---
title: "Focal lengths of achromatic lens pair"
subject: "Engineering Physics"
chapter: 5
key: "semester_1/engineering_physics:achromatic-lenses-equivalent-thirty-values"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/achromatic-lenses-equivalent-thirty-values
site: BCT Engineering
---

# Focal lengths of achromatic lens pair

Engineering Physics · Chapter 5: Geometrical Optics

- Asked in: 2071 Chaitra, 5 marks
- Video: [Achromatism / Achromatic Lens (Geometrical Optics Lecture :34) by Anshu Kapoor - for B.Sc, B.Tech](https://www.youtube.com/watch?v=YoPmK8aAcpg) — Anshu Kapoor
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/achromatic-lenses-equivalent-thirty-values
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-5

## Question

Two thin lenses of focal length f1 and f2 separated by a distance 'x' have an equivalent focal length 30 cm and both lenses are of same material. The combination satisfies the condition of achromatism and minimum spherical aberration. Find the values of f1 and f2.

## Short answer

Achromatism and minimum spherical aberration for identical materials give x = (f1+f2)/2 = f1 − f2, so f1 = 3f2. Equivalent focal length F = 1.5 f2, hence f2 = 20 cm and f1 = 60 cm (separation x = 40 cm).

## Answer

**Conditions** for two lenses of the same material separated by *x*:

1. Achromatism: x = (f₁ + f₂) / 2
2. Minimum spherical aberration: x = f₁ − f₂ (taking f₁ > f₂)

**Step 1: relation between f₁ and f₂**
```text
(f₁ + f₂)/2 = f₁ − f₂   ->   f₁ = 3 f₂
```

**Step 2: equivalent focal length**
```text
F = f₁f₂ / (f₁ + f₂ − x)
  = f₁f₂ / ((f₁ + f₂)/2)
  = 2 f₁ f₂ / (f₁ + f₂)
  = 2 (3f₂)(f₂) / (4 f₂)
  = 1.5 f₂
```

**Step 3: values for F = 30 cm**
```text
f₂ = 30 / 1.5 = 20 cm
f₁ = 3 × 20  = 60 cm
x  = (60 + 20)/2 = 40 cm
```
**Check:** F = 2 × 60 × 20 / 80 = 30 cm; x = f₁ − f₂ = 40 cm.

**Answer: f₁ = 60 cm, f₂ = 20 cm (separation x = 40 cm).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which two conditions are combined?
   A. x = 0 and x = f1
   B. x = f1 + f2 and x = f1 f2
   C. x = (f1 + f2)/2 and x = f1 − f2
   D. x = f1/2 and x = f2/2
   Answer: C. Achromatism and minimum spherical aberration together. (Key point: Two conditions)

2. What does equating them give?
   A. f1 = 4f2
   B. f1 = 3f2
   C. f1 = f2/3
   D. f1 = 2f2
   Answer: B. (f1 + f2)/2 = f1 − f2 gives f1 = 3f2. (Key point: Ratio)

3. For F = 30 cm, what are f1 and f2?
   A. 60 cm and 20 cm
   B. 90 cm and 30 cm
   C. 45 cm and 15 cm
   D. 30 cm and 10 cm
   Answer: A. F = 1.5 f2 = 30 gives f2 = 20 cm and f1 = 60 cm; x = 40 cm. (Key point: Values)

## Related in Engineering Physics

- [Separation of two achromatic lenses](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Atwo-lenses-equivalent-focal-fifty) — Engineering Physics, chapter 5
- [Condition for achromatism of two lenses](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Aachromatism-condition-dispersive-powers) — Engineering Physics, chapter 5
- [Principal points of achromatic lens pair](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Aachromatic-lens-pair-principal-points) — Engineering Physics, chapter 5
- [Image by two lenses 30 cm and 40 cm](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Atwo-lenses-thirty-forty-image) — Engineering Physics, chapter 5, asked 3×
- [Equivalent focal length and principal points](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Atwo-lenses-ten-four-principal) — Engineering Physics, chapter 5
- [Cardinal points and equivalent focal length](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Acardinal-points-equivalent-focal-length) — Engineering Physics, chapter 5

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