PhysicsISC Class 12 · Ray Optics and Optical Instruments

Lens Formula and Power of a Lens: Lens Maker's Formula, Combinations and Liquids

Updated 15 October 2026 · Fact-checked

For a thin lens, using the Cartesian sign convention, $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ and magnification $m = \frac{v}{u}$. The lens maker's formula gives $\frac{1}{f} = (n - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$. Power $P = \frac{1}{f}$ with f in metres, in dioptres; thin lenses in contact add: $P = P_1 + P_2$.

Measure from the lens: against the light is negative (u), along the light is positive (v); here m = v/u is negative, so the image is real and inverted.

Illustration drawn for QuestionBuddy.

Understand Lens Formula, Lens Maker's Formula and Power

Sign convention (Cartesian). Distances are measured from the optical centre; those in the direction of the incident light are positive, those against it negative. A real object has u < 0. A convex lens has f > 0, a concave lens f < 0. (Any one consistent convention is allowed in ISC numericals.)

Step 1: refraction at one spherical surface. For light going from medium $n_{1}$ to $n_{2}$ across a surface of radius R, with a real image (the syllabus case: convex towards the rarer medium),
$\frac{n _{2}}{v} - \frac{n _{1}}{u} = \frac{n _{2} - n _{1}}{R} .$

Step 2: lens maker's formula. A thin lens of index n in air has two surfaces, radii $R_{1}$ and $R_{2}$ . The first surface forms an image $I_{1}$ at $v_{1}$ : $\frac{n}{v _{1}} - \frac{1}{u} = \frac{n - 1}{R _{1}}$ . This acts as the object for the second surface (glass to air): $\frac{1}{v} - \frac{n}{v _{1}} = \frac{1 - n}{R _{2}}$ . Adding the two equations eliminates $v_{1}$ :
$\frac{1}{v} - \frac{1}{u} = (n - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}}) .$
For an object at infinity, v = f, so
$\frac{1}{f} = (n - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}}) and hence \frac{1}{v} - \frac{1}{u} = \frac{1}{f} .$
For a biconvex lens $R_{1} > 0$ , $R_{2} < 0$ ; for a plano-convex lens one radius is infinite.

Magnification. $m = \frac{h ^{'}}{h} = \frac{v}{u}$ . Negative m: real, inverted image; positive m: virtual, erect image.

Power. $P = \frac{1}{f}$ with f in metres; unit dioptre (D) = m⁻¹. Convex lenses have positive power, concave lenses negative. A lens of short focal length bends light more strongly.

Thin lenses in contact. The first lens forms an image at $v_{1}$ : $\frac{1}{v _{1}} - \frac{1}{u} = \frac{1}{f _{1}}$ . This is the object for the second lens: $\frac{1}{v} - \frac{1}{v _{1}} = \frac{1}{f _{2}}$ . Adding:
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f _{1}} + \frac{1}{f _{2}} \Rightarrow \frac{1}{F} = \frac{1}{f _{1}} + \frac{1}{f _{2}}, P = P_{1} + P_{2} .$
The magnification of a combination is the product $m = m_{1} m_{2}$ .

Lens in a liquid. Replace n by the relative index $\frac{n _{g}}{n _{l}}$ :
$\frac{1}{f _{l}} = (\frac{n _{g}}{n _{l}} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}}) .$

$n_{l} < n_{g}$ (e.g. glass in water): $f_{l}$ is longer, same type of lens.
$n_{l} = n_{g}$ : no refraction; the lens becomes invisible ( $f \to \infty$ ).
$n_{l} > n_{g}$ : the sign of f reverses; a convex lens diverges and a concave lens converges.

Experiment: focal length by no-parallax (lens formula). Measure several pairs of object and image distances. Plotting $\frac{1}{v}$ (dependent variable, y-axis) against $\frac{1}{∣ u ∣}$ (x-axis) gives a straight line of slope −1 cutting both axes at $\frac{1}{f}$ . Moving the object closer to F increases v and the magnification $\frac{v}{u}$ (asked in ISC 2024 Q15).

Lens Formula, Lens Maker's Formula and Power: formulas and key facts

Lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$; Cartesian signs.
Magnification: $m = \frac{h ^{'}}{h} = \frac{v}{u}$; Negative: real, inverted.
Single surface: $\frac{n _{2}}{v} - \frac{n _{1}}{u} = \frac{n _{2} - n _{1}}{R}$; Starting point.
Lens maker's formula: $\frac{1}{f} = (n - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$; n relative to the surrounding medium.
Power: $P = \frac{1}{f ( m )}$; Dioptre, D.
Lenses in contact: $\frac{1}{F} = \frac{1}{f _{1}} + \frac{1}{f _{2}}, P = P_{1} + P_{2}$; Thin lenses, touching.
Lens in a liquid: $\frac{f _{l}}{f _{a}} = \frac{( n _{g} - 1 )}{( \frac{n _{g}}{n _{l}} - 1 )}$; Same radii.

How to solve Lens Formula, Lens Maker's Formula and Power questions

Write the sign convention once, substitute with signs, and only then simplify.

1Put the given values with signs: real object u < 0; convex f > 0, concave f < 0.
2Solve $\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$ ; interpret the sign of v (positive: real image on the far side).
3Magnification $m = \frac{v}{u}$ ; image height $h^{'} = mh$ .
4For lens maker: decide the signs of $R_{1}$ and $R_{2}$ from the direction of light (biconvex: $R_{1} > 0$ , $R_{2} < 0$ ).
5For combinations: add powers in dioptres (f in metres) and convert back: $F = \frac{1}{P}$ .
6For a lens in a liquid: use $\frac{n _{g}}{n _{l}}$ in place of n; compare $n_{l}$ with $n_{g}$ to predict the behaviour.

Lens Formula, Lens Maker's Formula and Power 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.

Burning glass and the Sun's image

Suppose you hold a convex lens of focal length 20 cm in sunlight to focus the Sun on a sheet of paper. The Sun subtends about $9.3 \times 1 0^{- 3}$ rad.

Where is the bright spot, and how large is it?

Try it first, then show the solutionSolution

Rays from the distant Sun are parallel, so the image forms at the focal plane, 20 cm behind the lens.
Diameter of the image $= f θ = 200 mm \times 9.3 \times 1 0^{- 3} \approx 1.86$ mm.

Answer: At 20 cm, a spot about 1.9 mm across; all the light through the lens is concentrated there, so the paper can char.

Memory hook: Parallel in, focus out.

Adding a clip-on lens

Imagine a short-sighted student whose spectacles have power −1.5 D, who clips a +4 D magnifier over them to read fine print.

Find the power and focal length of the combination.

Try it first, then show the solutionSolution

$P = - 1.5 + 4 = + 2.5$ D.
$F = \frac{1}{2.5}$ m = 40 cm, converging.

Answer: +2.5 D; 40 cm, a converging combination.

Memory hook: Powers add; focal lengths do not.

Tricks to remember Lens Formula, Lens Maker's Formula and Power

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

v minus u
Lens: $\frac{1}{v} - \frac{1}{u}$ ; mirror: $\frac{1}{v} + \frac{1}{u}$ .
Use it when: Not mixing the formulas.
n minus one, R1 minus R2
Lens maker: "n minus 1 times (1 by R1 minus 1 by R2)".
Use it when: Formula recall.
Metres for dioptres
P = 100/f when f is in cm.
Use it when: Power numericals.
Liquid flips when it out-bends glass
$n_{l} > n_{g}$ : converging becomes diverging.
Use it when: Lens-in-liquid questions.

Lens Formula, Lens Maker's Formula and Power: 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

A concave lens has a focal length of 50 cm. Its power is:
(a) +2 D
(b) −2 D
(c) +0.5 D
(d) −0.5 D

Try it first, then show the solutionSolution

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

Answer: (b).

Example 2: Virtual image by a convex lens · 2 marks · Short answerBoard level

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

Try it first, then show the solutionSolution

$\frac{1}{v} = \frac{1}{15} + \frac{1}{- 10} = - \frac{1}{30}$ , so v = −30 cm (same side as the object). (1)
$m = \frac{- 30}{- 10} = + 3$ : virtual, erect, three times enlarged. (1)

Answer: v = −30 cm; m = +3; virtual and erect.

Example 3: Derivation: lenses in contact · 3 marks · Long answerHarder board level

Derive an expression for the focal length of two thin lenses in contact. Hence show that their powers add.

Try it first, then show the solutionSolution

Lens 1 forms an image at $v_{1}$ : $\frac{1}{v _{1}} - \frac{1}{u} = \frac{1}{f _{1}}$ . (1)
Being thin and in contact, this image is the object for lens 2: $\frac{1}{v} - \frac{1}{v _{1}} = \frac{1}{f _{2}}$ . (1)
Adding: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f _{1}} + \frac{1}{f _{2}} = \frac{1}{F}$ ; with $P = \frac{1}{f}$ , $P = P_{1} + P_{2}$ . (1)

Answer: $\frac{1}{F} = \frac{1}{f _{1}} + \frac{1}{f _{2}}$ ; $P = P_{1} + P_{2}$ .

Example 4: Plano-convex lens in air and in a liquid · 3 marks · Short answerHarder board level

A plano-convex lens of glass (n = 1.5) has a curved surface of radius 12 cm. Find its focal length and power in air, and its focal length in a liquid of refractive index 1.25.

Try it first, then show the solutionSolution

$\frac{1}{f} = 0.5 (\frac{1}{12} - 0)$ , so f = 24 cm. (1)
$P = \frac{100}{24} \approx + 4.17$ D. (1)
In the liquid: $\frac{n _{g}}{n _{l}} = 1.2$ ; $\frac{1}{f _{l}} = 0.2 \times \frac{1}{12}$ , so $f_{l} = 60$ cm. (1)

Answer: 24 cm; ≈ +4.17 D; 60 cm.

Example 5: Finding a hidden lens · 3 marks · Short answerHOTS / competency

A convex lens of focal length 12 cm is in contact with an unknown thin lens. The combination forms a real image 60 cm from it of an object placed 30 cm away. Find the focal length and power of the unknown lens.

Try it first, then show the solutionSolution

Combination: $\frac{1}{F} = \frac{1}{60} - \frac{1}{- 30} = \frac{1}{20}$ , F = 20 cm. (1)
$\frac{1}{f _{2}} = \frac{1}{20} - \frac{1}{12} = - \frac{1}{30}$ , so $f_{2}$ = −30 cm (concave). (1)
$P_{2} = \frac{100}{- 30} \approx - 3.33$ D. (1)

Answer: −30 cm (concave); ≈ −3.33 D.

Example 6: No-parallax experiment · 4 marks · Case-basedHOTS / competency

Suppose a student uses the no-parallax method with a convex lens and records object and image distances (magnitudes) of (30 cm, 60 cm), (40 cm, 40 cm) and (60 cm, 30 cm).
(i) Find f from each reading. (1 mark)
(ii) Describe the graph of $\frac{1}{v}$ against $\frac{1}{∣ u ∣}$ and how f is read from it. (1 mark)
(iii) As the object is moved from 60 cm to 30 cm, what happens to the magnification? (1 mark)
(iv) Find the power of the lens. (1 mark)

Try it first, then show the solutionSolution

(i) $\frac{1}{f} = \frac{1}{v} + \frac{1}{∣ u ∣}$ : $\frac{1}{60} + \frac{1}{30} = \frac{1}{20}$ , $\frac{1}{40} + \frac{1}{40} = \frac{1}{20}$ , $\frac{1}{30} + \frac{1}{60} = \frac{1}{20}$ : f = 20 cm each time.
(ii) A straight line of slope −1 cutting both axes at $\frac{1}{f} = 0.05$ cm⁻¹; f is the reciprocal of either intercept.
(iii) |m| = $\frac{v}{∣ u ∣}$ rises from 0.5 to 1 to 2: it increases as the object approaches F.
(iv) $P = \frac{1}{0.20 m} = + 5$ D.

Answer: (i) 20 cm; (ii) slope −1, intercepts 1/f; (iii) increases; (iv) +5 D.

Where students lose marks in Lens Formula, Lens Maker's Formula and Power

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

Mistake: Wrong sign or unit in a combination
The ISC 2025 examiners (Q6) saw sign-convention errors and focal lengths without proper units for a converging and a diverging lens in contact.
Wrong working
Lenses of +2.5 D and −1.5 D: $P = 2.5 + 1.5 = 4$ D, F = 0.25.
Right working
$P = 2.5 - 1.5 = 1.0$ D; F = 1 m (100 cm).
Do this: Add powers with their signs; give F with a unit.
Usually costs: 1 to 2 marks.
Mistake: Power with f in centimetres
In 2024 Q15(iii) some candidates confused the power relation with R = 2f; and P = 1/f gives dioptres only when f is in metres.
Do this: $P = \frac{1}{f}$ with f in metres (or $\frac{100}{f}$ with f in cm).
Usually costs: 1 mark.
Mistake: Verdict without reason
Also 2024 Q15: candidates said "correct" or "incorrect" without the reason.
Do this: State the rule, e.g. " $m = \frac{v}{u}$ , so magnification increases as u decreases".
Usually costs: 1 mark per part.
Mistake: Not seeing why f increases in water
In 2025 Q1(A)(vi) the link between f and refractive index was unclear.
Do this: f depends on $\frac{n _{g}}{n _{l}} - 1$ ; in water this is smaller, so f is larger.
Usually costs: 1 mark.

Exam tips

State the sign convention in one line at the start of every numerical.
In the lens maker's derivation, draw both surfaces and the intermediate image $I_{1}$ .
Keep powers to two decimal places and quote D.
For "nature of image", give real/virtual, erect/inverted and size.

Practice questions on Lens Formula, Lens Maker's Formula and Power

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 SI unit of the power of a lens is:
(a) watt
(b) dioptre
(c) metre
(d) joule
Hint
P = 1/f.
Answer
(b), equal to m⁻¹.
Q2 · 2 marks · Short answerBoard level
An object is 20 cm from a concave lens of focal length 20 cm. Find the image position and magnification.
Hint
f = −20 cm.
Answer
v = −10 cm; m = +0.5 (virtual, erect, diminished).
Q3 · 1 mark · Short answerHarder board level
Thin lenses of power +5 D and −3 D are in contact. Find the power and focal length of the combination.
Hint
Add powers.
Answer
+2 D; 50 cm.
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): A convex glass lens can act as a diverging lens.
Reason (R): This happens when it is immersed in a liquid of refractive index greater than that of the glass.
(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
Sign of $\frac{n _{g}}{n _{l}} - 1$ .
Answer
(a).
Q5 · 2 marks · Short answerHarder board level
A biconvex lens (n = 1.5) has radii of curvature 30 cm each. Find its focal length and power.
Hint
$R_{1} = + 30$ , $R_{2} = - 30$ .
Answer
30 cm; ≈ +3.33 D.
Q6 · 3 marks · Short answerHOTS / competency
A biconcave lens (n = 1.6) has radii 20 cm each. Find its focal length in air and in a liquid of refractive index 1.8. Comment.
Hint
Relative index 0.889.
Answer
Air: f ≈ −16.7 cm. Liquid: $\frac{1}{f} = (0.889 - 1) (- \frac{2}{20})$ , f ≈ +90 cm: the concave lens now converges.
Q7 · 3 marks · Short answerHOTS / competency
An object 20 cm from a convex lens gives an image three times its size. Find the focal length if the image is (a) real, (b) virtual.
Hint
m = −3 or +3.
Answer
(a) v = +60 cm, f = 15 cm; (b) v = −60 cm, f = 30 cm.
Q8 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a phone camera has a lens of focal length 4 mm and photographs a person 1.7 m tall standing 2 m away.
(i) Find the image distance. (1 mark)
(ii) Find the magnification. (1 mark)
(iii) Find the height of the image on the sensor. (1 mark)
(iv) Find the power of the lens. (1 mark)
Hint
u = −2000 mm.
Answer
(i) v ≈ 4.008 mm, just beyond F. (ii) m ≈ −0.002. (iii) ≈ 3.4 mm, inverted. (iv) +250 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.Lens formula?Show answer
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ .
2.Lens maker's formula?Show answer
$\frac{1}{f} = (n - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$ .
3.f = 50 cm: power?Show answer
+2 D.
4.Powers of lenses in contact?Show answer
They add: $P = P_{1} + P_{2}$ .
5.Glass lens in a liquid of the same n?Show answer
It becomes invisible; f is infinite.

Lens Formula, Lens Maker's Formula and Power: frequently asked questions

Why does a lens's focal length increase in water?

Bending depends on the difference in refractive index between lens and surroundings. Glass and water differ less than glass and air, so the lens bends light less and f is larger.

Why are powers added instead of focal lengths?

Because the reciprocals \frac{1}{f} add for thin lenses in contact. Power is defined as \frac{1}{f}, so it adds directly.

Can I use the New Cartesian convention or another one?

Yes. The ISC syllabus allows any one sign convention, provided it is used consistently.

What does the ISC 2027 syllabus need here?

Refraction at a spherical surface (one case), derivation of the lens maker's formula and lens formula, magnification, power P = 1/f in dioptres, thin lenses in contact (derivation), lenses in a liquid, and lens-mirror combinations.

How was this asked recently?

ISC 2024 Q15 (3 marks, no-parallax experiment and power), ISC 2025 Q6 (2 marks, +2.0 D and −1.5 D in contact) and 2025 Q1(A)(vi) (lens in water).