Chapter 1 · Light – Reflection and Refraction

1.2 Image Formation, the Mirror Formula and Magnification

Where is the image, how big is it, and is it real? Two rays answer by drawing; the mirror formula $\frac1v+\frac1u=\frac1f$ and the magnification $m=-\frac vu$ answer by calculation.

A dentist tilts a small round mirror behind your teeth and sees each tooth larger than life. A bus driver glances at a bulging mirror and sees the whole road behind, every car shrunk. A torch throws a beam across a dark field. All three use spherical mirrors from Section 1.1 — the difference is only where the object sits compared with the focus and the centre of curvature.

In this section we find images in two ways. First by drawing: two of the standard rays from the top of the object meet at the top of the image. Then by calculation: with a careful sign convention, one short formula gives the image position and a second gives its size. You will be able to predict the image for any object position — and check every prediction in the live ray-diagram lab.

Images Formed by a Concave Mirror

Place a small upright object $AB$ on the principal axis with its foot $A$ on the axis. Draw two standard rays from the top $B$: for example the ray parallel to the axis (which reflects through $F$) and the ray to the pole (which reflects at an equal angle below the axis). Where the reflected rays cross is $B'$, the image of $B$; the image of $A$ lies straight above or below it on the axis. If the reflected rays spread apart, extend them backwards behind the mirror (dashed): they meet at a virtual image.

📘 Definition 1.3 — Real and virtual images
A real image is formed where reflected (or refracted) rays actually meet; it can be caught on a screen and, for a single mirror, it is always inverted. A virtual image is formed where rays only appear to meet when traced backwards; it cannot be caught on a screen and, for a single mirror, it is always erect.

Moving the object step by step towards a concave mirror gives six cases. Try each one in Interactive 1.

Position of objectPosition of imageSize of imageNature
At infinityAt $F$Highly diminished, point-sizedReal, inverted
Beyond $C$Between $F$ and $C$DiminishedReal, inverted
At $C$At $C$Same sizeReal, inverted
Between $C$ and $F$Beyond $C$EnlargedReal, inverted
At $F$At infinityHighly enlargedReal, inverted
Between $F$ and $P$Behind the mirrorEnlargedVirtual, erect

Notice the pattern: as the object comes in from far away towards $F$, the real image runs out from $F$ towards infinity and grows. Once the object passes inside $F$, the rays can no longer meet in front and the image jumps behind the mirror, virtual and erect.

Images Formed by a Convex Mirror

For a convex mirror the reflected rays always spread out, so the image is always behind the mirror, virtual and erect. Only two cases are needed.

Position of objectPosition of imageSize of imageNature
At infinityAt $F$, behind the mirrorHighly diminished, point-sizedVirtual, erect
Anywhere between infinity and $P$Between $P$ and $F$, behind the mirrorDiminishedVirtual, erect
🧭 Uses of spherical mirrors — and why
  • Concave: torches, vehicle headlights and searchlights (the bulb at $F$ gives a strong parallel beam); shaving and make-up mirrors (face inside $F$ gives an enlarged, erect image); dentists' mirrors (enlarged view of a tooth); solar cookers and furnaces (sunlight gathered at $F$).
  • Convex: rear-view and side mirrors of vehicles, mirrors at blind road bends and in shops — the image is always erect, and the bulge reflects light from a much wider field of view than a plane mirror of the same size.
Interactive 1 — The Mirror Ray-Diagram Lab

Drag the tip of the green object (or use the slider) to move it along the principal axis. Two standard rays from its tip — parallel to the axis, and to the pole — reflect and meet at the image. The image is drawn from the mirror formula, so you can watch the rays agree with it. Real images are solid red; virtual images are dashed, with the rays traced back behind the mirror. Press Play to sweep the object from far away right up to the mirror.

4515
$u$—
$v$—
$m=-v/u$—
image—
object is—

The New Cartesian Sign Convention

To calculate with mirrors we must say not only how far the object and image are from the mirror, but on which side. We do this with signs, exactly as on a number line, with the pole as the origin.

📘 Definition 1.4 — New Cartesian sign convention
  1. The object is placed to the left of the mirror, so light falls on the mirror from the left.
  2. All distances along the principal axis are measured from the pole $P$.
  3. Distances measured in the direction of the incident light (to the right of $P$) are positive; distances against it (to the left of $P$) are negative.
  4. Heights measured upwards from the principal axis are positive; downwards, negative.

Consequences. The object distance $u$ is always negative. The focal length is negative for a concave mirror ($F$ is in front) and positive for a convex mirror ($F$ is behind). A real image (in front) has $v$ negative; a virtual image (behind) has $v$ positive.

📐 Law 1.4 — The mirror formula and magnification

For a spherical mirror of small aperture, with the object distance $u$, image distance $v$ and focal length $f$ all measured using the New Cartesian sign convention,

$$\frac1v+\frac1u=\frac1f .$$

The magnification $m$ is the ratio of the height of the image $h'$ to the height of the object $h$:

$$m=\frac{h'}{h}=-\frac vu .$$

$m$ negative ⟹ real and inverted; $m$ positive ⟹ virtual and erect; $|m|\gt1$ enlarged, $|m|\lt1$ diminished, $|m|=1$ same size.

🧭 Problem-solving strategy — mirror numericals
  1. Write the sign convention: object on the left, distances from $P$, light travels left to right.
  2. Put signs on the data: $u$ negative; $f$ negative for concave, positive for convex. Keep all lengths in the same unit.
  3. Rearrange: $\dfrac1v=\dfrac1f-\dfrac1u$. Substitute with brackets around negative numbers, and combine the fractions.
  4. Find $m=-\dfrac vu$ and, if needed, $h'=mh$.
  5. Interpret every sign in words: where the image is, real or virtual, erect or inverted, its size. Check against the table above.
✏️ Example 1 — A real, enlarged image
A $2.0\ \text{cm}$ tall object stands $25\ \text{cm}$ in front of a concave mirror of focal length $15\ \text{cm}$. Find the position, size and nature of the image.
Signs New Cartesian convention: $u=-25\ \text{cm}$, $f=-15\ \text{cm}$ (concave), $h=+2.0\ \text{cm}$.
Formula $$\frac1v=\frac1f-\frac1u=\frac{1}{-15}-\frac{1}{-25}=-\frac{1}{15}+\frac{1}{25}=\frac{-5+3}{75}=-\frac{2}{75},$$ so $v=-37.5\ \text{cm}$.
Size $m=-\dfrac vu=-\dfrac{-37.5}{-25}=-1.5$, and $h'=mh=-1.5\times2.0=-3.0\ \text{cm}$.
Meaning The image is $37.5\ \text{cm}$ in front of the mirror, real, inverted and $1.5$ times the size ($3.0\ \text{cm}$ tall). Check: the object ($25\ \text{cm}$) is between $F$ ($15\ \text{cm}$) and $C$ ($30\ \text{cm}$), so the table predicts an enlarged real image beyond $C$. ✓
✏️ Example 2 — The shaving mirror
A face is $10\ \text{cm}$ from a concave mirror of focal length $20\ \text{cm}$. Find the image position and magnification.
Signs $u=-10\ \text{cm}$, $f=-20\ \text{cm}$.
Formula $$\frac1v=\frac{1}{-20}-\frac{1}{-10}=-\frac1{20}+\frac{2}{20}=\frac1{20},\qquad v=+20\ \text{cm}.$$
Size $m=-\dfrac{+20}{-10}=+2.0$.
Meaning $v$ positive: the image is $20\ \text{cm}$ behind the mirror, virtual and erect, twice the size — exactly what you want for shaving. The face is inside $F$, as the table requires. ✓
✏️ Example 3 — A bus in a rear-view mirror
A convex rear-view mirror has a radius of curvature of $3.0\ \text{m}$. A bus $3.0\ \text{m}$ tall is $5.0\ \text{m}$ behind it. Locate the image and find its height.
Signs Convex: $f=+\dfrac R2=+1.5\ \text{m}$; $u=-5.0\ \text{m}$; $h=+3.0\ \text{m}$.
Formula $$\frac1v=\frac{1}{1.5}-\frac{1}{-5.0}=0.6667+0.2000=0.8667\ \text{m}^{-1},\qquad v=+1.15\ \text{m}.$$
Size $m=-\dfrac{1.154}{-5.0}=+0.23$, so $h'=0.23\times3.0\approx0.69\ \text{m}$.
Meaning The image is $1.15\ \text{m}$ behind the mirror — between $P$ and $F$ — virtual, erect and less than a quarter of the size. The driver's brain reads "small" as "far away", which is why such mirrors carry the warning that objects are closer than they appear.
✏️ Example 4 — Finding the focal length from an experiment
A candle is $30\ \text{cm}$ in front of a spherical mirror. A sharp image of the flame is caught on a screen $60\ \text{cm}$ in front of the mirror. Find the focal length, the type of mirror and the magnification.
Signs The image is on a screen, so it is real and in front: $u=-30\ \text{cm}$, $v=-60\ \text{cm}$.
Formula $$\frac1f=\frac1v+\frac1u=\frac{1}{-60}+\frac{1}{-30}=-\frac{1+2}{60}=-\frac1{20},\qquad f=-20\ \text{cm}.$$
Meaning $f$ is negative, so the mirror is concave, with $R=40\ \text{cm}$. $m=-\dfrac{-60}{-30}=-2$: the flame's image is inverted and twice as tall.
✏️ Example 5 — Designing for a magnification
Where should an object be placed in front of a concave mirror of focal length $12\ \text{cm}$ to obtain an image three times the size of the object if the image is (a) real, (b) virtual?
(a) Signs Real images are inverted: $m=-3$, so $-\dfrac vu=-3$ and $v=3u$. With $f=-12\ \text{cm}$: $$\frac{1}{3u}+\frac1u=\frac{4}{3u}=-\frac1{12}\ \Rightarrow\ u=-16\ \text{cm},\ v=-48\ \text{cm}.$$
(b) Signs Virtual images are erect: $m=+3$, so $v=-3u$: $$\frac{1}{-3u}+\frac1u=\frac{2}{3u}=-\frac1{12}\ \Rightarrow\ u=-8\ \text{cm},\ v=+24\ \text{cm}.$$
Check (a) $16\ \text{cm}$ lies between $F$ ($12$) and $C$ ($24$) ✓; (b) $8\ \text{cm}$ lies inside $F$ ✓. Substituting back: $\tfrac{1}{-48}+\tfrac{1}{-16}=-\tfrac{4}{48}=-\tfrac1{12}$ and $\tfrac1{24}+\tfrac{1}{-8}=-\tfrac{2}{24}=-\tfrac1{12}$. ✓
✏️ Example 6 — The image runs away
An object moves from $60\ \text{cm}$ to $30\ \text{cm}$ in front of a concave mirror of focal length $20\ \text{cm}$. How far does the image move, and how does its size change?
Start $u=-60$, $f=-20$: $\dfrac1v=-\dfrac1{20}+\dfrac1{60}=-\dfrac{2}{60}$, so $v=-30\ \text{cm}$ and $m=-0.5$.
End $u=-30$: $\dfrac1v=-\dfrac1{20}+\dfrac1{30}=-\dfrac1{60}$, so $v=-60\ \text{cm}$ and $m=-2$.
Meaning The object moved $30\ \text{cm}$ towards the mirror; the real image moved $30\ \text{cm}$ away from it (from $30$ to $60\ \text{cm}$) and grew from half-size to double-size. Object and image always move in opposite directions along the axis for a real image.
⚠️ Common mistakes

1. Dropping a sign. $f$ of a concave mirror is negative even though "focal length $15\ \text{cm}$" is written without a sign. Put signs on the data before you substitute.

2. Missing brackets. $\dfrac1f-\dfrac1u$ with $u=-25$ is $\dfrac1f+\dfrac1{25}$. Writing the brackets first prevents a double-negative slip.

3. Inverting too early. $\dfrac1v=-\dfrac{2}{75}$ gives $v=-\dfrac{75}{2}$, not $-\dfrac{2}{75}$ and not $-\dfrac1{15}-\dfrac1{25}$ turned upside down term by term.

4. Forgetting the minus in $m=-\dfrac vu$. For mirrors the formula has a minus sign (the lens formula in Section 1.4 does not).

🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — optional: where the formulas come from

In JEE Main, NEET and JEE Advanced 2026; the CBSE Class 10 syllabus uses the mirror formula but does not require its derivation.

Take a concave mirror and a real image, and use lengths without signs. The ray to the pole makes equal angles with the axis, so the triangles formed by object, pole and image are similar: $\dfrac{|h'|}{h}=\dfrac{|v|}{|u|}$. The parallel ray meets the mirror at height $h$ and passes through $F$; for a small aperture the triangles on either side of $F$ are similar too: $\dfrac{|h'|}{h}=\dfrac{|v|-|f|}{|f|}$. Equating and dividing by $|u||v||f|$ gives $\dfrac1{|u|}+\dfrac1{|v|}=\dfrac1{|f|}$. With $u=-|u|$, $v=-|v|$, $f=-|f|$ this is exactly $\dfrac1v+\dfrac1u=\dfrac1f$, and the image is inverted, so $m=-\dfrac{|v|}{|u|}=-\dfrac vu$. The sign convention lets the same formula cover every other case.

Interactive 2 — The Whole Image Table on One Graph

The graph plots the image distance $v$ against the object distance $u$ from $\dfrac1v=\dfrac1f-\dfrac1u$, both with their signs. Below the $u$-axis ($v\lt0$) the image is real and in front; above it ($v\gt0$) it is virtual and behind. The coloured bands are the zones of the table. Watch the point as the object moves in: the image races off to infinity at $F$ and reappears behind the mirror. The dashed line $v=u$ crosses the curve at $C$.

20−50
$u$—
$v$—
$m$—
image—

Exercises

I
Images by drawing. A concave mirror has $f=10\ \text{cm}$. Describe the position, size and nature of the image when the object is at the given distance in front of it, and draw a ray diagram for each.
  • 1 $30\ \text{cm}$
  • 2 $20\ \text{cm}$
  • 3 $15\ \text{cm}$
  • 4 $10\ \text{cm}$
  • 5 $6\ \text{cm}$
  • 6 Describe the image formed by a convex mirror for an object placed anywhere in front of it.
  • 7 A mirror always gives an erect image smaller than the object, wherever the object is. Which mirror is it?
  • 8 Why do dentists use concave mirrors rather than plane mirrors?
1 Beyond $C$ ($C$ at $20\ \text{cm}$): image between $F$ and $C$, real, inverted, diminished (formula: $v=-15\ \text{cm}$, $m=-0.5$).
2 At $C$: image at $C$, real, inverted, same size ($v=-20\ \text{cm}$, $m=-1$).
3 Between $C$ and $F$: image beyond $C$, real, inverted, enlarged ($v=-30\ \text{cm}$, $m=-2$).
4 At $F$: reflected rays are parallel; the image is at infinity, real, inverted, highly enlarged.
5 Between $F$ and $P$: image behind the mirror, virtual, erect, enlarged ($v=+15\ \text{cm}$, $m=+2.5$).
6 Behind the mirror between $P$ and $F$, virtual, erect and diminished.
7 A convex mirror.
8 With the tooth inside $F$ a concave mirror gives an erect, enlarged virtual image, so small details are easier to see.
II
The mirror formula. Use the New Cartesian sign convention throughout. Find $v$ and $m$ (or $f$ where asked) and state the nature of the image.
  • 1 Concave, $f=10\ \text{cm}$, object $30\ \text{cm}$ in front.
  • 2 Concave, $f=10\ \text{cm}$, object $15\ \text{cm}$ in front.
  • 3 Concave, $f=10\ \text{cm}$, object $5\ \text{cm}$ in front.
  • 4 Convex, $f=10\ \text{cm}$, object $10\ \text{cm}$ in front.
  • 5 Convex, $R=40\ \text{cm}$, object $30\ \text{cm}$ in front.
  • 6 Concave, $R=36\ \text{cm}$, object $27\ \text{cm}$ in front.
  • 7 An object $20\ \text{cm}$ in front of a mirror gives a real image $30\ \text{cm}$ in front. Find $f$ and the type of mirror.
  • 8 An object $6\ \text{cm}$ in front of a mirror gives a virtual image $12\ \text{cm}$ behind it. Find $f$ and the type of mirror.
  • 9 An object $12\ \text{cm}$ in front of a mirror gives a virtual image $4\ \text{cm}$ behind it. Find $f$ and the type of mirror.
  • 10 An object is at the focus of a concave mirror of focal length $25\ \text{cm}$. What does the formula say about $v$?
Check: type in $u$ and $f$ with their signs and compare $v$ and $m$ with your answer
$v$—
$m$—
image—
1 $u=-30$, $f=-10$: $\tfrac1v=-\tfrac1{10}+\tfrac1{30}=-\tfrac{2}{30}$, $v=-15\ \text{cm}$; $m=-0.5$. Real, inverted, half size, $15\ \text{cm}$ in front.
2 $\tfrac1v=-\tfrac1{10}+\tfrac1{15}=-\tfrac1{30}$, $v=-30\ \text{cm}$; $m=-2$. Real, inverted, enlarged.
3 $\tfrac1v=-\tfrac1{10}+\tfrac15=\tfrac1{10}$, $v=+10\ \text{cm}$; $m=+2$. Virtual, erect, enlarged, $10\ \text{cm}$ behind.
4 $f=+10$: $\tfrac1v=\tfrac1{10}+\tfrac1{10}=\tfrac15$, $v=+5\ \text{cm}$; $m=+0.5$. Virtual, erect, half size.
5 $f=+20$: $\tfrac1v=\tfrac1{20}+\tfrac1{30}=\tfrac{5}{60}$, $v=+12\ \text{cm}$; $m=+0.4$. Virtual, erect, diminished.
6 $f=-18$: $\tfrac1v=-\tfrac1{18}+\tfrac1{27}=\tfrac{-3+2}{54}=-\tfrac1{54}$, $v=-54\ \text{cm}$; $m=-2$. Real, inverted, enlarged.
7 $u=-20$, $v=-30$: $\tfrac1f=-\tfrac1{30}-\tfrac1{20}=-\tfrac{5}{60}$, $f=-12\ \text{cm}$: concave.
8 $u=-6$, $v=+12$: $\tfrac1f=\tfrac1{12}-\tfrac16=-\tfrac1{12}$, $f=-12\ \text{cm}$: concave ($m=+2$).
9 $u=-12$, $v=+4$: $\tfrac1f=\tfrac14-\tfrac1{12}=\tfrac{2}{12}$, $f=+6\ \text{cm}$: convex ($m=+\tfrac13$).
10 $\tfrac1v=\tfrac{1}{-25}-\tfrac{1}{-25}=0$: no finite $v$ — the reflected rays are parallel and the image is at infinity.
III
Magnification and image height.
  • 1 A $4.0\ \text{cm}$ object is $30\ \text{cm}$ in front of a concave mirror of focal length $20\ \text{cm}$. Find the image height.
  • 2 A $5.0\ \text{cm}$ object is $20\ \text{cm}$ in front of a convex mirror of focal length $15\ \text{cm}$. Find the image distance and height.
  • 3 A concave mirror gives $m=-3$ for an object $20\ \text{cm}$ in front of it. Find $v$ and $f$.
  • 4 A convex mirror gives $m=+0.25$ for an object $30\ \text{cm}$ in front of it. Find $v$ and $f$.
  • 5 A mirror forms an erect image three times as large as the object. Which mirror is it, and where is the object?
  • 6 Can a convex mirror ever give $m=-1$? Explain.
  • 7 Where must an object be placed in front of a concave mirror of focal length $15\ \text{cm}$ to give a real image of the same size?
1 $\tfrac1v=-\tfrac1{20}+\tfrac1{30}=-\tfrac1{60}$, $v=-60\ \text{cm}$; $m=-2$; $h'=-8.0\ \text{cm}$ (inverted, $8.0\ \text{cm}$ tall).
2 $\tfrac1v=\tfrac1{15}+\tfrac1{20}=\tfrac{7}{60}$, $v=+8.57\ \text{cm}$; $m=\tfrac{8.57}{20}=+0.43$; $h'=+2.1\ \text{cm}$ (erect, virtual).
3 $-\tfrac vu=-3$ gives $v=3u=-60\ \text{cm}$; $\tfrac1f=-\tfrac1{60}-\tfrac1{20}=-\tfrac{4}{60}$, $f=-15\ \text{cm}$.
4 $v=-mu=+7.5\ \text{cm}$; $\tfrac1f=\tfrac1{7.5}-\tfrac1{30}=\tfrac{4-1}{30}=\tfrac1{10}$, $f=+10\ \text{cm}$.
5 An enlarged erect image needs a concave mirror with the object between $P$ and $F$.
6 No. For a convex mirror $u\lt0$ and $f\gt0$, so $\tfrac1v=\tfrac1f+\tfrac1{|u|}\gt0$: $v$ is positive and $m=-\tfrac vu$ is always positive (and less than 1).
7 At $C$: $u=2f=-30\ \text{cm}$, i.e. $30\ \text{cm}$ in front ($v=-30\ \text{cm}$, $m=-1$).
IV
Applications and challenges.
  • 1 Make-up mirror. A face $15\ \text{cm}$ from a concave mirror should see an erect image twice its size. Find the focal length and radius of curvature needed.
  • 2 Dentistry. A dentist holds a concave mirror of focal length $2.0\ \text{cm}$ at $1.5\ \text{cm}$ from a tooth. Find the image position and magnification.
  • 3 Road safety. A convex mirror at a hill bend has $R=2.0\ \text{m}$. A car $1.5\ \text{m}$ tall is $9.0\ \text{m}$ away. Find the image distance and height.
  • 4 Projection. A concave mirror must throw a real image of a lamp filament, magnified 4 times, onto a screen $2.0\ \text{m}$ from the mirror. Where must the filament be, and what focal length is needed?
  • 5 An object moves from $40\ \text{cm}$ to $25\ \text{cm}$ in front of a concave mirror of focal length $20\ \text{cm}$. How far does the image move, and how does $m$ change?
  • 6 Challenge. Prove from the mirror formula that a convex mirror always forms its image between $P$ and $F$.
1 $u=-15\ \text{cm}$, $m=+2$ so $v=-mu=+30\ \text{cm}$. $\tfrac1f=\tfrac1{30}+\tfrac1{-15}=-\tfrac1{30}$: $f=-30\ \text{cm}$ (concave, focal length $30\ \text{cm}$), $R=60\ \text{cm}$.
2 $u=-1.5$, $f=-2.0$: $\tfrac1v=-0.500+0.667=0.167\ \text{cm}^{-1}$, $v=+6.0\ \text{cm}$ (behind); $m=-\tfrac{6.0}{-1.5}=+4.0$, virtual and erect.
3 $f=+1.0\ \text{m}$, $u=-9.0\ \text{m}$: $\tfrac1v=1+\tfrac19=\tfrac{10}{9}$, $v=+0.90\ \text{m}$; $m=+0.10$, $h'=0.15\ \text{m}$.
4 Real, so $m=-4$ and $v=-200\ \text{cm}$; $u=\tfrac{v}{4}=-50\ \text{cm}$ (from $m=-\tfrac vu$). $\tfrac1f=-\tfrac1{200}-\tfrac1{50}=-\tfrac5{200}$: $f=-40\ \text{cm}$. Filament $50\ \text{cm}$ in front of a concave mirror of focal length $40\ \text{cm}$.
5 At $40\ \text{cm}$ (at $C$): $v=-40\ \text{cm}$, $m=-1$. At $25\ \text{cm}$: $\tfrac1v=-\tfrac1{20}+\tfrac1{25}=-\tfrac1{100}$, $v=-100\ \text{cm}$, $m=-4$. The image moves $60\ \text{cm}$ away from the mirror and becomes four times larger.
6 With $u=-|u|$ and $f\gt0$: $\tfrac1v=\tfrac1f+\tfrac1{|u|}$. This is positive, so $v\gt0$ (behind the mirror), and it is larger than $\tfrac1f$, so $v\lt f$. Hence $0\lt v\lt f$: the image lies between $P$ and $F$.
🎯 Exam corner — board and entrance-exam practice BoardNEETJEE Main

E1. Board An object $3.0\ \text{cm}$ high is placed $18\ \text{cm}$ in front of a concave mirror of focal length $12\ \text{cm}$. Find the position, nature and size of the image. [3 marks]

Convention New Cartesian: $u=-18\ \text{cm}$, $f=-12\ \text{cm}$, $h=+3.0\ \text{cm}$.
Working $\tfrac1v=-\tfrac1{12}+\tfrac1{18}=\tfrac{-3+2}{36}=-\tfrac1{36}$, so $v=-36\ \text{cm}$; $m=-\tfrac{-36}{-18}=-2$; $h'=-6.0\ \text{cm}$.
Answer $36\ \text{cm}$ in front of the mirror; real, inverted, $6.0\ \text{cm}$ tall. Exam tip: examiners award a mark just for writing the sign convention and the signed data — do it every time.

E2. Board Draw a ray diagram for an object placed between infinity and the pole of a convex mirror. An object $2.0\ \text{cm}$ tall is $10\ \text{cm}$ from a convex mirror of radius of curvature $30\ \text{cm}$. Find the position and height of the image. [5 marks]

Diagram From the top of the object draw a ray parallel to the axis; after reflection it appears to come from $F$ behind the mirror (dashed back-extension). Draw a second ray aimed at $C$; it returns along itself. The dashed extensions meet behind the mirror between $P$ and $F$: a small, erect, virtual image.
Convention $f=+\tfrac{30}{2}=+15\ \text{cm}$, $u=-10\ \text{cm}$, $h=+2.0\ \text{cm}$.
Working $\tfrac1v=\tfrac1{15}+\tfrac1{10}=\tfrac{5}{30}$, $v=+6.0\ \text{cm}$; $m=-\tfrac{6.0}{-10}=+0.60$; $h'=+1.2\ \text{cm}$.
Answer $6.0\ \text{cm}$ behind the mirror, virtual, erect, $1.2\ \text{cm}$ tall. Exam tip: virtual rays must be dashed and real rays solid — this is checked in every ray-diagram question.

E3. Board Case study. Trucks carry a convex mirror of radius of curvature $2.0\ \text{m}$ beside the driver. (i) Why convex and not plane? (ii) A car is $6.0\ \text{m}$ behind the mirror. Find the image distance. (iii) Find the magnification. (iv) Why does the car seem farther away than it is? [4 marks]

(i) It always gives an erect image and a much wider field of view.
(ii) $f=+1.0\ \text{m}$, $u=-6.0\ \text{m}$: $\tfrac1v=1+\tfrac16=\tfrac76$, $v=+0.86\ \text{m}$ behind the mirror.
(iii) $m=-\tfrac{0.857}{-6.0}=+0.14$.
(iv) The image is only about one-seventh of the size, and we judge distance partly by apparent size. Exam tip: for convex mirrors, a positive $v$ smaller than $f$ is a built-in check.

E4. NEET An object is $10\ \text{cm}$ in front of a concave mirror of focal length $15\ \text{cm}$. The image is (A) real, $30\ \text{cm}$ in front (B) virtual, $30\ \text{cm}$ behind, three times enlarged (C) virtual, $6\ \text{cm}$ behind (D) real, $6\ \text{cm}$ in front

Working $u=-10$, $f=-15$: $\tfrac1v=-\tfrac1{15}+\tfrac1{10}=\tfrac1{30}$, $v=+30\ \text{cm}$; $m=+3$.
Answer (B). Exam tip: object inside $F$ of a concave mirror ⟹ virtual and enlarged; that alone eliminates (A), (C) and (D).

E5. JEE Main A concave mirror forms a real image three times the size of the object on a screen. The object and the image are $80\ \text{cm}$ apart. The focal length of the mirror is (A) $20\ \text{cm}$ (B) $30\ \text{cm}$ (C) $40\ \text{cm}$ (D) $60\ \text{cm}$

Idea Real image, $m=-3$, so $v=3u$; both are in front, so $|v|-|u|=80\ \text{cm}$.
Working $3|u|-|u|=80$ gives $|u|=40\ \text{cm}$, $|v|=120\ \text{cm}$. $\tfrac1f=-\tfrac1{120}-\tfrac1{40}=-\tfrac4{120}$, $f=-30\ \text{cm}$.
Answer (B) $30\ \text{cm}$. Exam tip: for $m=-n$, $|f|=\dfrac{n\,|u|}{n+1}$; here $\tfrac{3\times40}{4}=30$.
🔬 Research–Teaching NexusResearch-tutored · optional, about 40 minutes

Research question. How much does a convex road mirror shrink an approaching car, and how could that mislead a driver judging its distance?

Why it matters. On the hairpin bends of India's ghat and hill roads — the Nilgiris, the Western Ghats, the Himalayan highways — large convex mirrors let drivers see round blind corners. In the United States, vehicle safety rules require passenger-side convex mirrors to carry the printed warning "Objects in mirror are closer than they appear". Both are about the same physics: a convex mirror trades image size for field of view.

What to do.

  1. Take a road mirror with $R=3.0\ \text{m}$ and a car $1.5\ \text{m}$ tall. Using the mirror formula, find $v$, $m$ and the image height $h'$ when the car is $5$, $10$, $20$ and $40\ \text{m}$ away.
  2. Plot $m$ against the car's distance. How does $m$ change when the distance doubles?
  3. A plane mirror gives $m=1$ at every distance. Estimate how far away a car seems if a driver judges distance by assuming $m=1$ — that is, compare the image height with the height of a real car.
  4. Look for a convex mirror near your school or home (a shop, a parking ramp, a road bend) and estimate its radius of curvature by checking where the image of a far object appears.

Tools: calculator or spreadsheet; graph paper.

Hand in: A table of $v$, $m$ and $h'$; your graph; and a short recommendation to a road authority on where convex mirrors help and where they may mislead.

Going further. Find out why some side mirrors are "aspheric" — flat near the centre and more curved at the edge — and what problem that design solves.

Instructor note: with $f=+1.5\ \text{m}$: at $5\ \text{m}$, $v=1.15\ \text{m}$, $m=0.23$, $h'=0.35\ \text{m}$; at $10\ \text{m}$, $v=1.30\ \text{m}$, $m=0.13$, $h'=0.20\ \text{m}$; at $20\ \text{m}$, $v=1.40\ \text{m}$, $m=0.070$, $h'=0.10\ \text{m}$; at $40\ \text{m}$, $v=1.45\ \text{m}$, $m=0.036$, $h'=0.054\ \text{m}$. For distances much larger than $f$, $m$ roughly halves when the distance doubles.

🧪 Self-check quiz

0/5 correct

Q1. In the New Cartesian sign convention, the focal length of a concave mirror is

Q2. An object is placed at the centre of curvature of a concave mirror. The image is

Q3. A mirror gives a magnification of $+0.5$. The image is

Q4. For a concave mirror with $f=-20\ \text{cm}$ and $u=-40\ \text{cm}$, the image distance is

Q5. A virtual, erect and enlarged image can be formed by

← 1.1 Reflection of Light and Spherical Mirrors1.3 Refraction of Light and the Refractive Index →