ScienceCBSE Class 10 · Light – Reflection and Refraction

Lens Formula, Magnification and Power of a Lens

Q: What is the lens formula?

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

Q: What is 1 dioptre?

The power of a lens whose focal length is 1 metre.

Q: How is lens magnification different from mirror magnification?

For lenses m = v/u; for mirrors m = −v/u. Both equal h'/h.

Q: Why is the power of a concave lens negative?

Its focal length is negative in the sign convention, and P = 1/f.

Q: Do I need to derive the lens formula?

No. The 2026-27 syllabus says the derivation is not required.

Updated 14 October 2026 · Fact-checked

For spherical lenses, 1/v - 1/u = 1/f links object distance u, image distance v and focal length f, measured from the optical centre with the New Cartesian sign convention. Magnification is m = h'/h = v/u. Power, P = 1/f with f in metres, is measured in dioptres: positive for convex, negative for concave lenses.

Measure from O: object side u is negative, a real image on the far side has v positive; m = v/u is negative for inverted images, and power P = 1/f (f in metres)

Illustration drawn for QuestionBuddy.

Understand Lens Formula, Magnification and Power of a Lens

Sign convention for lenses. The same New Cartesian rules apply as for mirrors, with one change: distances are measured from the optical centre (O) of the lens.

The object is placed to the left, so u is negative.
Distances to the right of O (the direction light travels) are positive; to the left, negative.
A convex lens (converging) has its principal focus on the far side: f is positive. A concave lens (diverging) has f negative.
For a lens, a real image forms on the other side (right), so v is positive; a virtual image forms on the same side as the object, so v is negative. Note this is the opposite of mirrors.

Lens formula. $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ . It holds for both kinds of lens and all object positions if signs are used correctly. Derivation is not required.

Magnification. $m = \frac{h ^{'}}{h} = \frac{v}{u}$ (there is no minus sign for lenses).

m negative: real, inverted image (convex lens, object beyond F).
m positive: virtual, erect image (concave lens always; convex lens with object between F and O).
|m| > 1 magnified; |m| < 1 diminished.

Power of a lens. A lens that bends light more strongly has a shorter focal length. Power measures this: $P = \frac{1}{f}$ , with f in metres. The SI unit is the dioptre (D): 1 D is the power of a lens of focal length 1 m. A convex lens has positive power; a concave lens has negative power. An optician's prescription of +2.0 D means a convex lens of focal length +0.5 m.

Lenses in contact. When thin lenses are placed in contact, the net power is the sum of their powers: $P = P_{1} + P_{2} + P_{3} + \dots$ . Opticians use this to make up a required power by combining lenses; designers of instruments use it too.

Sense-check with ray diagrams. For a convex lens: object beyond 2F₁ gives a diminished real image between F₂ and 2F₂; at 2F₁ a same-size image at 2F₂; between F₁ and 2F₁ a magnified image beyond 2F₂; between F₁ and O a virtual, erect, magnified image on the same side (a magnifying glass). A concave lens always gives a virtual, erect, diminished image between F₁ and O. If your calculated v and m disagree with this picture, recheck the signs.

Lens Formula, Magnification and Power of a Lens: formulas and key facts

Lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$; Derivation not required. Note the minus sign.
Magnification (lens): $m = \frac{h ^{'}}{h} = \frac{v}{u}$; No minus sign, unlike mirrors.
Power: $P = \frac{1}{f ( in m )}$; Unit dioptre (D); convex +, concave −.
Lenses in contact: $P = P_{1} + P_{2} + P_{3} + \dots$; Net focal length f = 1/P.
Signs for lenses: $u < 0; f_{convex} > 0; f_{concave} < 0$; Real image: v > 0; virtual image: v < 0.

How to solve Lens Formula, Magnification and Power of a Lens questions

Use this routine for every lens numerical.

1Write u, f with signs: u negative; f positive for convex, negative for concave. Convert cm to m only for power.
2Rearrange: $\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$ , or $\frac{1}{u} = \frac{1}{v} - \frac{1}{f}$ .
3Combine fractions with an LCM, keep signs, invert.
4Find $m = \frac{v}{u}$ and $h^{'} = mh$ if heights are asked.
5Interpret: v positive means real (other side); negative means virtual (same side). m sign gives inverted or erect.
6For power, change f to metres and use $P = \frac{1}{f}$ ; for combinations add powers.
7Check by substituting back into the lens formula.

Lens Formula, Magnification and Power of a Lens 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.

Power of lenses and combination

Suppose an optician has a convex lens of focal length 25 cm and a concave lens of focal length 50 cm in the clinic and wants to know what single lens they are equivalent to when held in contact.

Find the power of each lens, the net power of the combination, and its focal length. Is the combination converging or diverging?

Try it first, then show the solutionSolution

Convex: $f = + 25$ cm = +0.25 m; $P_{1} = \frac{1}{0.25} = + 4$ D.
Concave: $f = - 50$ cm = −0.50 m; $P_{2} = \frac{1}{- 0.50} = - 2$ D.
Net power $P = P_{1} + P_{2} = + 4 - 2 = + 2$ D.
Net focal length $f = \frac{1}{P} = \frac{1}{2} = + 0.5$ m = +50 cm.
Positive power: the combination acts as a converging (convex) lens.

Answer: +4 D and −2 D; net +2 D; f = +50 cm; converging.

Memory hook: Add the dioptres, flip for the focal length.

Tricks to remember Lens Formula, Magnification and Power of a Lens

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

Mirror plus, lens minus
Mirror: 1/v + 1/u. Lens: 1/v − 1/u. Both = 1/f.
Use it when: Not mixing the two formulae.
Lens m has no minus
Lens magnification m = v/u straight; mirror m = −v/u.
Use it when: Magnification sign.
Power in metres
'P loves metres': always convert f to metres before 1/f.
Use it when: Avoiding power answers 100 times too small.

Lens Formula, Magnification and Power of a Lens: solved examples

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

Example 1: Convex lens, object between F and 2F · 3 marks · Short answerBoard level

An object is placed 15 cm from a convex lens of focal length 10 cm. Find the position, nature and magnification of the image.

Try it first, then show the solutionSolution

$u = - 15$ cm, $f = + 10$ cm (½ mark).
$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{10} - \frac{1}{15} = \frac{3 - 2}{30} = \frac{1}{30}$ , so $v = + 30$ cm (1 mark).
$m = \frac{v}{u} = \frac{30}{- 15} = - 2$ (1 mark).
Image 30 cm on the other side: real, inverted, magnified twice. Check: $\frac{1}{30} - \frac{1}{- 15} = \frac{1 + 2}{30} = \frac{1}{10}$ (½ mark).

Answer: v = +30 cm; real, inverted, magnified; m = −2.

Example 2: Concave lens · 3 marks · Short answerHarder board level

An object is 30 cm in front of a concave lens of focal length 20 cm. Find the image position and magnification and describe the image.

Try it first, then show the solutionSolution

$u = - 30$ cm, $f = - 20$ cm (½ mark).
$\frac{1}{v} = - \frac{1}{20} - \frac{1}{30} = \frac{- 3 - 2}{60} = - \frac{1}{12}$ , so $v = - 12$ cm (1 mark).
$m = \frac{v}{u} = \frac{- 12}{- 30} = + 0.4$ (1 mark).
Image 12 cm from the lens on the object's side: virtual, erect, diminished. Check: $- \frac{1}{12} + \frac{1}{30} = \frac{- 5 + 2}{60} = - \frac{1}{20}$ (½ mark).

Answer: v = −12 cm; virtual, erect, diminished; m = +0.4.

Example 3: Image height and power · 3 marks · Short answerHarder board level

A 3.0 cm tall object is placed 30 cm from a convex lens of focal length 20 cm. Find (a) the image distance, (b) the image height, (c) the power of the lens.

Try it first, then show the solutionSolution

(a) $\frac{1}{v} = \frac{1}{20} - \frac{1}{30} = \frac{3 - 2}{60} = \frac{1}{60}$ , $v = + 60$ cm (1 mark).
(b) $m = \frac{60}{- 30} = - 2$ ; $h^{'} = - 2 \times 3.0 = - 6.0$ cm: real, inverted (1 mark).
(c) $f = + 0.20$ m; $P = \frac{1}{0.20} = + 5$ D (1 mark).

Answer: (a) +60 cm (b) −6.0 cm (inverted) (c) +5 D.

Example 4: Find the focal length from a screen set-up · 5 marks · Long answerHOTS / competency

A convex lens forms a real image 4 times the size of an object on a screen. The object and screen are 100 cm apart. Find the object distance, image distance, focal length and power of the lens.

Try it first, then show the solutionSolution

Real image: m = −4 = v/u, so v = −4u. Let u = −x; then v = +4x (1 mark).
Object–screen distance: x + 4x = 5x = 100 cm, so x = 20 cm: u = −20 cm, v = +80 cm (1 mark).
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{80} + \frac{1}{20} = \frac{1 + 4}{80} = \frac{1}{16}$ , so f = +16 cm (1 mark).
P = $\frac{1}{0.16}$ = +6.25 D (1 mark).
Check: m = 80/(−20) = −4 ✓; 1/80 − 1/(−20) = 5/80 = 1/16 ✓ (1 mark).

Answer: u = −20 cm, v = +80 cm, f = +16 cm, P = +6.25 D.

Example 5: Working backwards with a concave lens · 3 marks · Short answerHarder board level

A concave lens of focal length 20 cm forms an image 15 cm from it on the same side as the object. Find the object distance and the magnification.

Try it first, then show the solutionSolution

$f = - 20$ cm; virtual image on the object side, so $v = - 15$ cm (½ mark).
$\frac{1}{u} = \frac{1}{v} - \frac{1}{f} = - \frac{1}{15} + \frac{1}{20} = \frac{- 4 + 3}{60} = - \frac{1}{60}$ , so $u = - 60$ cm (1½ marks).
$m = \frac{v}{u} = \frac{- 15}{- 60} = + 0.25$ . Check: $- \frac{1}{15} + \frac{1}{60} = - \frac{3}{60} = - \frac{1}{20}$ (1 mark).

Answer: Object 60 cm in front of the lens; m = +0.25.

Where students lose marks in Lens Formula, Magnification and Power of a Lens

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

Mistake: Using the mirror formula for a lens
The two formulae look alike.
Wrong working
1/v + 1/u = 1/f for a lens.
Right working
1/v − 1/u = 1/f for a lens.
Do this: Write 'lens: minus' in the margin before solving.
Usually costs: The whole numerical.
Mistake: Power with f in centimetres
Students forget the unit change.
Wrong working
f = 25 cm, P = 1/25 = 0.04 D.
Right working
f = 0.25 m, P = 1/0.25 = +4 D.
Do this: Convert to metres first.
Usually costs: 1 mark.
Mistake: Adding a minus to lens magnification
Carried over from mirrors.
Wrong working
m = −v/u = −30/(−15) = +2, erect.
Right working
m = v/u = 30/(−15) = −2, inverted.
Do this: Lens m = v/u.
Usually costs: 1 mark and the nature mark.
Mistake: Wrong sign of f for a concave lens
Students treat all focal lengths as positive.
Wrong working
Concave lens, f = +20 cm.
Right working
Concave lens, f = −20 cm.
Do this: Converging (convex) +, diverging (concave) −.
Usually costs: The answer and nature marks.

Exam tips

Write the sign convention values first (u = −…, f = +… or −…).
Keep the answer in the unit given (cm) for distances, but metres for power.
Interpret every answer: real or virtual, erect or inverted, magnified or diminished.
For combinations, add powers, not focal lengths.
Draw a rough ray diagram to check your numbers when time allows.

Practice questions on Lens Formula, Magnification and Power of a Lens

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
The power of a concave lens of focal length 50 cm is:
(a) +2 D (b) −2 D (c) +0.02 D (d) −0.5 D
Hint
Convert to metres; concave is negative.
Answer
(b) −2 D, since P = 1/(−0.50 m).
Q2 · 1 mark · Assertion-reasonBoard level
Assertion (A): The power of a convex lens is positive.
Reason (R): The focal length of a convex lens is taken as positive in the sign convention.
(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
P = 1/f.
Answer
(a).
Q3 · 3 marks · Short answerHarder board level
A convex lens of focal length 12 cm is used as a magnifying glass with an object 8 cm away. Find the image position and magnification and describe the image.
Hint
Object inside F.
Answer
u = −8 cm, f = +12 cm. 1/v = 1/12 − 1/8 = (2 − 3)/24 = −1/24, v = −24 cm (same side, virtual). m = −24/−8 = +3: erect and three times magnified. Check: −1/24 + 1/8 = 2/24 = 1/12.
Q4 · 4 marks · Case-basedHOTS / competency
Read the passage and answer.
Suppose a student focuses the image of a distant tree on a screen using a lens and finds the sharp image 20 cm from the lens. She then places a candle 30 cm from the same lens.
(a) What type of lens is it and what is its focal length? (1 mark)
(b) Find its power. (1 mark)
(c) Find where the screen must be placed for a sharp image of the candle, and the magnification. (2 marks)
Hint
Rays from a distant object focus at F.
Answer
(a) Convex (it forms a real image); f = +20 cm (image of a distant object forms at F). (b) P = 1/0.20 = +5 D. (c) u = −30: 1/v = 1/20 − 1/30 = 1/60, v = +60 cm on the other side; m = 60/(−30) = −2: real, inverted, magnified.
Q5 · 3 marks · Short answerHOTS / competency
Two thin lenses of powers +3.5 D and −2.5 D are placed in contact. Find the focal length of the combination and state its nature. Then find the image distance for an object 50 cm away from this combination.
Hint
Add powers first.
Answer
P = +3.5 − 2.5 = +1.0 D, f = +1.0 m = +100 cm, converging. u = −50 cm: 1/v = 1/100 − 1/50 = −1/100, v = −100 cm: a virtual, erect image 100 cm on the object side (object is inside F); m = −100/−50 = +2.
Q6 · 2 marks · Very short answerHarder board level
Where should an object be placed in front of a convex lens of focal length 15 cm to get a real image of the same size? How far apart are the object and the image?
Hint
Same size real image: object at 2F.
Answer
At 2F: u = −30 cm. Then 1/v = 1/15 − 1/30 = 1/30, v = +30 cm, m = −1. Object and image are 60 cm apart (4f).

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.Magnification for a lens?Show answer
$m = \frac{v}{u} = \frac{h ^{'}}{h}$ .
3.SI unit of power?Show answer
Dioptre (D).
4.Power of a lens of focal length +0.5 m?Show answer
+2 D.
5.Sign of power of a concave lens?Show answer
Negative.
6.Net power of lenses in contact?Show answer
Sum of individual powers.

Lens Formula, Magnification and Power of a Lens: frequently asked questions

What is the lens formula?

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

What is 1 dioptre?

The power of a lens whose focal length is 1 metre.

How is lens magnification different from mirror magnification?

For lenses m = v/u; for mirrors m = −v/u. Both equal h'/h.

Why is the power of a concave lens negative?

Its focal length is negative in the sign convention, and P = 1/f.

Do I need to derive the lens formula?

No. The 2026-27 syllabus says the derivation is not required.