ScienceCBSE Class 10 · Light – Reflection and Refraction

Mirror Formula, Magnification and Sign Convention

Updated 14 October 2026 · Fact-checked

For spherical mirrors, object distance u, image distance v and focal length f are related by 1/v + 1/u = 1/f, and magnification is m = h'/h = -v/u. All distances are measured from the pole using the New Cartesian sign convention: along the incident light positive, against it negative, upward heights positive.

Understand Mirror Formula, Magnification and Sign Convention

Why a sign convention? With mirrors, the object and image can be in front of or behind the mirror, and the image can be upright or upside down. A fixed set of signs lets one formula cover every case. Class 10 uses the New Cartesian Sign Convention:
1. The pole (P) of the mirror is the origin; the principal axis is the x-axis.
2. The object is always placed to the left of the mirror, so light falls on the mirror from the left.
3. Distances measured to the right of the pole (in the direction of incident light) are positive; distances to the left (against the incident light) are negative.
4. Heights measured upwards from the principal axis are positive; downwards are negative.

Consequences you will use every time:

the object distance u is always negative (object in front, on the left);
a concave mirror has its focus in front, so f is negative; a convex mirror has its focus behind, so f is positive;
a real image forms in front of the mirror, so v is negative; a virtual image forms behind it, so v is positive;
the radius of curvature is twice the focal length: $R = 2 f$ .

Mirror formula. $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ . It is valid for all spherical mirrors and all object positions, as long as you substitute values with their signs. (Its derivation is not required in your syllabus.)

Magnification. It tells how much bigger or smaller the image is: $m = \frac{h ^{'}}{h}$ , where h is object height and h' image height. It also equals $m = - \frac{v}{u}$ .

Negative m: image is real and inverted.
Positive m: image is virtual and erect.
|m| > 1 magnified; |m| < 1 diminished; |m| = 1 same size.

A convex mirror always gives a virtual, erect, diminished image, so its m is always positive and less than 1. A concave mirror gives real inverted images (m negative) for objects beyond F, and a virtual erect magnified image (m positive, greater than 1) for an object between P and F.

A quick sense-check. After you calculate, compare with the ray-diagram result: for a concave mirror, an object beyond C gives a diminished image between F and C; an object at C gives a same-size image at C; between C and F gives a magnified image beyond C; between P and F gives a virtual image behind the mirror. If your numbers disagree with the picture, look for a sign error.

Mirror Formula, Magnification and Sign Convention: formulas and key facts

Mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$; Substitute u, v, f with signs. Derivation not required.
Magnification: $m = \frac{h ^{'}}{h} = - \frac{v}{u}$; m negative: real, inverted; m positive: virtual, erect.
Radius and focal length: $R = 2 f$; Concave: f and R negative; convex: positive.
Signs at a glance: $u < 0; f_{concave} < 0; f_{convex} > 0$; Real image: v < 0; virtual image: v > 0.

How to solve Mirror Formula, Magnification and Sign Convention questions

Use this routine for every mirror numerical.

1Draw a quick sketch with the object on the left and mark which mirror it is.
2Write u, f (or R) with signs: u negative; f negative for concave, positive for convex; f = R/2.
3Rearrange the formula for the unknown: $\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$ , or $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ .
4Use a common denominator (LCM) and keep the signs; then invert to get v.
5Find $m = - \frac{v}{u}$ and, if needed, $h^{'} = m \times h$ .
6Interpret: sign of v (front or behind), sign of m (real/inverted or virtual/erect), size of |m|.
7Check by substituting back into the mirror formula, and compare with the ray-diagram expectation.

Mirror Formula, Magnification and Sign Convention in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Convex mirror as a rear-view mirror

Suppose a scooter's rear-view mirror is convex with a radius of curvature of 2.40 m, and a car is 6.00 m behind it.

Find the position, nature and magnification of the car's image, and explain why a convex mirror is chosen for this job.

Try it first, then show the solutionSolution

$f = \frac{R}{2} = + 1.20$ m (convex: positive); $u = - 6.00$ m.
$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{1.20} + \frac{1}{6.00} = 0.833 + 0.167 = 1.00$ , so $v = + 1.00$ m.
$m = - \frac{v}{u} = - \frac{1.00}{- 6.00} = + 0.17$ (about $\frac{1}{6}$ ).
Check: $\frac{1}{1.00} + \frac{1}{- 6.00} = 1 - 0.167 = 0.833 = \frac{1}{1.20}$ .
v positive: image 1.00 m behind the mirror, virtual; m positive and less than 1: erect and diminished. A convex mirror always gives an erect, diminished image and covers a wider field of view, so the rider sees more of the traffic behind.

Answer: Image 1.00 m behind the mirror; virtual, erect, diminished; m ≈ +0.17. Convex mirrors give erect images and a wider field of view.

Memory hook: Convex keeps it upright and small, so you see the whole road behind.

Tricks to remember Mirror Formula, Magnification and Sign Convention

Short memory aids. Say each one aloud once, then use it on a question the same day.

Against the light is minus
Anything measured back towards where the light came from (left) is negative, so u is always negative.
Use it when: Never forgetting the sign of u.
Cave in, f minus
Concave caves in towards the object: focus in front, f negative. Convex bulges out: focus behind, f positive.
Use it when: Sign of f.
Minus m, real thing
Negative magnification = real and inverted; positive = virtual and erect.
Use it when: Reading the nature of the image from m.

Mirror Formula, Magnification and Sign Convention: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Concave mirror, object between C and F · 3 marks · Short answerBoard level

An object is placed 25 cm in front of a concave mirror of focal length 15 cm. Find the position, nature and magnification of the image.

Try it first, then show the solutionSolution

$u = - 25$ cm, $f = - 15$ cm (½ mark).
$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = - \frac{1}{15} + \frac{1}{25} = \frac{- 5 + 3}{75} = - \frac{2}{75}$ , so $v = - 37.5$ cm (1 mark).
$m = - \frac{v}{u} = - \frac{- 37.5}{- 25} = - 1.5$ (1 mark).
Image 37.5 cm in front of the mirror; real, inverted, magnified 1.5 times. Check: $- \frac{1}{37.5} - \frac{1}{25} = - 0.0267 - 0.04 = - 0.0667 = - \frac{1}{15}$ (½ mark).

Answer: v = −37.5 cm; real, inverted, magnified; m = −1.5.

Example 2: Image height · 3 marks · Short answerHarder board level

A 4.0 cm tall object stands 30 cm in front of a concave mirror of focal length 20 cm. Find the image distance, the image height and describe the image.

Try it first, then show the solutionSolution

$u = - 30$ cm, $f = - 20$ cm, $h = + 4.0$ cm (½ mark).
$\frac{1}{v} = - \frac{1}{20} + \frac{1}{30} = \frac{- 3 + 2}{60} = - \frac{1}{60}$ , so $v = - 60$ cm (1 mark).
$m = - \frac{v}{u} = - \frac{- 60}{- 30} = - 2$ ; $h^{'} = mh = - 2 \times 4.0 = - 8.0$ cm (1 mark).
Image 60 cm in front of the mirror (beyond C), real, inverted, 8.0 cm tall (twice the size). Check: $- \frac{1}{60} - \frac{1}{30} = - \frac{3}{60} = - \frac{1}{20}$ (½ mark).

Answer: v = −60 cm; h' = −8.0 cm; real, inverted, magnified (m = −2).

Example 3: Object between P and F · 3 marks · Short answerHarder board level

A dentist holds a concave mirror of focal length 12 cm at 8 cm from a tooth. Find the position and magnification of the image and state its nature. Why is a concave mirror useful here?

Try it first, then show the solutionSolution

$u = - 8$ cm, $f = - 12$ cm (½ mark).
$\frac{1}{v} = - \frac{1}{12} + \frac{1}{8} = \frac{- 2 + 3}{24} = \frac{1}{24}$ , so $v = + 24$ cm (1 mark).
$m = - \frac{24}{- 8} = + 3$ (½ mark).
Image 24 cm behind the mirror: virtual, erect, three times larger. Check: $\frac{1}{24} - \frac{1}{8} = \frac{1 - 3}{24} = - \frac{1}{12}$ . A concave mirror with the object inside F gives an enlarged erect image, so the dentist sees the tooth clearly (1 mark).

Answer: v = +24 cm; m = +3; virtual, erect, magnified.

Example 4: Working backwards from magnification · 3 marks · Short answerHOTS / competency

A concave mirror of focal length 15 cm forms a real image three times the size of the object. Find the object and image distances.

Try it first, then show the solutionSolution

Real image means m is negative: $m = - 3 = - \frac{v}{u}$ , so $v = 3 u$ (1 mark).
Substitute in $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ : $\frac{1}{3 u} + \frac{1}{u} = \frac{4}{3 u} = - \frac{1}{15}$ , so $3 u = - 60$ , $u = - 20$ cm (1 mark).
$v = 3 u = - 60$ cm. Check: $- \frac{1}{60} - \frac{1}{20} = - \frac{1 + 3}{60} = - \frac{1}{15}$ (1 mark).

Answer: Object 20 cm and image 60 cm in front of the mirror.

Example 5: Finding focal length · 3 marks · Short answerHarder board level

When an object is 30 cm in front of a spherical mirror, a real image forms 20 cm in front of it. (a) Which kind of mirror is it? (b) Find its focal length and radius of curvature. (c) Find the magnification.

Try it first, then show the solutionSolution

(a) Only a concave mirror forms real images (½ mark).
(b) $u = - 30$ , $v = - 20$ : $\frac{1}{f} = - \frac{1}{20} - \frac{1}{30} = - \frac{3 + 2}{60} = - \frac{1}{12}$ , so $f = - 12$ cm; $R = 2 f = - 24$ cm (1½ marks).
(c) $m = - \frac{- 20}{- 30} = - \frac{2}{3} \approx - 0.67$ : real, inverted, diminished (1 mark).

Answer: (a) Concave. (b) f = −12 cm, R = −24 cm. (c) m ≈ −0.67.

Where students lose marks in Mirror Formula, Magnification and Sign Convention

The slips examiners see most often, and what to write instead.

Mistake: Dropping signs
Students substitute magnitudes only.
Wrong working
Concave, f = 15, u = 25: 1/v = 1/15 − 1/25.
Right working
f = −15, u = −25: 1/v = −1/15 + 1/25.
Do this: Write each quantity with its sign before substituting.
Usually costs: The answer mark and the nature mark.
Mistake: Forgetting to invert at the end
Students stop at 1/v.
Wrong working
v = −2/75 cm.
Right working
1/v = −2/75, so v = −75/2 = −37.5 cm.
Do this: Write '1/v =' on the line before inverting.
Usually costs: 1 mark.
Mistake: Wrong formula for magnification
Students use m = v/u without the minus.
Wrong working
m = v/u = (−37.5)/(−25) = +1.5, erect.
Right working
m = −v/u = −1.5, real and inverted.
Do this: Remember the minus sign in the mirror magnification.
Usually costs: 1 mark and the nature mark.
Mistake: Not interpreting the result
Students give numbers only.
Wrong working
v = −60 cm, m = −2.
Right working
Image 60 cm in front of the mirror, real, inverted, twice the size.
Do this: Always translate signs into words: position, real or virtual, erect or inverted, size.
Usually costs: ½ to 1 mark.

Exam tips

Write the sign convention line 'u = −…, f = −… (concave)' at the start; it earns a step mark.
Use fractions with an LCM rather than decimals to avoid rounding errors.
Verify your v by substituting back; it takes ten seconds and catches sign errors.
Give units (cm or m) and keep all distances in the same unit.
State the nature of the image from the signs of v and m.

Practice questions on Mirror Formula, Magnification and Sign Convention

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
According to the New Cartesian sign convention, the focal length of a convex mirror is taken as:
(a) always negative (b) always positive (c) positive or negative depending on the object (d) zero
Hint
Where is the focus of a convex mirror?
Answer
(b) always positive; the focus lies behind the mirror.
Q2 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The magnification produced by a convex mirror is always positive.
Reason (R): A convex mirror always forms a virtual and erect image.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Positive m means erect.
Answer
(a).
Q3 · 2 marks · Very short answerBoard level
An object is placed 20 cm in front of a convex mirror of focal length 20 cm. Find the image position and magnification.
Hint
f = +20 cm, u = −20 cm.
Answer
$\frac{1}{v} = \frac{1}{20} + \frac{1}{20} = \frac{1}{10}$ , v = +10 cm (behind the mirror). m = −(10)/(−20) = +0.5: virtual, erect, half size. Check: 1/10 − 1/20 = 1/20.
Q4 · 3 marks · Short answerHarder board level
Where should an object be placed in front of a concave mirror of radius of curvature 20 cm so that a real image of the same size is formed? Verify with the mirror formula and find m.
Hint
Same size real image forms at C.
Answer
R = −20 cm, f = −10 cm. Place the object at C: u = −20 cm. $\frac{1}{v} = - \frac{1}{10} + \frac{1}{20} = - \frac{1}{20}$ , so v = −20 cm. m = −(−20)/(−20) = −1: real, inverted, same size.
Q5 · 4 marks · Case-basedHOTS / competency
Read the passage and answer.
Suppose a student uses a concave mirror of focal length 10 cm on an optical bench. She moves a candle from 40 cm to 15 cm in front of the mirror and records where a sharp image forms on a screen.
(a) Find the image distance when the candle is at 40 cm. (1 mark)
(b) Find the image distance when it is at 15 cm. (1 mark)
(c) Find the magnification in both cases and describe how the image changes as the candle moves closer. (2 marks)
Hint
f = −10 cm. Use 1/v = 1/f − 1/u.
Answer
(a) u = −40: 1/v = −1/10 + 1/40 = (−4 + 1)/40 = −3/40, v = −13.3 cm. (b) u = −15: 1/v = −1/10 + 1/15 = (−3 + 2)/30 = −1/30, v = −30 cm. (c) At 40 cm: m = −(−13.3)/(−40) = −0.33, real, inverted, diminished. At 15 cm: m = −(−30)/(−15) = −2, real, inverted, magnified. As the candle moves towards F, the image moves away from the mirror and grows larger.
Q6 · 3 marks · Short answerHOTS / competency
A student calculates for a concave mirror (f = 10 cm, object at 5 cm): '1/v = 1/10 − 1/5 = −1/10, so v = −10 cm, a real image.' (a) Identify the sign errors. (b) Solve correctly and state the nature of the image.
Hint
Both f and u for a concave mirror with a real object are negative.
Answer
(a) f should be −10 cm and u −5 cm; the student used positive values. (b) 1/v = 1/f − 1/u = −1/10 + 1/5 = 1/10, v = +10 cm: the image is 10 cm behind the mirror, virtual and erect. m = −10/(−5) = +2, magnified. Check: 1/10 − 1/5 = −1/10 = 1/f.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Mirror formula?Show answer
$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ .
2.Magnification formula for mirrors?Show answer
$m = \frac{h ^{'}}{h} = - \frac{v}{u}$ .
3.Sign of f for a concave mirror?Show answer
Negative.
4.What does a negative m mean?Show answer
Real and inverted image.
5.Relation between R and f?Show answer
R = 2f.
6.Origin in the New Cartesian convention?Show answer
The pole of the mirror.

Mirror Formula, Magnification and Sign Convention: frequently asked questions

What is the New Cartesian sign convention for mirrors?

The pole is the origin. Distances in the direction of incident light (to the right) are positive and against it negative. Heights above the principal axis are positive and below negative. The object is placed on the left.

Is the object distance always negative?

Yes, for a real object placed in front of the mirror, as in all Class 10 questions.

How do I know if the image is real or virtual from the answer?

For a mirror, negative v (in front) means real and positive v (behind) means virtual. Negative m means inverted; positive m means erect.

Do I need to derive the mirror formula?

No. The 2026-27 syllabus says the derivation is not required. You need to apply it.

Why does a convex mirror always give a diminished image?

Its focus is behind the mirror; for any real object the formula gives a positive v smaller than |u|, so m is positive and less than 1.