PhysicsCBSE Class 12 · Ray Optics and Optical Instruments

Refraction at Spherical Surfaces and Lenses Class 12: Lens Maker's Formula

Q: What is the lens maker's formula used for?

It tells how the focal length depends on the refractive index of the lens material (relative to the surrounding medium) and on the radii of curvature of its two faces. Lens makers use it to grind lenses of a required focal length.

Q: Why does a lens's focal length change in water?

The lens maker's formula uses the relative index n_lens/n_medium. In water this ratio is closer to 1, so the lens bends light less and its focal length increases.

Q: What is the sign convention for R?

R is positive if the centre of curvature lies on the side where the light goes after refraction, and negative if it lies on the incident side. For a biconvex lens, R₁ > 0 and R₂ < 0.

Q: Why are lenses combined?

Combining lenses gives a required power, increases magnification in a compact system, and helps reduce image defects. For thin lenses in contact the powers simply add.

Q: What does a negative magnification mean?

The image is inverted, which for a single lens means it is real and formed on the opposite side from the object.

Updated 15 October 2026 · Fact-checked

At a spherical surface, n₂/v − n₁/u = (n₂ − n₁)/R. Applying it to both faces of a thin lens gives the lens maker's formula 1/f = (n₂₁ − 1)(1/R₁ − 1/R₂) and the thin lens formula 1/v − 1/u = 1/f. Power P = 1/f in dioptre; thin lenses in contact add powers.

Understand Refraction at Spherical Surfaces and Lenses

A glass lens is just two curved glass-air surfaces placed close together. If you know how one curved surface bends light, you can build the lens formulas step by step.

Sign convention (Cartesian). Measure all distances from the pole (or optical centre). Distances in the direction of the incident light are positive; against it, negative. Heights above the axis are positive. An object on the left (light travelling left to right) has negative u.

Refraction at one spherical surface. A surface of radius of curvature R separates media of indices $n_{1}$ (where the light comes from) and $n_{2}$ . Using Snell's law for small angles (paraxial rays) and the geometry of the surface, an object at distance u forms an image at v with $\frac{n _{2}}{v} - \frac{n _{1}}{u} = \frac{n _{2} - n _{1}}{R}$ . R is positive if the centre of curvature is on the side where light goes after refraction. This single formula covers convex and concave surfaces, real and virtual images.

Lens maker's formula. For a thin lens of index $n_{2}$ in a medium of index $n_{1}$ :

Surface 1 (radius $R_{1}$ ) forms an image $I_{1}$ of the object: $\frac{n _{2}}{v _{1}} - \frac{n _{1}}{u} = \frac{n _{2} - n _{1}}{R _{1}}$ .
$I_{1}$ acts as the object for surface 2 (radius $R_{2}$ ); the lens is thin, so the same distance $v_{1}$ is used: $\frac{n _{1}}{v} - \frac{n _{2}}{v _{1}} = \frac{n _{1} - n _{2}}{R _{2}}$ .
Adding: $\frac{n _{1}}{v} - \frac{n _{1}}{u} = (n_{2} - n_{1}) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$ .
For an object at infinity, v = f: $\frac{1}{f} = (n_{21} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$ , with $n_{21} = \frac{n _{2}}{n _{1}}$ . This is the lens maker's formula, and comparing the two results gives the thin lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ .

For a biconvex lens, $R_{1} > 0$ and $R_{2} < 0$ , so f > 0 (converging). In a liquid whose index exceeds the glass's, $n_{21} < 1$ and a convex lens diverges.

Magnification and power. Linear magnification $m = \frac{h ^{'}}{h} = \frac{v}{u}$ : negative for a real inverted image, positive for a virtual erect one. The power of a lens, $P = \frac{1}{f}$ (f in metres), measures its ability to converge light; its unit is the dioptre (D = m⁻¹). Converging lenses have positive power, diverging lenses negative.

Lenses in contact. Two thin lenses in contact: the first forms an image that is the object for the second. Adding their lens equations gives $\frac{1}{F} = \frac{1}{f _{1}} + \frac{1}{f _{2}}$ , i.e. $P = P_{1} + P_{2}$ , and the total magnification is $m = m_{1} m_{2}$ . Combinations let designers correct defects and make compact powerful systems.

Refraction at Spherical Surfaces and Lenses: formulas and key facts

Refraction at a spherical surface: $\frac{n _{2}}{v} - \frac{n _{1}}{u} = \frac{n _{2} - n _{1}}{R}$; Light goes from medium n₁ to n₂; Cartesian signs.
Lens maker's formula: $\frac{1}{f} = (n_{21} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$; n₂₁ = n_lens/n_medium.
Thin lens formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$; Same for convex and concave lenses with signs.
Magnification: $m = \frac{h ^{'}}{h} = \frac{v}{u}$; m < 0: real, inverted; m > 0: virtual, erect.
Power: $P = \frac{1}{f ( in m )}$; Unit dioptre (D).
Thin lenses in contact: $\frac{1}{F} = \frac{1}{f _{1}} + \frac{1}{f _{2}}, P = P_{1} + P_{2}, m = m_{1} m_{2}$; Use signs for diverging lenses.

How to solve Refraction at Spherical Surfaces and Lenses questions

Use this method for lens and spherical-surface numericals.

1Draw the axis with light going left to right; put signs on every distance: object distance u negative for a real object on the left.
2For a single surface, identify n₁ (incident side) and n₂, and the sign of R (positive if the centre lies on the outgoing side).
3For lenses, use the lens maker's formula with $n_{21} = \frac{n _{l e n s}}{n _{m e d i u m}}$ and signed $R_{1}$ , $R_{2}$ .
4Solve $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ for the unknown; then $m = \frac{v}{u}$ .
5For lenses in contact, add powers (or 1/f values) with signs.
6Interpret: v positive means a real image on the far side; v negative means a virtual image on the object side.
7Express power in dioptres with f in metres.

Refraction at Spherical Surfaces and Lenses 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.

Converging lens as a reading glass

Suppose a person's reading glasses have a power of +2.5 D, and they hold a book 25 cm from the lens.

(a) Find the focal length of the lens. (b) Where is the image of the print formed? (c) Find the magnification and describe the image.

Try it first, then show the solutionSolution

$f = \frac{1}{2.5}$ m = 40 cm (converging).
$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{40} + \frac{1}{- 25} = - \frac{3}{200}$ , so v ≈ −66.7 cm.
$m = \frac{v}{u} = \frac{- 66.7}{- 25} \approx 2.67$ : a virtual, erect image about 2.7 times larger, on the same side as the book.

Answer: (a) 40 cm; (b) about 66.7 cm from the lens on the book's side (virtual); (c) m ≈ 2.67, virtual, erect, enlarged.

Memory hook: Object inside f of a convex lens: a big, upright, virtual image.

Focusing a camera lens

Imagine a phone camera lens of focal length 4 mm, photographing a friend standing 2 m away.

(a) How far behind the lens must the sensor be? (b) How far must the lens move from its 'infinity' position to focus? (c) Why is such a tiny movement enough?

Try it first, then show the solutionSolution

$\frac{1}{v} = \frac{1}{4} + \frac{1}{- 2000}$ (in mm), giving v ≈ 4.008 mm.
For a distant object v = f = 4 mm, so the lens moves about 0.008 mm = 8 μm.
Because u is very large compared with f, v differs from f only slightly; tiny actuators can make such movements.

Answer: (a) about 4.008 mm; (b) about 8 μm; (c) the object is very far compared with f.

Memory hook: Far objects focus almost exactly at f.

Tricks to remember Refraction at Spherical Surfaces and Lenses

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

New over v minus old over u
n₂/v − n₁/u = (n₂ − n₁)/R: the medium the light goes into goes with v.
Use it when: Single surface formula.
Lens maker = (n − 1) × curvature difference
1/f = (n − 1)(1/R₁ − 1/R₂); for a symmetric biconvex lens it is 2(n − 1)/R.
Use it when: Quick focal length calculations.
Powers add, focal lengths don't
In contact: P = P₁ + P₂; never add f directly.
Use it when: Combination questions.
Minus m, real and down
Negative magnification: real, inverted; positive: virtual, upright.
Use it when: Describing images.

Refraction at Spherical Surfaces and Lenses: solved examples

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

Example 1: Image by a convex lens · 2 marks · Very short answerBoard level

An object is placed 36 cm in front of a convex lens of focal length 24 cm. Find the position, nature and magnification of the image.

Try it first, then show the solutionSolution

$\frac{1}{v} = \frac{1}{24} + \frac{1}{- 36} = \frac{1}{72}$ , so v = +72 cm. (1)
$m = \frac{v}{u} = \frac{72}{- 36} = - 2$ : real, inverted, twice the size, 72 cm on the other side. (1)

Answer: v = 72 cm, real and inverted, m = −2.

Example 2: Derivation: lens maker's formula · 5 marks · Long answerBoard level

Derive the lens maker's formula for a thin convex lens, starting from the formula for refraction at a single spherical surface. State the assumptions.

Try it first, then show the solutionSolution

Assumptions: thin lens, paraxial rays, small aperture; lens of index n₂ in a medium of index n₁; Cartesian sign convention. (1)
Surface 1 (radius R₁) forms an image at v₁: $\frac{n _{2}}{v _{1}} - \frac{n _{1}}{u} = \frac{n _{2} - n _{1}}{R _{1}}$ . (1)
This image is a virtual object for surface 2 (radius R₂), at the same distance since the lens is thin: $\frac{n _{1}}{v} - \frac{n _{2}}{v _{1}} = \frac{n _{1} - n _{2}}{R _{2}}$ . (1)
Adding: $\frac{n _{1}}{v} - \frac{n _{1}}{u} = (n_{2} - n_{1}) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$ . (1)
For u = −∞, v = f: $\frac{1}{f} = (n_{21} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$ , and hence $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ . (1)

Answer: $\frac{1}{f} = (n_{21} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})$ .

Example 3: Refraction at a curved glass surface · 3 marks · Short answerHarder board level

An object in air is 30 cm from the convex surface (radius of curvature 10 cm) of a long glass rod of refractive index 1.5. Find the position and nature of the image.

Try it first, then show the solutionSolution

n₁ = 1, n₂ = 1.5, u = −30 cm, R = +10 cm (centre inside the glass). (1)
$\frac{1.5}{v} - \frac{1}{- 30} = \frac{0.5}{10}$ , so $\frac{1.5}{v} = 0.05 - 0.0333 = 0.0167$ . (1)
v = +90 cm: a real image inside the glass, 90 cm from the surface. (1)

Answer: A real image 90 cm inside the rod.

Example 4: Lens in water · 3 marks · Short answerHarder board level

A biconvex lens of glass (n = 1.5) has both radii of curvature 30 cm. Find its focal length (a) in air and (b) in water (n = 4/3). (c) Would it converge in a liquid of n = 1.6?

Try it first, then show the solutionSolution

(a) $\frac{1}{f} = 0.5 (\frac{1}{30} + \frac{1}{30}) = \frac{1}{30}$ : f = 30 cm. (1)
(b) $n_{21} = \frac{1.5}{4/3} = 1.125$ : $\frac{1}{f} = 0.125 \times \frac{2}{30} = \frac{1}{120}$ , f = 120 cm. (1)
(c) $n_{21} = \frac{1.5}{1.6} < 1$ , so f is negative: the lens diverges. (1)

Answer: (a) 30 cm; (b) 120 cm; (c) no, it becomes diverging.

Example 5: Combination of lenses · 5 marks · Long answerHOTS / competency

A convex lens of focal length 20 cm is placed in contact with a concave lens of focal length 30 cm.
(a) Find the power and focal length of the combination. (2)
(b) An object is placed 40 cm from the combination. Find the image position and magnification. (2)
(c) Check (b) by letting each lens act in turn. (1)

Try it first, then show the solutionSolution

(a) $P = \frac{1}{0.2} + \frac{1}{- 0.3} = 5 - 3.33 = 1.67$ D. (1)
F = 60 cm (converging). (1)
(b) $\frac{1}{v} = \frac{1}{60} + \frac{1}{- 40} = - \frac{1}{120}$ : v = −120 cm (virtual, on the object side). (1)
$m = \frac{- 120}{- 40} = 3$ : virtual, erect, 3 times larger. (1)
(c) Convex lens alone: $\frac{1}{v _{1}} = \frac{1}{20} - \frac{1}{40}$ , v₁ = 40 cm. For the concave lens this is the object: $\frac{1}{v} = - \frac{1}{30} + \frac{1}{40} = - \frac{1}{120}$ , v = −120 cm. Same result. (1)

Answer: (a) about 1.67 D, F = 60 cm; (b) v = −120 cm, m = 3; (c) confirmed.

Example 6: Designing a plano-convex lens · 4 marks · Case-basedHarder board level

Read the passage and answer.

An optician needs a plano-convex lens of focal length 40 cm, made of glass of refractive index 1.5.
(a) Find the radius of curvature of the curved face. (1)
(b) Find its power. (1)
(c) It is placed in contact with a lens of power −0.5 D. Find the focal length of the combination. (1)
(d) Does the answer to (a) depend on which face meets the light first? (1)

Try it first, then show the solutionSolution

(a) For a plano-convex lens, $\frac{1}{f} = (n - 1) \frac{1}{R}$ : R = 0.5 × 40 = 20 cm. (1)
(b) $P = \frac{1}{0.4} = + 2.5$ D. (1)
(c) P = 2.5 − 0.5 = +2 D: F = 50 cm. (1)
(d) No: reversing a thin lens swaps and changes the signs of R₁ and R₂, but $\frac{1}{R _{1}} - \frac{1}{R _{2}}$ stays the same. (1)

Answer: (a) 20 cm; (b) +2.5 D; (c) 50 cm; (d) no.

Where students lose marks in Refraction at Spherical Surfaces and Lenses

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

Mistake: Dropping signs of u, v, R or f.
Using magnitudes as in Class 10 tricks.
Wrong working
1/v = 1/24 + 1/36
Right working
1/v = 1/24 + 1/(−36) = 1/72
Do this: Write every distance with its sign in the first line.
Usually costs: The whole numerical.
Mistake: Using R₂ = +20 cm for the second face of a biconvex lens.
Forgetting the second centre of curvature is on the incident side.
Do this: For a biconvex lens R₁ > 0 and R₂ < 0.
Usually costs: 1 to 2 marks.
Mistake: Adding focal lengths for lenses in contact.
Confusing with power.
Wrong working
F = 20 + (−30) = −10 cm
Right working
1/F = 1/20 − 1/30, F = 60 cm
Do this: Add powers or reciprocals.
Usually costs: Most of the question.
Mistake: Using n_glass instead of n_glass/n_medium in a liquid.
Forgetting the medium.
Do this: Use n₂₁ = n_lens/n_medium in the lens maker's formula.
Usually costs: 1 mark.
Mistake: Power in dioptres with f in cm.
Unit slip.
Do this: P = 100/f when f is in cm.
Usually costs: ½ to 1 mark.

Exam tips

State the sign convention and assumptions at the start of every derivation.
Lens maker's formula and lens numericals carried 3- to 5-mark questions in recent board papers (for example 2023 55/5/1 Q34, 2025 55/1/1 Q30 and 2026 55/1/1 Q32).
Show both surfaces separately in the lens maker's derivation and name the intermediate image.
Interpret the sign of v and m in words in the final line.
For combinations, check the answer by letting the lenses act one after the other.

Practice questions on Refraction at Spherical Surfaces and Lenses

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 40 cm is
(a) +2.5 D (b) −2.5 D (c) +0.4 D (d) −0.4 D
Hint
P = 1/f in metres, negative for diverging.
Answer
(b) −2.5 D.
Q2 · 2 marks · Very short answerBoard level
An object is 15 cm from a concave lens of focal length 10 cm. Find the image distance and magnification.
Hint
f = −10 cm.
Answer
$\frac{1}{v} = - \frac{1}{10} - \frac{1}{15} = - \frac{1}{6}$ : v = −6 cm; m = 0.4 (virtual, erect, diminished).
Q3 · 1 mark · Very short answerHarder board level
Find the focal length of a biconvex lens with radii 30 cm and 30 cm, made of glass of refractive index 1.6.
Hint
1/f = (n − 1)(2/R).
Answer
$\frac{1}{f} = 0.6 \times \frac{2}{30} = 0.04$ : f = 25 cm.
Q4 · 1 mark · Very short answerHarder board level
Lenses of power +4 D and −2.5 D are placed in contact. Find the power and focal length of the combination.
Hint
Powers add.
Answer
P = +1.5 D; F ≈ 66.7 cm (converging).
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): A convex glass lens can behave as a diverging lens when immersed in a suitable liquid.
Reason (R): If the liquid's refractive index exceeds that of the glass, the factor (n₂₁ − 1) in the lens maker's formula becomes negative.
(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
n₂₁ = n_lens/n_liquid.
Answer
(a).
Q6 · 3 marks · Short answerHOTS / competency
A small air bubble inside a glass sphere (n = 1.5, radius 6 cm) is 2 cm from the centre. Where does it appear when viewed from the side of the sphere nearer to it?
Hint
Light goes from glass to air; u = −4 cm from the near surface; R = −6 cm.
Answer
$\frac{1}{v} - \frac{1.5}{- 4} = \frac{1 - 1.5}{- 6}$ , so $\frac{1}{v} = 0.0833 - 0.375 = - 0.2917$ : v ≈ −3.43 cm. The bubble appears about 3.4 cm inside the surface (nearer than its real depth of 4 cm).
Q7 · 2 marks · Very short answerHOTS / competency
A thin biconvex lens of focal length f is cut into two halves (a) along its principal axis and (b) perpendicular to its principal axis (into two plano-convex halves). Find the focal length of each half in each case.
Hint
Which radii change?
Answer
(a) Each half keeps both curved faces: focal length f (only the brightness of the image falls). (b) Each half has one curved face and one flat face: $\frac{1}{f ^{'}} = (n - 1) \frac{1}{R}$ = half of $\frac{1}{f}$ , so f' = 2f.

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 refraction at a spherical surface?Show answer
n₂/v − n₁/u = (n₂ − n₁)/R.
2.Lens maker's formula?Show answer
1/f = (n₂₁ − 1)(1/R₁ − 1/R₂).
3.Thin lens formula?Show answer
1/v − 1/u = 1/f.
4.SI unit of power of a lens?Show answer
Dioptre (m⁻¹).
5.Power of two thin lenses in contact?Show answer
P = P₁ + P₂.
6.Magnification of a lens?Show answer
m = v/u.

Refraction at Spherical Surfaces and Lenses: frequently asked questions

What is the lens maker's formula used for?

It tells how the focal length depends on the refractive index of the lens material (relative to the surrounding medium) and on the radii of curvature of its two faces. Lens makers use it to grind lenses of a required focal length.

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

The lens maker's formula uses the relative index n_lens/n_medium. In water this ratio is closer to 1, so the lens bends light less and its focal length increases.

What is the sign convention for R?

R is positive if the centre of curvature lies on the side where the light goes after refraction, and negative if it lies on the incident side. For a biconvex lens, R₁ > 0 and R₂ < 0.

Why are lenses combined?

Combining lenses gives a required power, increases magnification in a compact system, and helps reduce image defects. For thin lenses in contact the powers simply add.

What does a negative magnification mean?

The image is inverted, which for a single lens means it is real and formed on the opposite side from the object.