lens maker formula

Lens Maker Formula | Ray Optics Notes
Ray Optics · Complete Notes

Lens Maker Formula

From two spherical refracting surfaces to focal length, power, lenses in media, colour dependence, lens cutting, combinations and exam-ready numericals.

Class 12 PhysicsCBSENEETJEEIB · A-Level · AP
1

Thin Lens and Lens Maker Formula

A thin lens is bounded by two refracting surfaces whose separation is negligible compared with the object distance, image distance and radii of curvature.

Optical centre OL

A paraxial ray passing through the optical centre emerges nearly undeviated.

Radii R₁ and R₂

R₁ belongs to the first surface met by light; R₂ belongs to the second. Their signs follow the Cartesian convention.

Lens power

P = 1/f when f is in metres. A converging lens has positive power; a diverging lens has negative power.

Lens maker formula in a medium1/f = (μlens/μmedium − 1)(1/R₁ − 1/R₂)
What the formula says: focal length is determined by surface curvature and relative refractive index, not by the object position.
2

Convex Lens Maker Formula Derivation

A point object O is imaged successively by the two spherical surfaces. The first surface forms an intermediate image I′; I′ is the object for the second surface.

Two-surface construction

O I′ I P₁ P₂ C₁ C₂ A₁ A₂ R₁ R₂ normal A₁C₁normal A₂C₂ surface 1 alone would form I′ μ₁ → μ₂: bends towards normal μ₂ → μ₁: bends away from normal medium μ₁lens μ₂medium μ₁
Construction check: At A₁ the ray bends towards A₁C₁. At A₂ it bends away from A₂C₂. The dotted continuation of the first-surface ray locates I′, while the two actual emergent rays meet at final image I.

First surface: μ₁ → μ₂

−μ₁/u + μ₂/v′ = (μ₂ − μ₁)/R₁

u is measured from P₁; v′ is the intermediate-image distance from P₁.

Second surface: μ₂ → μ₁

−μ₂/v′ + μ₁/v = (μ₁ − μ₂)/R₂

For a thin lens, the distance of I′ from P₂ is taken as v′. Its sign is automatically handled by the Cartesian convention.

Add the equations

−μ₁/u + μ₂/v′ − μ₂/v′ + μ₁/v = (μ₂ − μ₁)/R₁ + (μ₁ − μ₂)/R₂

Cancel the intermediate-image terms

μ₁/v − μ₁/u = (μ₂ − μ₁)(1/R₁ − 1/R₂)

Object at infinity

For u → −∞, v = f:

1/f = (μ₂/μ₁ − 1)(1/R₁ − 1/R₂)
Lens in air: μ₁ = 1 and lens index = μ1/f = (μ − 1)(1/R₁ − 1/R₂)
3

Convex Lens Special Cases

Equiconvex lens

R₁ = +R, R₂ = −R

1/f = 2(μ − 1)/R
f = R/[2(μ − 1)]

f is positive because both curved surfaces contribute positive converging power.

Plano-convex

R₁ = +R, R₂ = ∞

1/f = (μ − 1)/R
f = R/(μ − 1)

Plano-convex reversed

R₁ = ∞, R₂ = −R

1/f = (μ − 1)/R

In air, reversing a thin plano-convex lens does not change its paraxial focal length.

Lens cut into two equal halves

CutGeometry after cuttingFocal length / powerAperture and brightness
Perpendicular to principal axisTwo plano-convex lensesFor an equiconvex lens, each half has f′ = 2f and P′ = P/2.Nearly same aperture; each isolated half has lower surface power.
Along principal axisTwo half-aperture lenses with both original curvaturesf′ = f and P′ = P.Aperture area halves, so image position/size stay the same but brightness decreases.
NEET/JEE checkpoint: a longitudinal half-lens forms a complete image, not half an image. Fewer rays reach the screen, so the complete image is dimmer.

Combination of thin lenses in contact

1/F = 1/f₁ + 1/f₂ + 1/f₃ + …    and    P = P₁ + P₂ + P₃ + …

This follows by applying the thin-lens equation successively with negligible separation between lenses.

4

Lens in a Medium

1/fm = (μlens/μmedium − 1)(1/R₁ − 1/R₂)

Lens in air

μmedium ≈ 1. Relative index is largest in common classroom situations, so the magnitude of power is comparatively high.

Glass lens in water

1/fw = (μg/μw − 1)K

The relative index decreases, so power decreases and |f| increases.

Matched indices

If μlens = μmedium, then P = 0 and f → ∞. The boundary becomes optically invisible.

Relative indexConvex shapeConcave shape
μlens/μmedium > 1Converging, f > 0Diverging, f < 0
= 1No powerNo power
< 1Diverging, f < 0Converging, f > 0
5

Effect of Colour and Wavelength

1/fv = (μv − 1)(1/R₁ − 1/R₂)
1/fr = (μr − 1)(1/R₁ − 1/R₂)

For normal dispersion, μv > μr, hence fv < fr. Violet focuses closer; red focuses farther.

Fᵥ Fᵣ Incident Light → Refracted Violet Ray Refracted Red Ray Direction of Light → red focuses farther
Chromatic aberration: a white object does not have one common focus because different wavelengths have different refractive indices. Achromatic combinations reduce this defect.
6

Concave Lens Maker Formula

O I′ I P₁ P₂ C₁ C₂ A₁ A₂ R₁ R₂ normal A₁C₁normal A₂C₂ μ₁ → μ₂: bends towards normal μ₂ → μ₁: bends away from normal dotted extensions locate I′ and I medium μ₁lens μ₂medium μ₁
Construction check: The first surface produces virtual I′. At the second surface the ray bends away from A₂C₂, so the emergent rays diverge. Their dotted backward extensions meet the axial ray at final virtual image I.

First surface

−μ₁/u + μ₂/v′ = (μ₂ − μ₁)/R₁

Second surface

−μ₂/v′ + μ₁/v = (μ₁ − μ₂)/R₂

Add and cancel v′

μ₁/v − μ₁/u = (μ₂ − μ₁)(1/R₁ − 1/R₂)

Object at infinity

1/f = (μ₂/μ₁ − 1)(1/R₁ − 1/R₂)
Equiconcave lens: R₁ = −R and R₂ = +R, so 1/f = −2(μ − 1)/R. Hence f is negative and the lens is diverging when μlens > μmedium.
7

Lens Behaviour in Different Media

Shape alone does not decide convergence. The sign of (μlens/μmedium − 1) reverses the optical behaviour.

A · μL > μm

FDirection of Light →Converges

Convex converges; concave diverges.

B · μL = μm

Direction of Light →No deviation

No refractive power; f → ∞.

C · μL < μm

FDirection of Light →Diverges

Convex diverges; concave converges.

Important: an air bubble of convex shape inside glass behaves optically like a concave lens because its refractive index is lower than that of the surrounding glass.
8

Applications

Spectacles

Select material and curvature to produce the prescribed positive or negative power.

Cameras

Multi-element groups combine powers while controlling aberrations and field curvature.

Microscopes

Short focal-length objectives require high curvature and carefully chosen optical glass.

Telescopes

Long-focus objectives reduce angular aberrations and form sharp distant images.

Human eye

The crystalline lens changes curvature to adjust optical power during accommodation.

Lens design

Designers choose R₁, R₂ and glass index to achieve a target focal length.

Material choice

High-index materials provide a given power with flatter, thinner surfaces.

Achromats

Crown and flint elements combine opposite dispersions to reduce colour fringing.

9

Solved Numericals

Each solution follows Given → Formula → Substitution → Calculation → Final Answer.

1[CBSE] An equiconvex glass lens has μ = 1.5 and radius 20 cm. Find its focal length in air.

View solution
GivenR₁ = +20 cm, R₂ = −20 cm, μ = 1.5.Formula1/f = (μ − 1)(1/R₁ − 1/R₂)Substitution1/f = 0.5(1/20 + 1/20)Calculation1/f = 1/20 cm⁻¹.Final Answerf = +20 cm.
Converging lens of power +5 D.

2[NEET] The lens in Question 1 is immersed in water of index 4/3. Find its new focal length.

View solution
Givenμrel = 1.5/(4/3) = 1.125; R₁ = +20 cm, R₂ = −20 cm.Formula1/fw = (μrel − 1)(1/R₁ − 1/R₂)Substitution1/fw = 0.125(2/20)Calculation1/fw = 1/80 cm⁻¹.Final Answerfw = +80 cm.
Focal length increases four times in water.

3[JEE Main] A plano-convex lens of index 1.6 has curved radius 30 cm. Find focal length and power.

View solution
Givenμ = 1.6, R = 30 cm.Formulaf = R/(μ − 1)Substitutionf = 30/0.6 cm.Calculationf = 50 cm = 0.50 m; P = 1/0.50.Final Answerf = +50 cm, P = +2 D.

4[JEE Advanced] An equiconvex lens of focal length 20 cm is cut perpendicular to its principal axis. Find the focal length and power of each half.

View solution
GivenOriginal f = 20 cm; cut creates two plano-convex lenses.FormulaOriginal power is due to two equal curved surfaces; each half retains one.SubstitutionP′ = P/2, hence 1/f′ = 1/(2f).Calculationf′ = 40 cm; P′ = 1/0.40 = 2.5 D.Final AnswerEach half has f′ = +40 cm and P′ = +2.5 D.

5[IB / A-Level] Two thin lenses of powers +5 D and −2 D are in contact. Find equivalent focal length.

View solution
GivenP₁ = +5 D, P₂ = −2 D.FormulaP = P₁ + P₂; F = 1/P.SubstitutionP = 5 − 2 = 3 D.CalculationF = 1/3 m = 0.333 m.Final AnswerF ≈ +33.3 cm.

6[AP Physics] A lens has fv = 19.0 cm and fr = 20.0 cm. Calculate longitudinal chromatic aberration.

View solution
Givenfv = 19.0 cm, fr = 20.0 cm.FormulaLongitudinal chromatic aberration = fr − fv.Substitution20.0 − 19.0 cm.Calculation1.0 cm.Final AnswerRed and violet foci are separated by 1.0 cm.
10

PYQs and Exam Practice

CBSEICSE

Board style

  1. Derive the lens maker formula for a thin lens in a medium.
  2. Explain why a convex lens has a longer focal length in water.
  3. Compare lateral and longitudinal cutting of a lens.
NEETJEE Main

Objective style

  1. Find power after immersion in a liquid.
  2. Identify when a convex lens becomes diverging.
  3. Calculate equivalent power of lenses in contact.
JEE Advanced

Advanced style

  1. Combine cutting, immersion and lens-contact concepts.
  2. Find an unknown liquid index for zero lens power.
  3. Compare red and violet image shifts quantitatively.
IBIGCSE

Explanation style

  1. Discuss assumptions in the thin-lens derivation.
  2. Explain chromatic aberration from dispersion data.
  3. Design an experiment to measure focal length in water.
A-Level

Structured problems

  1. Use two surface equations to eliminate I′.
  2. Interpret signs of R₁ and R₂.
  3. Evaluate uncertainty in calculated focal length.
AP Physics

Modelling style

  1. Graph power against relative refractive index.
  2. Predict focal shift with wavelength.
  3. Justify an achromatic two-lens design.
11

Case Studies

1 · Lens in water

Passage: A glass lens is moved from air to water without changing its shape.

MCQ: Its power (A) rises (B) falls (C) stays same. Answer: B.

Assertion–Reason: Focal length increases because relative index approaches unity. Both true; reason explains assertion.

2 · Spectacles

Passage: A myopic eye needs −2.5 D correction.

MCQ: Lens type? Concave. Focal length? −0.40 m.

A–R: Negative power diverges incoming rays. Both true.

3 · Camera design

Passage: A designer uses high-index glass to reduce lens curvature.

MCQ: For fixed power, higher μ permits (A) flatter surfaces (B) steeper surfaces. Answer: A.

Solution: The index factor rises, so the curvature factor may fall.

4 · Chromatic aberration

Passage: Violet and red rays focus at different axial points.

MCQ: Which focuses nearer? Violet.

A–R: μv > μr, so Pv > Pr. Both true.

5 · Lens combination

Passage: A +4 D lens touches a −1 D lens.

MCQ: Net power? +3 D.

Solution: Powers add algebraically for thin lenses in contact.

6 · Lens cutting

Passage: A lens is cut along a plane containing its principal axis.

MCQ: Focal length? Unchanged. Brightness? Reduced.

A–R: Both curvatures remain, but aperture decreases. Both true.

Σ

Complete Formula Sheet

Lens in medium1/f = (μL/μm − 1)(1/R₁ − 1/R₂)
Lens in air1/f = (μ − 1)(1/R₁ − 1/R₂)
PowerP = 1/f(m)   dioptre
Lenses in contact1/F = Σ(1/fᵢ),   P = ΣPᵢ
Equiconvexf = R/[2(μ − 1)]
Equiconcavef = −R/[2(μ − 1)]
✓

Sign Convention, Comparisons and Exam Tips

Sign convention

  • Real object: u < 0
  • Real image: v > 0
  • Centre right of surface: R > 0
  • Centre left of surface: R < 0
  • Measure from the relevant pole

Convex vs concave

  • In air: convex f > 0
  • In air: concave f < 0
  • Index reversal reverses behaviour
  • Equal indices give zero power
  • Shape is not the only criterion

Common mistakes

  • Using absolute radii without signs
  • Using μlens instead of relative index
  • Taking power in cm⁻¹
  • Assuming half lens makes half image
  • Forgetting violet has shorter f
Quick revision: write the relative index first, assign R₁ and R₂ in the direction light travels, calculate curvature factor separately, and convert focal length to metres before finding power.
One-to-One Physics Guidance

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Lens Maker Formula · Ray Optics Notes · Kumar Physics Classes
Lens Maker Formula – Fully Solved Numerical Problems
Ray Optics · Jaipur/Jodhpur Coaching Edition

Lens Maker Formula – Fully Solved Numerical Problems

Thirteen exam-focused problems with explicit substitutions, sign convention, medium effects, cutting cases and professional SVG ray geometry.

CBSENEETJEE MainJEE AdvancedLens in Medium13 Solved Problems
1

Equiconcave lens: find refractive index and convergence condition

Concept + Medium Reversal

Question

The magnitude of focal length of an equiconcave lens is 3/4 times the radius of either surface. Find μ. When will it converge?

Given Data

R₁ = −R, R₂ = +R, |f| = 3R/4. In air an equiconcave lens has f < 0.

Formula Used

1/f = (μ − 1)(1/R₁ − 1/R₂) = −2(μ − 1)/R

Step-by-Step Solution

  1. |f| = R/[2(μ − 1)]
  2. 3R/4 = R/[2(μ − 1)]
  3. 3/4 = 1/[2(μ − 1)]
  4. 6(μ − 1) = 4 ⇒ μ = 5/3.
μ = 5/3 ≈ 1.667. It converges when μmedium > μlens.
Concept Box: A concave shape becomes converging if its relative index μlens/μmedium is less than 1.
Common Mistake: Writing f = +3R/4 for an equiconcave lens in air. The given ratio is a magnitude.
Exam Tip: Decide the sign of f from lens behaviour before substituting.

Equiconcave geometry and sign convention

O F₁F₂ C₁C₂ R₁ = −RR₂ = +R Incident rays →diverges in air
2

Range of μ when |f| of an equiconvex lens exceeds R

Inequality

Question

Write the lens maker formula and obtain the range of μ for which an equiconvex lens in air has focal length greater than its radius in magnitude.

Given Data

R₁ = +R, R₂ = −R, lens in air, |f| > R. For ordinary glass μ > 1.

Formula Used

f = R/[2(μ − 1)]

Step-by-Step Solution

  1. R/[2(μ − 1)] > R
  2. 1/[2(μ − 1)] > 1
  3. 2(μ − 1) < 1
  4. μ < 3/2. Together with μ > 1:
1 < μ < 3/2
Concept Box: A smaller refractive-index contrast produces weaker power and therefore a longer focal length.
Common Mistake: Forgetting the physical lower bound μ > 1 for a lens in air.
Exam Tip: Cancel positive R only after stating R > 0 as a magnitude.

Equiconvex lens: long focal length

OF₂F₁C₁C₂R₁=+RR₂=−Rf > R
3

Biconvex lens with f = R/2

Direct Lens Maker

Question

A biconvex lens has a focal length equal to half the radius of curvature of either surface. Find the refractive index of the material of the lens.

Given Data

R₁=+R, R₂=−R, f=R/2.

Formula Used

1/f = 2(μ−1)/R

Step-by-Step Solution

  1. 2/R = 2(μ−1)/R
  2. μ−1=1
  3. μ=2.
Refractive index μ = 2.00
Concept Box: High μ gives high power for unchanged curvature.
Common Mistake: Using R₂=+R for a biconvex lens.
Exam Tip: For an equiconvex lens, memorize f=R/[2(μ−1)] only after understanding signs.

Biconvex geometry

OF₂F₁C₁C₂+R−R
4

Reconstructed Figure 9.78: thin lens with C₁ and C₂

Figure Assumption Declared

Question

Figure 9.78 shows a thin lens with centres of curvature C₁ and C₂. Find its focal length. Take μ = 1.5.

Important: The original Figure 9.78 was not supplied. The diagram below is a transparent reconstruction using R₁=+20 cm and R₂=−10 cm. For any other figure values, use the general expression shown.

Given Data

μ=1.5, reconstructed R₁=+20 cm, R₂=−10 cm.

Formula Used

1/f=(1.5−1)(1/R₁−1/R₂)

Step-by-Step Solution

  1. 1/f=0.5[1/20−1/(−10)]
  2. 1/f=0.5(1/20+2/20)=3/40
  3. f=40/3 cm.
For the reconstructed figure: f = +13.33 cm
General answer: f = 1/{0.5(1/R₁−1/R₂)}
Concept Box: C₁ and C₂ determine the signs and magnitudes of R₁ and R₂.
Common Mistake: Claiming a unique number when a referenced figure is missing.
Exam Tip: Read radius from each pole to its own centre, not from the optical centre.

Reconstructed Figure 9.78

OF₂F₁C₁C₂R₁=+20 cmR₂=−10 cmIncident rays →
5

Converging lens moved from air to liquid μ=1.3

Immersion

Question

A converging lens has a focal length of 20 cm in air. It is made of a material of refractive index 1.6. If it is immersed in a liquid of refractive index 1.3, what will be its new focal length?

Given Data

fair=20 cm, μL=1.6, μm=1.3.

Formula Used

Pm/Pair=[μL/μm−1]/(μL−1)

Step-by-Step Solution

  1. μrel=1.6/1.3=16/13
  2. μrel−1=3/13
  3. fm=fair(0.6)/(3/13)
  4. fm=20×2.6=52 cm.
New focal length = +52 cm
Concept Box: Liquid reduces index contrast, so the lens remains converging but becomes weaker.
Common Mistake: Using μL−μm instead of μL/μm−1.
Exam Tip: Use a power ratio when radii are unchanged.

Before and after immersion

OOF airF liquidold FC₁C₂Air: f=20 cmLiquid μ=1.3: f=52 cmμ lens=1.6
6

Plano-convex lens: radius 15 cm

Plano-Convex

Question

A plano-convex lens (μ = 1.5) has a curved surface of radius 15 cm. What is its focal length?

Given Data

μ=1.5, R₁=+15 cm, R₂=∞.

Formula Used

1/f=(μ−1)(1/R−0)

Step-by-Step Solution

  1. 1/f=0.5/15=1/30 cm⁻¹
  2. f=+30 cm.
f = +30 cm
Concept Box: A plane surface has zero curvature and contributes no optical power.
Common Mistake: Writing R₂=0; a plane has R₂=∞.
Exam Tip: Reversing a thin plano-convex lens in air keeps the same focal length.

Plano-convex lens

OF₂F₁C₁C₂ at ∞R₁=+15 cmR₂=∞
7

Plano-convex lens: find radius from f=18 cm

Inverse Problem

Question

A plano-convex lens (μ = 1.5) has a focal length of 18 cm in air. Calculate the radius of curvature of the spherical surface.

Given Data

μ=1.5, f=18 cm, one surface plane.

Formula Used

R=f(μ−1)

Step-by-Step Solution

  1. R=18×(1.5−1)
  2. R=18×0.5=9 cm.
Radius of curved surface = 9 cm
Concept Box: For a plano-convex lens, f is twice R when μ=1.5.
Common Mistake: Dividing by μ−1 instead of multiplying when solving for R.
Exam Tip: Check dimensions: R and f must both be lengths.

Radius recovered from focal length

OF₂F₁C₁C₂ at ∞R₁=+9 cmR₂=∞f=18 cm
8

Concavo-convex lens moved into water

Meniscus + Immersion

Question

The focal length of a concavo-convex lens of radii of curvature 5 cm and 10 cm is 20 cm. What will be its focal length in water? Given μwater = 4/3.

Given Data

Converging meniscus: R₁=+5 cm, R₂=+10 cm, fair=20 cm, μw=4/3.

Formula Used

1/20=(μ−1)(1/5−1/10)=(μ−1)/10 ⇒ μ=1.5

Step-by-Step Solution

  1. μrel=1.5/(4/3)=1.125
  2. 1/fw=0.125(1/5−1/10)
  3. 1/fw=0.0125 cm⁻¹
  4. fw=80 cm.
Focal length in water = +80 cm
Concept Box: The meniscus remains converging because its convex surface has greater curvature.
Common Mistake: Giving opposite signs to both radii of a meniscus whose centres lie on the same side.
Exam Tip: Draw the meniscus and mark C₁/C₂ before assigning signs.

Concavo-convex lens in water

OF₂F₁C₁C₂R₁=+5R₂=+10Water μ=4/3Lens μ=1.5air focuswater focus shifts away
9

Convex lens of index 1.5 in three liquids

Three Behaviour Regimes

Question

A convex lens of focal length f and refractive index 1.5 is immersed in a liquid of refractive index:
(i) 1.6
(ii) 1.3
(iii) 1.5
What changes happen to the focal length of the lens in the three cases?

Given Data

fair=f, μL=1.5. Therefore curvature factor K=2/f.

Formula Used

1/f′=(μL/μm−1)K, where K=2/f

Step-by-Step Solution

  1. μm=1.6: μrel−1=15/16−1=−1/16 ⇒ 1/f′=(−1/16)(2/f)=−1/(8f) ⇒ f′=−8f.
  2. μm=1.3: μrel−1=15/13−1=2/13 ⇒ 1/f′=4/(13f) ⇒ f′=13f/4.
  3. μm=1.5: μrel=1 ⇒ power=0 ⇒ f′=∞.
(i) −8f   (ii) 13f/4   (iii) ∞
Concept Box: Shape does not fix behaviour; relative refractive index does.
Common Mistake: Assuming a convex lens always converges.
Exam Tip: Compare μL and μm before calculating.

(i) μm=1.6: diverging

OF₁F₂C₁C₂f′=−8f

(ii) μm=1.3: weak convergence

OF₂F₁C₁C₂f′=13f/4

(iii) μm=1.5: no power

OF₁ at ∞F₂ at ∞C₁C₂f′=∞
10

Plano-convex lens: find R, then immerse in μ=1.6 liquid

Index Reversal

Question

The focal length of a plano-convex lens is 20 cm in air. Refractive index of glass is 1.5. Calculate:
(i) the radius of curvature of the lens surface
(ii) its focal length when immersed in liquid of refractive index 1.6.

Given Data

fair=20 cm, μL=1.5, μm=1.6.

Formula Used

R=f(μ−1);   1/fm=(μL/μm−1)(1/R)

Step-by-Step Solution

  1. R=f(μ−1)=20×0.5=10 cm.
  2. μrel=1.5/1.6=0.9375.
  3. 1/fm=(−0.0625)(1/10).
  4. 1/fm=−1/160 cm⁻¹, so fm=−160 cm.
R = 10 cm; focal length in liquid = −160 cm
Concept Box: The convex lens becomes diverging because the liquid has higher refractive index.
Common Mistake: Reporting +160 cm despite relative index below unity.
Exam Tip: Predict the sign before calculating the magnitude.

Plano-convex lens: convergence changes to divergence

OOF₂F₁C₁C₂∞C₁C₂∞Air: f=+20 cmLiquid: f=−160 cmμm=1.6 > μL=1.5
11

Glass convex lens: focal length in water

Power Ratio

Question

The focal length of a glass convex lens in air is 15 cm. Calculate its focal length when it is totally immersed in water. Given μwater = 4/3 and μglass = 1.5.

Given Data

fair=15 cm, μg=1.5, μw=4/3.

Formula Used

fw/fair=(μg−1)/(μg/μw−1)

Step-by-Step Solution

  1. μrel=1.5/(4/3)=1.125
  2. fw/15=0.5/0.125=4
  3. fw=60 cm.
Focal length in water = +60 cm
Concept Box: The focal length becomes four times because the refractive-index factor becomes one-fourth.
Common Mistake: Using water index as the lens index.
Exam Tip: Curvature factor cancels in the ratio method.

Focus shift from 15 cm to 60 cm

OOF airF waterC₁C₂C₁C₂Air: 15 cmWater: 60 cm
12

Equiconvex lens and cuts along AB and CD

Lens Cutting

Question

The radius of curvature of each surface of a convex lens is 20 cm and the refractive index of the material of the lens is 3/2.
(i) Calculate its focal length.
(ii) If this lens is cut along the plane AB, what will be the focal length of each of the two halves?
(iii) What happens if the lens is cut along CD?

Given Data

R₁=+20 cm, R₂=−20 cm, μ=3/2. Principal axis is horizontal. AB is the vertical cut perpendicular to the axis; CD is the horizontal cut along the axis.

Formula Used

1/f=(μ−1)(1/R₁−1/R₂)

Step-by-Step Solution

  1. 1/f=0.5(1/20+1/20)=1/20, so f=20 cm.
  2. Vertical cut AB: each half becomes plano-convex; one curved surface remains. Power halves, so f′=2f=40 cm.
  3. Horizontal cut CD: both curvatures remain. Focal length stays 20 cm; aperture and brightness reduce.
Original f=20 cm; AB cut: each half f=40 cm; CD cut: each half f=20 cm
Concept Box: Focal length depends on curvature, while brightness depends strongly on aperture.
Common Mistake: Saying a longitudinal half-lens forms half an image.
Exam Tip: A cut perpendicular to the axis changes a curved surface into a plane surface.

Original lens

OF₂F₁C₁C₂R₁=+20, R₂=−20f=20 cm

Vertical cut AB

ABOF₂F₁C₁C₂∞f=40 cm

Horizontal cut CD

CDOF₂F₁C₁C₂f=20 cm, dimmer
13

Equiconvex lens of μ=1.5 immersed in μ=1.7 liquid

Converging to Diverging

Question

A converging lens of refractive index 1.5 and focal length 15 cm in air has the same radii of curvature for both sides. If it is immersed in a liquid of refractive index 1.7, calculate the focal length of the lens in the liquid.

Given Data

fair=15 cm, μL=1.5, equal radii, μm=1.7.

Formula Used

K=1/[fair(μL−1)];   1/fm=(μL/μm−1)K

Step-by-Step Solution

  1. K=1/(15×0.5)=2/15 cm⁻¹.
  2. μrel=1.5/1.7=15/17.
  3. μrel−1=−2/17.
  4. 1/fm=(−2/17)(2/15)=−4/255.
  5. fm=−255/4=−63.75 cm.
Focal length in liquid = −63.75 cm
Concept Box: The physical convex shape becomes a diverging optical lens in a higher-index liquid.
Common Mistake: Keeping the focal length positive because the lens is convex.
Exam Tip: A negative final answer is the expected physical result when μm>μL.

Before and after immersion

OOF₂F₁C₁C₂C₁C₂Air: +15 cmLiquid: −63.75 cmμm=1.7, μL=1.5

How Refractive Index of Surrounding Medium Changes Lens Behaviour

ConditionRelative index factorConvex shapeConcave shapeFocal-length effect
μL>μmPositiveConvergingDivergingUsual signs
μL=μmZeroNo powerNo powerf→∞
μL<μmNegativeDivergingConvergingSigns reverse

Final Answer Summary

QTopicFinal Answer
1Equiconcave indexμ=5/3; converges if μmedium>μlens
2Range of μ1<μ<3/2
3Biconvexμ=2
4Reconstructed Fig. 9.78f=13.33 cm for R₁=20 cm, R₂=−10 cm
5Lens in μ=1.3+52 cm
6Plano-convex+30 cm
7Find radius9 cm
8Meniscus in water+80 cm
9Three liquids−8f; 13f/4; ∞
10Plano-convex immersionR=10 cm; f=−160 cm
11Glass lens in water+60 cm
12Lens cutting20 cm; AB→40 cm; CD→20 cm
13Lens in μ=1.7−63.75 cm

Formula Sheet

Lens maker in medium1/f=(μ₂/μ₁−1)(1/R₁−1/R₂)
Lens maker in air1/f=(μ−1)(1/R₁−1/R₂)
Equiconvexf=R/[2(μ−1)]
Equiconcavef=−R/[2(μ−1)]
Plano-convexf=R/(μ−1)
Medium power ratioPm/Pair=(μL/μm−1)/(μL−1)
PowerP=1/f(m) dioptre
Sign conventionR positive if centre lies in direction of incident light
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