real objects have $m=\dfrac F{F-U}\lt0$ — always inverted: false.
Answer (A), (B), (C). Exam tip: the two forms $m=\dfrac f{f-u}$ and $m=\dfrac{f-v}f$ settle most "which statements are true" items in one line.

E6. Board · Case study A truck has a convex rear-view mirror of radius of curvature $3.0$ m. A car $1.8$ m wide is $12$ m behind it. (i) (2 marks) Find the position of the image of the car and the width of the image. (ii) (1 mark) State the nature of the image. (iii) (1 mark) The car comes to $3.0$ m from the mirror. How does its image change?

(i) Cartesian convention: $f=+1.5$ m, $u=-12$ m. $\dfrac1v=\dfrac1{1.5}+\dfrac1{12}=\dfrac{8+1}{12}$, $v=+1.33$ m behind the mirror. $m=-\dfrac{v}{u}=\dfrac{1.33}{12}\approx0.111$, width $\approx0.20$ m.
(ii) Virtual, erect and diminished (as always for a convex mirror).
(iii) $u=-3.0$ m: $\dfrac1v=\dfrac1{1.5}+\dfrac13=1$, $v=+1.0$ m, $m=+\tfrac13$: the image moves towards the mirror and grows to $0.60$ m wide, still erect and virtual. Exam tip: in case studies, reuse $f$ from part (i) — do not recompute it.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. What radius of curvature should a convex safety mirror at a blind corner have, so that drivers see a wide stretch of road but approaching vehicles are not shrunk too much to judge?

Why it matters. Convex traffic mirrors are fixed at hairpin bends in hill roads from the Himalayas to the Alps, at car-park exits and at factory crossings. Road-safety agencies specify their size and curvature. Too flat a mirror shows too little of the road; too strongly curved a mirror shows everything but so small that drivers misjudge speed and distance.

What to do.

  1. Use $m=\dfrac{f}{f-u}$ with $f=R/2\gt0$ and $u=-d$. For $R=1.0$ m and $R=2.0$ m, find $m$ for a vehicle at $d=5$, $20$ and $50$ m. Tabulate.
  2. A rough estimate (a model you should state): for a mirror of aperture $D$ seen from far away, the normals at the two edges are tilted by $\pm D/2R$, so the field of view is about $2D/R$ radians. Find it for $D=0.60$ m and both values of $R$, in degrees.
  3. A driver can recognise a car whose image is at least about $1.5$ cm long when seen from $10$ m (an illustrative threshold). For a car $4.0$ m long, how far away can it be for each mirror?
  4. Recommend a radius of curvature and justify the trade-off in a short paragraph.

Tools: Calculator or spreadsheet; a shiny steel spoon (its back is a convex mirror) to try the effect.

Hand in: The table of magnifications, the field-of-view estimates, the recognition distances and your recommendation.

Going further. Measure the radius of curvature of the back of a spoon: hold a ruler at a known distance, measure the size of its image, and use the mirror equation in reverse.

Instructor note: $R=1.0$ m ($f=0.50$ m): $m=0.091,\ 0.024,\ 0.0099$ at $5,20,50$ m. $R=2.0$ m: $m=0.167,\ 0.048,\ 0.020$. Field of view $2D/R$: $1.2$ rad $\approx69^\circ$ and $0.60$ rad $\approx34^\circ$. Recognition: need $m\times4.0\ \text{m}\ge0.015$ m, i.e. $m\ge0.00375$: $\dfrac{f}{f+d}\ge0.00375\Rightarrow d\le f(1/0.00375-1)$, about $133$ m for $R=1.0$ m and $266$ m for $R=2.0$ m (the threshold is illustrative, so only the ratio matters: the flatter mirror shows cars twice as far away but with half the field of view).

🧪 Self-check quiz

0/5 correct

Q1. In the Cartesian sign convention, the focal length of a convex mirror of radius of curvature $30$ cm is

Q2. A mirror forms an image with magnification $m=-0.5$. The image is

Q3. Which mirror forms a virtual, erect, diminished image of a real object wherever the object is placed?

Q4. An object is $15$ cm in front of a concave mirror of focal length $10$ cm. The image is

Q5. The relation $f=R/2$ for a spherical mirror holds

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