PhysicsICSE Class 9 · Spherical Mirrors

Mirror Formula, Cartesian Sign Convention and Magnification: Solved Numericals (ICSE Class 9)

Q: What is the mirror formula?

1/v + 1/u = 1/f, where u is the object distance, v the image distance and f the focal length, all with Cartesian signs measured from the pole.

Q: What does the sign of magnification tell us?

A negative m means a real, inverted image; a positive m means a virtual, erect image.

Q: Why is the focal length of a concave mirror negative?

Its focus lies in front of the mirror, against the direction of the incident light, so by the Cartesian convention the distance is negative.

Updated 17 October 2026 · Fact-checked

With the Cartesian sign convention (distances measured from the pole, positive along the incident light), the mirror formula is 1/v + 1/u = 1/f and the magnification is m = −v/u = h₂/h₁. Concave mirror: f negative. Convex mirror: f positive. A real object has u negative; a real image has v negative; a virtual image has v positive.

Understand Mirror Formula, Cartesian Sign Convention and Magnification

Cartesian sign convention. Place the object on the left; light travels left to right.

All distances are measured from the pole $P$ .
Distances measured along the direction of the incident light (to the right of the pole) are positive; against it (to the left) are negative.
Heights above the principal axis are positive; below, negative.

Consequences.

Object distance $u$ is negative for a real object.
Concave mirror: $f$ and $R$ are negative (focus in front).
Convex mirror: $f$ and $R$ are positive (focus behind).
A real image (in front of the mirror) has $v$ negative; a virtual image (behind the mirror) has $v$ positive.

Mirror formula (no derivation): $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ , with $f = \frac{R}{2}$ .

Magnification. $m = \frac{h _{2}}{h _{1}} = - \frac{v}{u}$ , where $h_{1}$ is the object height and $h_{2}$ the image height.

$m$ negative: real, inverted image. $m$ positive: virtual, erect image.
$∣ m ∣ > 1$ : magnified; $∣ m ∣ < 1$ : diminished; $∣ m ∣ = 1$ : same size.

Solving steps.
1. Draw a quick sketch to predict the nature of the image.
2. Write the signs of $u$ and $f$ , then substitute in $\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$ .
3. Find $m = - \frac{v}{u}$ and the image height $h_{2} = m h_{1}$ .
4. Interpret the sign of $v$ and of $m$ ; state the result in words.

Types of problem.

Direct: given $u$ and $f$ (or $R$ ), find $v$ and $m$ .
Given magnification and $f$ : write $v = - m u$ and substitute.
Screen at a given distance: a screen can receive only a real image, so $v$ is negative.
Given object and image separation: use the signs of $u$ and $v$ to write the separation.

Check. A convex mirror always gives $v > 0$ and $0 < m < 1$ ; if your answer differs, recheck the signs.

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

Mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$; Cartesian signs.
Focal length: $f = \frac{R}{2}$; Concave negative, convex positive.
Magnification: $m = \frac{h _{2}}{h _{1}} = - \frac{v}{u}$; Negative: real, inverted.

How to solve Mirror Formula, Cartesian Sign Convention and Magnification questions

Sign first, number second.

1Assign signs: $u < 0$ ; $f < 0$ for concave, $f > 0$ for convex.
2Compute $\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$ and invert.
3Interpret $v$ : negative means real (in front), positive means virtual (behind).
4Compute $m = - v / u$ and $h_{2} = m h_{1}$ .
5Write the nature, position and size of the image in words.

Mirror Formula, Cartesian Sign Convention and Magnification 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.

A shaving mirror

Suppose a shaving mirror has radius of curvature 60 cm and you hold your face 15 cm from it.

Find the image distance and magnification.

Try it first, then show the solutionSolution

$f = - 30$ cm, $u = - 15$ cm.
$\frac{1}{v} = \frac{1}{- 30} - \frac{1}{- 15} = \frac{1}{30}$ , so $v = + 30$ cm (virtual, behind).
$m = - \frac{30}{- 15} = + 2$ : erect and twice as large.

Answer: v = +30 cm; m = +2.

Memory hook: Inside F: virtual and magnified.

Tricks to remember Mirror Formula, Cartesian Sign Convention and Magnification

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

Light direction is positive
Right is positive; real objects are on the left, so u is negative.
Use it when: Setting signs.
Concave is minus, convex is plus
f: concave negative, convex positive.
Use it when: Signs of f.
m negative means inverted
Minus sign in −v/u: real images flip.
Use it when: Interpreting m.

Mirror Formula, Cartesian Sign Convention and Magnification: solved examples

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

Example 1: Sign of the focal length · 1 mark · MCQBoard level

In the Cartesian sign convention the focal length of a concave mirror is:
(a) positive
(b) negative
(c) zero
(d) infinite

Try it first, then show the solutionSolution

The focus of a concave mirror is in front of it, against the direction of the incident light, so $f < 0$ .

Answer: (b) negative

Example 2: Radius to focal length · 2 marks · Short answerBoard level

Write the focal length with its sign for (a) a concave mirror of radius 60 cm and (b) a convex mirror of radius 60 cm.

Try it first, then show the solutionSolution

(a) $f = \frac{R}{2} = \frac{- 60}{2} = - 30$ cm. (1)
(b) $f = + 30$ cm. (1)

Answer: (a) −30 cm (b) +30 cm

Example 3: Real image formed by a concave mirror · 3 marks · Short answerHarder board level

An object 3 cm high is placed 27 cm in front of a concave mirror of focal length 18 cm. Find the image distance, magnification and the image height. State the nature of the image.

Try it first, then show the solutionSolution

$\frac{1}{v} = \frac{1}{- 18} - \frac{1}{- 27} = \frac{- 3 + 2}{54} = - \frac{1}{54}$ , so $v = - 54$ cm. (1)
$m = - \frac{v}{u} = - \frac{- 54}{- 27} = - 2$ . (1)
$h_{2} = - 2 \times 3 = - 6$ cm: real, inverted, 6 cm high. (1)

Answer: v = −54 cm; m = −2; image 6 cm high, inverted, real

Example 4: Convex mirror image · 3 marks · Short answerHarder board level

A 5 cm tall object is 50 cm from a convex mirror of focal length 25 cm. Find the image position, magnification and height.

Try it first, then show the solutionSolution

$f = + 25$ , $u = - 50$ : $\frac{1}{v} = \frac{1}{25} + \frac{1}{50} = \frac{3}{50}$ , so $v = + 16.7$ cm. (1)
$m = - \frac{16.7}{- 50} = + 0.33$ . (1)
$h_{2} = 0.33 \times 5 = 1.67$ cm: virtual, erect, diminished. (1)

Answer: v ≈ +16.7 cm; m ≈ +0.33; h ≈ 1.67 cm

Example 5: A dentist’s mirror · 4 marks · Case-basedHOTS / competency

A dentist uses a concave mirror of focal length 24 cm to get an erect image four times the size of a tooth.
(a) Find the object distance and the image distance.
(b) State the radius of curvature of the mirror with its sign.
(c) If a feature on the tooth is 3 mm across, find the size of its image.
(d) Why is $v$ positive here?

Try it first, then show the solutionSolution

(a) $m = + 4 = - \frac{v}{u}$ , so $v = - 4 u$ . $\frac{1}{- 4 u} + \frac{1}{u} = \frac{3}{4 u} = \frac{1}{f} = - \frac{1}{24}$ , giving $u = - 18$ cm and $v = + 72$ cm. (2)
(b) $R = 2 f = - 48$ cm. (½)
(c) $4 \times 3 = 12$ mm. (½)
(d) A virtual image lies behind the mirror, in the direction of the incident light. (1)

Answer: (a) u = −18 cm, v = +72 cm (b) −48 cm (c) 12 mm (d) virtual image behind the mirror

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

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

Mistake: Object distance taken positive
A real object has u negative.
Wrong working
u = +27 cm
Right working
u = −27 cm
Do this: Write signs before substituting.
Usually costs: The numerical.
Mistake: Concave mirror f taken positive
f is negative for concave.
Wrong working
f = +18 cm
Right working
f = −18 cm
Do this: Concave = minus.
Usually costs: The numerical.
Mistake: Interpreting a negative m as a diminished image
The sign gives orientation; the size is |m|.
Wrong working
m = −2: diminished.
Right working
m = −2: inverted and twice as large.
Do this: Read the sign and the magnitude separately.
Usually costs: The nature mark.

Exam tips

State the convention in one line before the numerical.
Always write "f = −…" for concave and "f = +…" for convex.
Finish by stating real or virtual, erect or inverted, and magnified or diminished.
Check that your answer agrees with the ray-diagram table.

Practice questions on Mirror Formula, Cartesian Sign Convention and Magnification

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
For a virtual image formed by a spherical mirror, the image distance v is:
(a) negative
(b) positive
(c) zero
(d) infinite
Hint
Behind the mirror.
Answer
(b) positive
Q2 · 1 mark · Very short answerBoard level
What is the sign of the object distance for a real object?
Hint
Against the light.
Answer
Negative.
Q3 · 2 marks · Short answerHarder board level
A concave mirror has f = −10 cm. An object is 15 cm in front of it. Find v and m.
Hint
1/v = 1/f − 1/u.
Answer
$\frac{1}{v} = - \frac{1}{10} + \frac{1}{15} = - \frac{1}{30}$ , $v = - 30$ cm; $m = - \frac{- 30}{- 15} = - 2$ .
Q4 · 3 marks · Short answerHarder board level
A real image is formed 36 cm in front of a concave mirror of focal length 12 cm. Find the object distance and the magnification.
Hint
Use v = −36, f = −12.
Answer
$\frac{1}{u} = \frac{1}{f} - \frac{1}{v} = - \frac{1}{12} + \frac{1}{36} = - \frac{1}{18}$ , $u = - 18$ cm; $m = - \frac{- 36}{- 18} = - 2$ .
Q5 · 2 marks · Assertion-reasonHarder board level
Assertion (A): The magnification of a real image formed by a spherical mirror is negative.
Reason (R): A real image is inverted.
(a) Both true, R explains A.
(b) Both true, R does not explain A.
(c) A true, R false.
(d) A false, R true.
Hint
m = −v/u.
Answer
(a)
Q6 · 3 marks · Short answerHOTS / competency
A concave mirror forms a real image on a screen that is 30 cm away from the object, and the image is twice as large as the object. Find the object distance and the focal length of the mirror.
Hint
Both u and v are negative: |v| = 2|u|, and |v| − |u| = 30.
Answer
$∣ v ∣ = 2∣ u ∣$ and $∣ v ∣ - ∣ u ∣ = 30$ , so $∣ u ∣ = 30$ cm, $∣ v ∣ = 60$ cm; $u = - 30$ , $v = - 60$ . $\frac{1}{f} = - \frac{1}{60} - \frac{1}{30} = - \frac{1}{20}$ , so $f = - 20$ cm.

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
1/v + 1/u = 1/f.
2.Magnification formula?Show answer
m = −v/u = h₂/h₁.
3.Sign of f for convex mirror?Show answer
Positive.
4.Sign of v for a real image?Show answer
Negative.
5.What does negative m show?Show answer
A real, inverted image.

Mirror Formula, Cartesian Sign Convention and Magnification: frequently asked questions

What is the mirror formula?

1/v + 1/u = 1/f, where u is the object distance, v the image distance and f the focal length, all with Cartesian signs measured from the pole.

What does the sign of magnification tell us?

A negative m means a real, inverted image; a positive m means a virtual, erect image.

Why is the focal length of a concave mirror negative?

Its focus lies in front of the mirror, against the direction of the incident light, so by the Cartesian convention the distance is negative.