PhysicsICSE Class 10 · Lenses

Power of a Lens: P = 1/f, the Dioptre and Signs for Convex and Concave Lenses (ICSE Class 10)

Updated 15 October 2026 · Fact-checked

The power of a lens measures how strongly it converges or diverges light: P = 1/f, with f in metres. Its 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 and a concave lens negative power. With f in centimetres, P = 100/f.

Understand Power of a Lens (Dioptre)

Two convex lenses can both converge light, but the one with the shorter focal length bends rays more sharply and brings them to a focus closer to the lens. The power of a lens measures this bending ability.

Definition. The power of a lens is the reciprocal of its focal length expressed in metres:

$P = \frac{1}{f (in m)} .$

Unit. The SI unit of power is the dioptre (D). One dioptre is the power of a lens whose focal length is 1 metre: $1 D = 1 m^{- 1}$ .

If the focal length is in centimetres:

$P = \frac{100}{f (in cm)} .$

Sign of power. Power carries the sign of the focal length:

Convex (converging) lens: $f$ positive, so P positive. Example: $f = + 50$ cm gives $P = + 2$ D.
Concave (diverging) lens: $f$ negative, so P negative. Example: $f = - 50$ cm gives $P = - 2$ D.

A prescription such as '−1.5 D' therefore names a concave lens of focal length $- \frac{1}{1.5}$ m $\approx - 67$ cm.

Large and small power.

A large positive power means a short focal length: a strongly converging lens (a magnifying glass needs a lens of short focal length, i.e. high power).
A small power means a long focal length: a weak lens. A flat sheet of glass has $f \to \infty$ and $P = 0$ .

Finding power from an experiment.
1. Focus a distant object (a window or tree) on a screen: the image distance ≈ $f$ ; then $P = \frac{100}{f}$ .
2. Or measure $u$ and $v$ for a near object, use the lens formula to get $f$ , then the power.

From power back to images. Given P, find $f = \frac{1}{P}$ m $= \frac{100}{P}$ cm, then use the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ as usual.

Common slips. Forgetting to convert cm to m (giving 0.04 D instead of 4 D); dropping the minus sign for a concave lens; leaving the answer as a fraction.

Power of a Lens (Dioptre): formulas and key facts

Power of a lens: $P = \frac{1}{f}$; f in metres; P in dioptres (D).
With f in centimetres: $P = \frac{100}{f}$; f in cm.
Focal length from power: $f = \frac{1}{P} m = \frac{100}{P} cm$; Sign of f follows the sign of P.

How to solve Power of a Lens (Dioptre) questions

Convert the focal length to metres (or use 100/f with f in cm), and keep the sign.

1If u and v are given, find f first with the lens formula (with signs).
2Convert: f in cm → divide 100 by f; f in m → take the reciprocal.
3Attach the sign: + for convex (converging), − for concave (diverging).
4Give the answer as a decimal with the unit D.
5For the reverse problem, $f = \frac{100}{P}$ cm, then use the lens formula for images.

Power of a Lens (Dioptre) 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.

Reading glasses

Suppose a pair of reading glasses from a chemist is labelled '+2.5'. This number is the power of each lens in dioptres.

What kind of lens is it, and what is its focal length?

Try it first, then show the solutionSolution

Positive power: a convex (converging) lens.
$f = \frac{100}{2.5} = 40$ cm.

Answer: Convex, with a focal length of 40 cm.

Memory hook: Plus is a converger; divide 100 by the label for f in cm.

A clip-on macro lens

Imagine a clip-on macro lens for a phone camera marked '+12 D'.

Find its focal length, and say why a high power suits close-up photography.

Try it first, then show the solutionSolution

$f = \frac{100}{12} \approx 8.3$ cm.
A short focal length bends rays strongly, letting the camera focus on objects only a few centimetres away.

Answer: About 8.3 cm; high power means strong bending for very close objects.

Memory hook: High power, short focus, close-ups.

Tricks to remember Power of a Lens (Dioptre)

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

Hundred over f
f in cm? Power = 100 ÷ f. f in m? Power = 1 ÷ f.
Use it when: Avoiding unit errors.
Sign follows the lens
Convex converges, positive power; concave caves out, negative power.
Use it when: Signs of P.
Short focus, strong lens
Power and focal length are a see-saw: one up, the other down.
Use it when: Comparing lenses.
1 D = 1 m
One dioptre is the lens with a one-metre focal length.
Use it when: Defining the dioptre.

Power of a Lens (Dioptre): solved examples

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

Example 1: Power of a concave lens · 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

Try it first, then show the solutionSolution

$f = - 50$ cm $= - 0.5$ m; $P = \frac{1}{- 0.5} = - 2$ D.

Answer: (b) −2 D

Example 2: Power from focal length · 2 marks · Short answerBoard level

Find the power of a convex lens of focal length 50 cm. Define the unit you use.

Try it first, then show the solutionSolution

$P = \frac{100}{f} = \frac{100}{50} = + 2$ D. (1)
One dioptre is the power of a lens of focal length 1 m. (1)

Answer: +2 D; 1 D = power of a lens of focal length 1 m

Example 3: From power to image · 3 marks · Short answerHarder board level

A lens has a power of −2.5 D. (a) Name the lens and find its focal length. (b) An object is placed 40 cm from it. Find the position and magnification of the image.

Try it first, then show the solutionSolution

(a) Negative power: concave; $f = \frac{100}{- 2.5} = - 40$ cm. (1)
(b) $\frac{1}{v} = - \frac{1}{40} + \frac{1}{- 40} = - \frac{2}{40}$ , so $v = - 20$ cm. (1)
$m = \frac{- 20}{- 40} = + 0.5$ : virtual, erect, half size. (1)

Answer: (a) Concave, −40 cm (b) 20 cm on the object's side; m = +0.5

Example 4: Comparing two convex lenses · 3 marks · Short answerHarder board level

Two converging lenses in a kit are marked +6 D and +1.5 D. (a) Work out the focal length of each. (b) Which bends light more strongly? (c) Which would make the better magnifying glass, and why?

Try it first, then show the solutionSolution

(a) $f_{1} = \frac{100}{6} \approx 16.7$ cm; $f_{2} = \frac{100}{1.5} \approx 66.7$ cm. (1)
(b) The +6 D lens (greater power, shorter focal length). (1)
(c) The +6 D lens: a magnifying glass needs a short focal length (high power). (1)

Answer: (a) About 16.7 cm and 66.7 cm (b) +6 D (c) +6 D, shorter focal length

Example 5: Power from u and v, then a prediction · 4 marks · Long answerHOTS / competency

An object 90 cm from a lens forms a sharp image on a screen 45 cm behind it. (a) Find the focal length and power of the lens. (b) The object is moved to 20 cm from the lens. Where is the new image, and what is its nature?

Try it first, then show the solutionSolution

(a) $u = - 90$ , $v = + 45$ : $\frac{1}{f} = \frac{1}{45} + \frac{1}{90} = \frac{3}{90}$ , $f = + 30$ cm. (1)
$P = \frac{100}{30} \approx + 3.33$ D. (1)
(b) $\frac{1}{v} = \frac{1}{30} - \frac{1}{20} = - \frac{1}{60}$ , so $v = - 60$ cm. (1)
$m = \frac{- 60}{- 20} = + 3$ : virtual, erect, magnified, on the object's side (object inside F). (1)

Answer: (a) +30 cm, about +3.33 D (b) 60 cm on the object's side; virtual, erect, 3 times

Example 6: Case study: sorting unlabelled lenses · 4 marks · Case-basedHOTS / competency

In a laboratory, a student finds three unlabelled lenses X, Y and Z. Focusing a distant window, X gives a sharp image on a screen 32 cm away and Y one 16 cm away. Z gives no real image; instead, with an object 30 cm from Z, a virtual image is seen 12 cm from it on the same side.
(a) Find the powers of X and Y.
(b) Find the focal length and power of Z, and name it.
(c) Which lens has the greatest bending power? Which should be used as a magnifying glass?

Try it first, then show the solutionSolution

(a) Distant object: image at F. $P_{X} = \frac{100}{32} \approx + 3.13$ D; $P_{Y} = \frac{100}{16} = + 6.25$ D. (1)
(b) $u = - 30$ , $v = - 12$ : $\frac{1}{f} = - \frac{1}{12} + \frac{1}{30} = - \frac{3}{60}$ , $f = - 20$ cm. (1)
$P_{Z} = \frac{100}{- 20} = - 5$ D: concave. (1)
(c) Y has the greatest bending power: 6.25 D is larger in magnitude than Z (5 D) or X (about 3.13 D). Y is also converging, so it is the one to use as a magnifying glass; Z, being concave, could never magnify. (1)

Answer: (a) About +3.13 D, +6.25 D (b) −20 cm, −5 D, concave (c) Y in both cases

Where students lose marks in Power of a Lens (Dioptre)

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

Mistake: Not converting the focal length to metres
P = 1/f needs f in metres; with f in cm use 100/f.
Wrong working
$f = 20$ cm: $P = \frac{1}{20} = 0.05$ D.
Right working
$P = \frac{100}{20} = 5$ D.
Do this: Write 'f = … m' before taking the reciprocal.
Usually costs: Reported by CISCE examiners for 2024 Q4(i).
Mistake: Leaving power as a fraction
Answers are expected as decimals with the unit.
Wrong working
$P = \frac{100}{24}$ .
Right working
$P \approx + 4.17$ D.
Do this: Divide out to two decimal places and add 'D'.
Usually costs: Also reported for 2024 Q4(i).
Mistake: Dropping the sign
The sign tells the examiner which kind of lens it is.
Wrong working
Concave lens, $f = 40$ cm: $P = 2.5$ D.
Right working
$P = - 2.5$ D.
Do this: Concave lens: write the minus sign with f from the first line.
Usually costs: The answer mark.
Mistake: Thinking larger focal length means greater power
Power is the reciprocal of focal length.
Wrong working
'The 1 m lens is more powerful than the 10 cm lens.'
Right working
'The 10 cm lens (+10 D) is more powerful than the 1 m lens (+1 D).'
Do this: Compute both powers before comparing.
Usually costs: Reasoning marks.

Exam tips

In 2024 Q4(i), after finding $f = 24$ cm, the expected power was $P = \frac{100}{24} \approx + 4.17$ D, as a decimal with sign and unit.
Always state the type of lens from the sign of the power when asked to 'name the lens'.
For 'define 1 dioptre', write: the power of a lens whose focal length is 1 metre.
In multi-step questions, carry f with its sign from the lens formula into $P = \frac{100}{f}$ .

Practice questions on Power of a Lens (Dioptre)

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
One dioptre is equal to:
(a) 1 m
(b) 1 m⁻¹
(c) 1 cm⁻¹
(d) 100 m
Hint
Power is a reciprocal of length.
Answer
(b) 1 m⁻¹
Q2 · 1 mark · Very short answerBoard level
A diverging lens has a focal length of magnitude 12.5 cm. What is its power?
Hint
Concave: f negative.
Answer
$P = \frac{100}{- 12.5} = - 8$ D.
Q3 · 2 marks · Short answerHarder board level
Which has the greater power: a convex lens of focal length 10 cm or one of focal length 1 m? Give both powers.
Hint
Shorter focal length, greater power.
Answer
10 cm lens: +10 D; 1 m lens: +1 D. The 10 cm lens has the greater power.
Q4 · 3 marks · Short answerHarder board level
A lens of power +2.5 D forms an image of an object placed 60 cm from it. Find the image distance and magnification.
Hint
$f = \frac{100}{2.5}$ cm.
Answer
$f = + 40$ cm; $\frac{1}{v} = \frac{1}{40} - \frac{1}{60} = \frac{1}{120}$ , $v = + 120$ cm; $m = \frac{120}{- 60} = - 2$ .
Q5 · 3 marks · Short answerHarder board level
A lens of power −3 D has an object 50 cm in front of it. Find the image distance and magnification.
Hint
$f = - \frac{100}{3}$ cm.
Answer
$\frac{1}{v} = - \frac{3}{100} - \frac{1}{50} = - \frac{5}{100}$ , $v = - 20$ cm; $m = \frac{- 20}{- 50} = + 0.4$ .
Q6 · 2 marks · Short answerHarder board level
(a) Why is the power of a concave lens taken as negative? (b) What does a lens of power 0 D mean?
Hint
Power carries the sign of f.
Answer
(a) A concave lens has a negative focal length (its focus is virtual, on the object's side), and $P = \frac{1}{f}$ . (b) Infinite focal length: it neither converges nor diverges light, like a flat glass sheet.
Q7 · 3 marks · Short answerHOTS / competency
An object 24 cm from a lens gives an erect image three times its size on the same side as the object. Find the focal length and power of the lens.
Hint
Erect, same side: $m = + 3$ , so $v = 3 u$ .
Answer
$v = 3 \times (- 24) = - 72$ cm; $\frac{1}{f} = - \frac{1}{72} + \frac{1}{24} = \frac{2}{72}$ , $f = + 36$ cm; $P = \frac{100}{36} \approx + 2.78$ D.

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.Formula for the power of a lens?Show answer
$P = \frac{1}{f}$ (f in metres).
2.SI unit of power of a lens?Show answer
Dioptre (D).
3.Sign of the power of a concave lens?Show answer
Negative.
4.Focal length of a +10 D lens?Show answer
10 cm.
5.Focal length of a −0.5 D lens?Show answer
−2 m.

Power of a Lens (Dioptre): frequently asked questions

What is meant by the power of a lens?

It is the ability of a lens to converge or diverge light, measured as the reciprocal of its focal length in metres: P = 1/f.

What is one dioptre?

One dioptre is the power of a lens whose focal length is one metre. 1 D = 1 m⁻¹.

Why is the power of a concave lens negative?

Its focal length is negative in the sign convention, because its focus is virtual and on the same side as the incoming light. Power has the same sign as the focal length.