--- title: "Principal points of achromatic lens pair" subject: "Engineering Physics" chapter: 5 key: "semester_1/engineering_physics:achromatic-lens-pair-principal-points" source: https://www.bctengineer.com/q/semester_1/engineering_physics/achromatic-lens-pair-principal-points site: BCT Engineering --- # Principal points of achromatic lens pair Engineering Physics · Chapter 5: Geometrical Optics - Asked in: 2069 Asar, 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-lens-pair-principal-points - Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-5 ## Question Two lenses of focal lengths 8cm and 4cm are placed at a certain distance apart. Calculate the position of principal points if they form an achromatic combination. ## Short answer For an achromatic pair of same material, separation x = (8+4)/2 = 6 cm. Equivalent focal length F = 5.33 cm. The first principal point is 8 cm right of the 8 cm lens (2 cm beyond the 4 cm lens); the second is 4 cm left of the 4 cm lens (2 cm right of the 8 cm lens). ## Answer For two lenses to form an achromatic combination with the same glass, the separation must be x = (f₁ + f₂)/2. **Given:** f₁ = 8 cm, f₂ = 4 cm. **Step 1: separation** ```text x = (8 + 4)/2 = 6 cm ``` **Step 2: equivalent focal length** ```text 1/F = 1/8 + 1/4 − 6/(8×4) = 0.125 + 0.25 − 0.1875 = 0.1875 F = 5.33 cm ``` **Step 3: principal points** ```text α = F x / f₂ = 5.33 × 6 / 4 = 8 cm (H₁ right of L1) β = F x / f₁ = 5.33 × 6 / 8 = 4 cm (H₂ left of L2) ``` ```text L1 H2 L2 H1 |------.-----------|----. 0 2.0 6.0 8.0 (cm from L1) ``` H₁ lies 8 − 6 = 2 cm beyond the second lens; H₂ lies 6 − 4 = 2 cm after the first lens, so the principal points are crossed. **Answer:** x = 6 cm, F = 5.33 cm; the first principal point is 8 cm to the right of the 8 cm lens (2 cm beyond the second lens); the second principal point is 4 cm to the left of the 4 cm lens (2 cm from the first lens). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For achromatism with f1 = 8, f2 = 4 cm (same glass), x is: A. 6 cm B. 4 cm C. 2 cm D. 12 cm Answer: A. x = (8 + 4)/2 = 6 cm. (Key point: Separation) 2. What is F? A. 2.67 cm B. 5.33 cm C. 4 cm D. 6 cm Answer: B. 1/F = 1/8 + 1/4 − 6/32 = 0.1875. (Key point: Equivalent focal length) 3. Where is H1, and what is unusual? A. At L2 B. 2 cm left of L1 C. 8 cm right of L1, 2 cm beyond L2, so the points are crossed D. 4 cm right of L1 Answer: C. α = 5.33 × 6 / 4 = 8 cm beyond 6 cm L2; H2 is at 2 cm from L1. (Key point: Crossed principal points) ## Related in Engineering Physics - [Focal lengths of achromatic lens pair](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Aachromatic-lenses-equivalent-thirty-values) — Engineering Physics, chapter 5 - [Principal points and image, two lenses](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Atwo-lenses-principal-points-image) — Engineering Physics, chapter 5 - [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 - [Principal points of 3 cm and 4 cm lenses](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Atwo-lenses-three-four-principal-points) — Engineering Physics, chapter 5 - [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 - [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×