PhysicsICSE Class 10 · Lenses

Lens Formula and Magnification: Sign Convention, 1/v − 1/u = 1/f and m = v/u (ICSE Class 10)

Q: What is the lens formula?

1/v − 1/u = 1/f, where u is the object distance, v the image distance and f the focal length, all measured from the optical centre with the New Cartesian sign convention.

Q: How do I know whether an image is real or virtual from my answer?

For a lens, a positive v means the image is on the far side (real); a negative v means it is on the same side as the object (virtual). A negative magnification also means real and inverted.

Q: Do I need to derive the lens formula for ICSE Class 10?

No. The CISCE syllabus asks only for direct numerical problems using the lens formula; the derivation is not required.

Updated 15 October 2026 · Fact-checked

The lens formula is 1/v − 1/u = 1/f, with all distances measured from the optical centre using the New Cartesian sign convention. Magnification is m = h′/h = v/u. For a real object u is negative; f is positive for a convex lens and negative for a concave lens. Negative m means a real, inverted image.

Understand Lens Formula and Magnification

The lens formula connects the object distance $u$ , the image distance $v$ and the focal length $f$ of a thin lens. The syllabus asks for direct numericals with it (its derivation is not required).

New Cartesian sign convention.
1. All distances are measured from the optical centre O along the principal axis.
2. Light is drawn travelling from left to right; distances measured in the direction of the incident light (to the right of O) are positive, and against it (to the left) are negative.
3. Heights above the principal axis are positive, below negative.

Consequences:

A real object is on the left, so ** $u$ is always negative**.
Convex lens: $f$ positive. Concave lens: $f$ negative.
A real image forms on the right: $v$ positive. A virtual image forms on the same side as the object: $v$ negative.

Lens formula.

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f} .$

Linear magnification.

$m = \frac{h ^{'}}{h} = \frac{v}{u} .$

$m$ negative: real and inverted image. $m$ positive: virtual and erect.
$∣ m ∣ > 1$ : magnified; $∣ m ∣ < 1$ : diminished; $∣ m ∣ = 1$ : same size.
Magnification is image : object (I : O), never O : I.

Useful checks.

Concave lens, real object: $v$ is always negative and $0 < m < 1$ (virtual, erect, diminished), wherever the object is.
Convex lens: object beyond $2 F$ gives $∣ m ∣ < 1$ ; at $2 F$ gives $m = - 1$ ; between $F$ and $2 F$ gives $∣ m ∣ > 1$ (all real); inside $F$ gives a virtual, erect, magnified image ( $m > 1$ ).
For a real image of a convex lens, a size ratio I : O tells you where the object is: $\frac{1}{4}$ means beyond $2 F$ , $\frac{3}{1}$ means between $F$ and $2 F$ .

Solving for f. If $u$ and $v$ are known, $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$ , or $f = \frac{uv}{u - v}$ . Then the sign of $f$ names the lens.

Using magnification to find a distance. From $m = \frac{v}{u}$ , $v = m u$ ; substituting into the lens formula gives $u = \frac{f ( 1 - m )}{m}$ , useful when a required image size is given.

Two positions with fixed object and screen. If the object and screen are a fixed distance $D$ apart, a convex lens gives a sharp image at two positions (if $D > 4 f$ ), with $u$ and $v$ interchanged; one image is magnified and the other diminished.

Lens Formula and Magnification: formulas and key facts

Lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$; New Cartesian signs; u negative for a real object.
Magnification: $m = \frac{h ^{'}}{h} = \frac{v}{u}$; Negative: real, inverted. Positive: virtual, erect.
Focal length from u and v: $f = \frac{uv}{u - v}$; Rearranged lens formula.
Object distance for a given m: $u = \frac{f ( 1 - m )}{m}$; From v = mu in the lens formula.

How to solve Lens Formula and Magnification questions

Write the signed values first, substitute once into the lens formula, then read the nature of the image from the signs.

1Write $u$ (negative for a real object), and $f$ (+ for convex, − for concave) or $v$ (+ real, − virtual) with signs.
2Substitute into $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ ; take a common denominator carefully.
3Invert to get the unknown; give the answer as a decimal with its sign and unit.
4Find $m = \frac{v}{u}$ and, if needed, $h^{'} = m \times h$ .
5State the nature: sign of $v$ (real/virtual), sign of $m$ (inverted/erect), size of $∣ m ∣$ (magnified/diminished).

Lens Formula 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.

Finding the focal length of a lens with a window

Suppose you hold a convex lens 2.1 m from a bright window, 1.4 m tall, and get a sharp image of it on a sheet of paper 15 cm behind the lens.

Find the focal length of the lens and the height of the window's image.

Try it first, then show the solutionSolution

$u = - 210$ cm, $v = + 15$ cm: $\frac{1}{f} = \frac{1}{15} + \frac{1}{210} = \frac{15}{210}$ , so $f = 14$ cm.
$m = \frac{15}{- 210} = - \frac{1}{14}$ ; $h^{'} = - \frac{140}{14} = - 10$ cm (inverted).

Answer: $f = 14$ cm; a real, inverted image 10 cm tall.

Memory hook: A far object focuses just beyond F.

A simple door viewer

Imagine a simple door viewer made of a single concave lens of focal length 2 cm, and a visitor 1.8 m tall standing 1 m in front of it.

Where is the image, and how tall is it?

Try it first, then show the solutionSolution

$\frac{1}{v} = - \frac{1}{2} + \frac{1}{- 100} = - \frac{51}{100}$ , so $v \approx - 1.96$ cm (virtual, same side).
$m = \frac{v}{u} = \frac{1}{51}$ ; $h^{'} = \frac{180}{51} \approx 3.5$ cm, erect.

Answer: A virtual, erect image about 2 cm from the lens and about 3.5 cm tall: the whole visitor fits in view.

Memory hook: Concave lenses shrink the world into view.

Tricks to remember Lens Formula and Magnification

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

V minus U equals F
'Very Useful Formula': 1/V − 1/U = 1/F, in alphabetical order backwards (v, u, f).
Use it when: Writing the lens formula (not the mirror formula with a plus).
Object always left, always negative
Light comes from the left, so a real object sits on the left: u is negative, every time.
Use it when: Sign convention.
Image over Object
m = I/O = v/u: 'I Over O'. Minus means upside down.
Use it when: Magnification sign and ratio.
Concave = Can't make real
A concave lens can't make a real image of a real object: v is negative, m between 0 and 1.
Use it when: Checking concave-lens answers.

Lens Formula and Magnification: solved examples

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

Example 1: Magnification of a concave lens · 1 mark · MCQBoard level

For a real object, the magnification produced by a concave lens is always:
(a) negative and greater than 1
(b) positive and less than 1
(c) positive and greater than 1
(d) negative and less than 1

Try it first, then show the solutionSolution

A concave lens always gives a virtual, erect, diminished image: $m$ positive and less than 1.

Answer: (b) positive and less than 1

Example 2: Image by a convex lens · 3 marks · Short answerBoard level

A convex lens with $f = 15$ cm has a small object standing 45 cm away on its axis. Calculate (a) the image distance, (b) the magnification, (c) the nature of the image.

Try it first, then show the solutionSolution

$u = - 45$ cm, $f = + 15$ cm. $\frac{1}{v} = \frac{1}{15} + \frac{1}{- 45} = \frac{3 - 1}{45} = \frac{2}{45}$ . (1)
(a) $v = + 22.5$ cm. (½)
(b) $m = \frac{22.5}{- 45} = - 0.5$ . (½)
(c) Real (v positive), inverted (m negative), diminished ( $∣ m ∣ < 1$ ). (1)

Answer: (a) 22.5 cm behind the lens (b) −0.5 (c) Real, inverted, diminished

Example 3: Image by a concave lens · 3 marks · Short answerHarder board level

A concave lens with $f = 12$ cm views an object 24 cm away. Locate the image, calculate the magnification and describe the image.

Try it first, then show the solutionSolution

$u = - 24$ cm, $f = - 12$ cm. $\frac{1}{v} = - \frac{1}{12} - \frac{1}{24} = - \frac{3}{24}$ . (1)
$v = - 8$ cm: 8 cm from the lens, on the same side as the object. (1)
$m = \frac{- 8}{- 24} = + \frac{1}{3}$ : virtual, erect, one-third the size. (1)

Answer: 8 cm in front of the lens; $m = + \frac{1}{3}$ ; virtual, erect, diminished

Example 4: Focal length from a screen image · 3 marks · Short answerHarder board level

A candle 20 cm in front of a lens gives a sharp image on a screen 30 cm behind the lens. Find (a) the focal length and type of the lens, (b) the magnification.

Try it first, then show the solutionSolution

$u = - 20$ cm, $v = + 30$ cm (real image on a screen). (½)
(a) $\frac{1}{f} = \frac{1}{30} - \frac{1}{- 20} = \frac{2 + 3}{60} = \frac{5}{60}$ , so $f = + 12$ cm: convex. (1½)
(b) $m = \frac{30}{- 20} = - 1.5$ . (1)

Answer: (a) +12 cm, convex (b) −1.5

Example 5: Real and virtual images twice the size · 4 marks · Long answerHOTS / competency

A convex lens has a focal length of 10 cm. Where must an object be placed to get (a) a real image twice its size, (b) a virtual image twice its size? Find the image distance in each case.

Try it first, then show the solutionSolution

(a) Real, inverted: $m = - 2$ , so $v = - 2 u$ . $\frac{1}{- 2 u} - \frac{1}{u} = \frac{1}{10}$ gives $- \frac{3}{2 u} = \frac{1}{10}$ , $u = - 15$ cm. (1)
Then $v = + 30$ cm. (1)
(b) Virtual, erect: $m = + 2$ , so $v = 2 u$ . $\frac{1}{2 u} - \frac{1}{u} = \frac{1}{10}$ gives $- \frac{1}{2 u} = \frac{1}{10}$ , $u = - 5$ cm. (1)
Then $v = - 10$ cm (same side as the object). (1)

Answer: (a) Object 15 cm from the lens, image 30 cm behind it (b) Object 5 cm from the lens, image 10 cm in front of it

Example 6: Case study: a camera lens · 4 marks · Case-basedHOTS / competency

A camera's convex lens has a focal length of 5 cm. It photographs a potted plant 60 cm tall standing 105 cm in front of the lens.
(a) How far behind the lens must the sensor be?
(b) Find the magnification.
(c) How tall is the image on the sensor, and which way up is it?
(d) To photograph a closer object, should the lens be moved towards or away from the sensor? Explain.

Try it first, then show the solutionSolution

(a) $\frac{1}{v} = \frac{1}{5} - \frac{1}{105} = \frac{21 - 1}{105} = \frac{20}{105}$ , so $v = 5.25$ cm. (1)
(b) $m = \frac{5.25}{- 105} = - 0.05$ . (1)
(c) $h^{'} = - 0.05 \times 60 = - 3$ cm: 3 cm tall, inverted. (1)
(d) Away: as $∣ u ∣$ decreases towards $f$ , $v$ increases, so the lens must be moved farther from the sensor. (1)

Answer: (a) 5.25 cm (b) −0.05 (c) 3 cm, inverted (d) Away from the sensor

Where students lose marks in Lens Formula and Magnification

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

Mistake: Wrong signs or interchanged u and v
The sign convention must be applied at the moment of substitution.
Wrong working
Concave lens, object 60 cm, image 20 cm: taking $u = + 60$ gives $\frac{1}{f} = - \frac{1}{20} - \frac{1}{60}$ , $f = - 15$ cm.
Right working
$u = - 60$ cm, $v = - 20$ cm: $\frac{1}{f} = - \frac{1}{20} + \frac{1}{60}$ , $f = - 30$ cm.
Do this: Write 'u = …, v = …, f = …' with signs before the formula.
Usually costs: Reported by CISCE examiners for 2024 Q4(i) and 2025 Q5(i).
Mistake: Taking magnification as O : I
m is image height ÷ object height.
Wrong working
Image 2 cm, object 5 cm: $m = 2.5$ .
Right working
$m = \frac{2}{5} = 0.4$ in size (negative if real).
Do this: Say 'I over O' as you write it.
Usually costs: Reported for 2024 Q1(xii).
Mistake: Not linking a size ratio to the object position
For a convex lens, a diminished real image means the object is beyond 2F.
Wrong working
Real image with I : O = 1 : 4, 'object between F and 2F'.
Right working
$∣ m ∣ < 1$ for a real image: object beyond $2 F$ .
Do this: Learn the table of object positions with $∣ m ∣$ .
Usually costs: Reported for 2025 Q1(vii).
Mistake: Leaving answers as fractions
Examiners expect decimals with units.
Wrong working
$v = \frac{45}{2}$ .
Right working
$v = + 22.5$ cm.
Do this: Divide out and add the unit and sign.
Usually costs: Accuracy marks (noted in 2024).

Exam tips

In 2024 Q4(i) (candle 36 cm from a lens, image on a screen 72 cm away), the expected working was $u = - 36$ cm, $v = + 72$ cm, $\frac{1}{f} = \frac{1}{72} + \frac{1}{36}$ , $f = 24$ cm.
In 2025 Q5(i) (concave lens, object 45 cm, image 30 cm), the expected answer was $f = - 90$ cm: keep the negative sign.
State the nature of the image from the signs of $v$ and $m$ , not from memory alone.
Write the formula you use before substituting; 'wrong formula' was a common error.

Practice questions on Lens Formula 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
In the New Cartesian sign convention for lenses, all distances are measured from:
(a) the principal focus
(b) the centre of curvature
(c) the optical centre
(d) the object
Hint
The point through which rays pass undeviated.
Answer
(c) the optical centre
Q2 · 2 marks · Short answerBoard level
With $f = 25$ cm, a convex lens has an object 50 cm from it. Calculate v and m, and state where the object is relative to the foci.
Hint
$u = - 50$ , $f = + 25$ .
Answer
$\frac{1}{v} = \frac{1}{25} - \frac{1}{50} = \frac{1}{50}$ , $v = + 50$ cm; $m = - 1$ . The object is at $2 F_{1}$ ; the image is at $2 F_{2}$ , real, inverted, same size.
Q3 · 3 marks · Short answerHarder board level
A concave lens ( $f = 20$ cm) has an object 60 cm away. Calculate the image distance and magnification and describe the image.
Hint
Both u and f are negative.
Answer
$\frac{1}{v} = - \frac{1}{20} - \frac{1}{60} = - \frac{4}{60}$ , $v = - 15$ cm; $m = \frac{- 15}{- 60} = + 0.25$ : virtual, erect, diminished, 15 cm on the object's side.
Q4 · 3 marks · Short answerHarder board level
A lens forms a virtual, erect image 120 cm from it, on the same side as an object placed 40 cm from it. Find the focal length and type of lens, and the magnification.
Hint
$u = - 40$ , $v = - 120$ .
Answer
$\frac{1}{f} = - \frac{1}{120} + \frac{1}{40} = \frac{2}{120}$ , $f = + 60$ cm: convex. $m = \frac{- 120}{- 40} = + 3$ .
Q5 · 2 marks · Short answerHarder board level
A concave lens forms an image 6 cm from it when an object is 18 cm from it. Find its focal length and the magnification.
Hint
Both u and v are negative.
Answer
$\frac{1}{f} = - \frac{1}{6} + \frac{1}{18} = - \frac{2}{18}$ , $f = - 9$ cm; $m = \frac{- 6}{- 18} = \frac{1}{3}$ .
Q6 · 3 marks · Short answerHarder board level
A convex lens forms an inverted image 15 cm tall of an object 5 cm tall on a screen 60 cm from the lens. Find (a) the magnification, (b) the object distance, (c) the focal length.
Hint
Inverted means $h^{'}$ is negative.
Answer
(a) $m = \frac{- 15}{5} = - 3$ . (b) $u = \frac{v}{m} = \frac{60}{- 3} = - 20$ cm. (c) $\frac{1}{f} = \frac{1}{60} + \frac{1}{20} = \frac{4}{60}$ , $f = 15$ cm.
Q7 · 4 marks · Long answerHOTS / competency
A candle and a screen are fixed 90 cm apart. A convex lens of focal length 20 cm is moved between them. (a) Find the two positions of the lens that give a sharp image. (b) Find the magnification in each position. (c) What do you notice about u and v?
Hint
$∣ u ∣ + v = 90$ and $\frac{1}{v} + \frac{1}{∣ u ∣} = \frac{1}{20}$ .
Answer
(a) From the two equations, $∣ u ∣ \cdot v = 20 \times 90 = 1800$ with $∣ u ∣ + v = 90$ : $∣ u ∣ = 30$ cm, $v = 60$ cm, or $∣ u ∣ = 60$ cm, $v = 30$ cm. So the lens is 30 cm or 60 cm from the candle. (b) $m = - 2$ and $m = - 0.5$ . (c) The values of $∣ u ∣$ and $v$ are interchanged.

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.Lens formula?Show answer
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ .
2.Sign of f for a concave lens?Show answer
Negative.
3.Magnification for an object at 2F of a convex lens?Show answer
−1.
4.What does a negative magnification tell you?Show answer
The image is real and inverted.
5.Sign of u for a real object?Show answer
Negative.

Lens Formula and Magnification: frequently asked questions

What is the lens formula?

1/v − 1/u = 1/f, where u is the object distance, v the image distance and f the focal length, all measured from the optical centre with the New Cartesian sign convention.

How do I know whether an image is real or virtual from my answer?

For a lens, a positive v means the image is on the far side (real); a negative v means it is on the same side as the object (virtual). A negative magnification also means real and inverted.

Do I need to derive the lens formula for ICSE Class 10?

No. The CISCE syllabus asks only for direct numerical problems using the lens formula; the derivation is not required.