---
title: "Cardinal points and equivalent focal length"
subject: "Engineering Physics"
chapter: 5
key: "semester_1/engineering_physics:cardinal-points-equivalent-focal-length"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/cardinal-points-equivalent-focal-length
site: BCT Engineering
---

# Cardinal points and equivalent focal length

Engineering Physics · Chapter 5: Geometrical Optics

- Asked in: 2069 Chaitra, 5 marks
- Video: [Cardinal points of two lens system](https://www.youtube.com/watch?v=G2jKYm_VX-A) — 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/cardinal-points-equivalent-focal-length
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-5

## Question

What are cardinal points of an optical system? Determine the equivalent focal length of a combination of two thin lenses separated by a finite distance.

## Short answer

Cardinal points are six points (focal, principal, nodal) on the axis of a coaxial optical system that simplify image formation. For two thin lenses of focal lengths f1, f2 separated by distance d, the equivalent focal length is F = f1 f2 / (f1 + f2 - d).

## Answer

#### Cardinal points
Cardinal points are six points on the axis of a coaxial optical system which let us find the image of any object without tracing rays through each surface.

1. **First and second focal points (F₁, F₂):** an object at F₁ gives parallel rays on emergence; parallel rays entering the system converge to F₂.
2. **First and second principal points (H₁, H₂):** the points where the principal planes cut the axis. The principal planes are conjugate planes with unit lateral magnification (+1).
3. **First and second nodal points (N₁, N₂):** a ray directed toward N₁ emerges parallel to its original direction as if from N₂. If the medium on both sides is the same (air), N₁ and N₂ coincide with H₁ and H₂.

#### Equivalent focal length of two thin lenses
Take lenses of focal lengths f₁, f₂ a distance *d* apart in air.

```text
 h1 ---|-------|---
       L1      L2      ray parallel to axis
       |<--d-->|       at heights h1 (L1), h2 (L2)
```
1. A ray parallel to the axis at height h₁ strikes L1. After L1 it converges toward the point at distance f₁ with slope −h₁/f₁.
2. At L2 (distance *d* away) its height is h₂ = h₁ − d·h₁/f₁ = h₁ (1 − d/f₁).
3. L2 deviates the ray further by h₂/f₂, so the final slope is
 −h₁/f₁ − h₂/f₂ = −h₁/f₁ − h₁(1 − d/f₁)/f₂.
4. For the equivalent lens of focal length F, the final slope is −h₁/F. Therefore

 1/F = 1/f₁ + 1/f₂ − d/(f₁f₂)

 **F = f₁f₂ / (f₁ + f₂ − d)**

Positions of the principal points follow from the same figure:
- H₁ is at α = F d / f₂ from L1 (on the side of L2).
- H₂ is at β = F d / f₁ from L2 (on the side of L1).

For d = 0 (lenses in contact) this reduces to 1/F = 1/f₁ + 1/f₂.

## Check yourself

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

1. What is special about the principal planes?
   A. Conjugate planes with unit lateral magnification (+1)
   B. They pass through the foci
   C. They are at infinity
   D. Magnification −1
   Answer: A. Principal planes are conjugate with m = +1. (Key point: Principal planes)

2. When do nodal points coincide with principal points?
   A. Never
   B. Always
   C. When the medium on both sides is the same
   D. Only for thick lenses
   Answer: C. Same medium on both sides (air). (Key point: Nodal points)

3. What is the equivalent focal length of two thin lenses a distance d apart?
   A. F = f1f2 / d
   B. 1/F = 1/f1 + 1/f2 − d/(f1f2)
   C. 1/F = 1/f1 + 1/f2 + d/(f1f2)
   D. F = f1 + f2 − d
   Answer: B. Derived from the slope of the ray after both lenses. (Key point: Formula)

4. H1 from L1 is:
   A. α = F f2 / d
   B. α = F d / f2
   C. α = d / F
   D. α = F d / f1
   Answer: B. H1 lies at F d/f2 from L1; H2 at F d/f1 from L2. (Key point: Principal point positions)

## Related in Engineering Physics

- [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
- [Ramsden's eyepiece cardinal points](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Aramsden-eyepiece-cardinal-points) — 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
- [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
- [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
- [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