--- title: "Ramsden's eyepiece cardinal points" subject: "Engineering Physics" chapter: 5 key: "semester_1/engineering_physics:ramsden-eyepiece-cardinal-points" source: https://www.bctengineer.com/q/semester_1/engineering_physics/ramsden-eyepiece-cardinal-points site: BCT Engineering --- # Ramsden's eyepiece cardinal points Engineering Physics · Chapter 5: Geometrical Optics - Asked in: 2074 Asoj, 5 marks - Video: [Ramsden Eyepiece](https://www.youtube.com/watch?v=ME_rt5Mu-ic) — Prof.Chishty Syed Qadeeruddin - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/engineering_physics/ramsden-eyepiece-cardinal-points - Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-5 ## Question In Ramsden's eyepiece a coaxial lens system is used. There are two lenses in air and are of equal focal length f separated by a distance 2f/3. Find positions of the cardinal points. ## Short answer F = 3f/4. H₁ is f/2 right of L1, H₂ is f/2 left of L2 (f/6 right of L1). F₁ is f/4 left of L1, F₂ is f/4 right of L2; nodal points coincide with principal points. ## Answer Take two thin lenses of equal focal length f separated by d = 2f/3, with L1 the field lens (facing the object) and L2 the eye lens. **Formulas** for two thin lenses in air (distances positive to the right): - Equivalent focal length: 1/F = 1/f₁ + 1/f₂ − d/(f₁f₂) - First principal point H₁ from L1: α = F d / f₂ (to the right of L1) - Second principal point H₂ from L2: β = F d / f₁ (to the left of L2) - Nodal points coincide with principal points (same medium on both sides). **Step 1: equivalent focal length** ```text 1/F = 1/f + 1/f − (2f/3)/f² = 2/f − 2/(3f) = 4/(3f) F = 3f/4 ``` **Step 2: principal points** ```text α = F d / f₂ = (3f/4)(2f/3)/f = f/2 H₁ is f/2 right of L1 β = F d / f₁ = f/2 H₂ is f/2 left of L2 ``` Since L2 is 2f/3 from L1, H₂ is at 2f/3 − f/2 = f/6 right of L1. **Step 3: focal points** ```text F₂ (second focal point) from L2 = F − β = 3f/4 − f/2 = f/4 (right of L2) F₁ (first focal point) from L1 = α − F = f/2 − 3f/4 = −f/4 (f/4 left of L1) ``` ```text F1 L1 H2 H1 L2 F2 |-----|-----|---------|------|------| -f/4 0 f/6 f/2 2f/3 11f/12 (distances from L1) ``` The focal points lie **outside** the lens system, f/4 from each lens, which is the reason a Ramsden eyepiece can be used with a cross-wire or scale placed at the first focal point. **Answer:** F = 3f/4; H₁ is f/2 behind L1; H₂ is f/2 in front of L2 (f/6 behind L1); F₁ is f/4 in front of L1; F₂ is f/4 behind L2; nodal points coincide with H₁ and H₂. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For two lenses of focal length f separated by 2f/3, what is F? A. f/2 B. 3f/4 C. 2f/3 D. 4f/3 Answer: B. 1/F = 2/f − 2/(3f) = 4/(3f), so F = 3f/4. (Key point: Equivalent focal length) 2. Where is H1 measured from L1? A. f/6 to the right B. f/4 to the left C. 2f/3 to the right D. f/2 to the right of L1 Answer: D. α = F d / f2 = (3f/4)(2f/3)/f = f/2. (Key point: Principal point H1) 3. Where are the focal points relative to the lenses? A. Inside the system between the lenses B. Outside, f/4 from each lens C. At the lenses D. At infinity Answer: B. F1 is f/4 left of L1 and F2 is f/4 right of L2, which lets a cross-wire be placed at F1. (Key point: Focal points) 4. Where are the nodal points? A. At the focal points B. Coincide with the principal points since the medium on both sides is air C. At the lens centres D. At infinity Answer: B. With the same medium on both sides, nodal points coincide with principal points. 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