---
title: "Principal points and image, two lenses"
subject: "Engineering Physics"
chapter: 5
key: "semester_1/engineering_physics:two-lenses-principal-points-image"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/two-lenses-principal-points-image
site: BCT Engineering
---

# Principal points and image, two lenses

Engineering Physics · Chapter 5: Geometrical Optics

- Asked in: 2067 Asar, 5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/two-lenses-principal-points-image
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-5

## Question

Two thin converging lenses of focal lengths 0.2m and 0.3m are placed coaxially 0.10m apart in air. An object is located 0.6m in front of the lens of smaller focal length. Find the position of the two principal points and that of the image.

## Short answer

Equivalent focal length F = 0.15 m. H1 is 0.05 m behind the first lens; H2 is 0.075 m in front of the second lens (0.025 m behind the first lens). The final image is 0.12 m beyond the second lens, real and inverted (magnification -0.3).

## Answer

**Given:** f₁ = 0.2 m (smaller), f₂ = 0.3 m, d = 0.10 m, object u = −0.6 m in front of the 0.2 m lens (first lens).

#### Equivalent focal length
```text
1/F = 1/0.2 + 1/0.3 − 0.1/(0.2×0.3)
    = 5 + 3.333 − 1.667 = 6.667 m⁻¹
F   = 0.15 m
```

#### Principal points
```text
α = F d / f₂ = 0.15 × 0.10 / 0.3 = 0.05 m   (H₁ is 5 cm right of L1)
β = F d / f₁ = 0.15 × 0.10 / 0.2 = 0.075 m  (H₂ is 7.5 cm left of L2)
```
H₂ is 0.10 − 0.075 = 0.025 m (2.5 cm) to the right of L1.

#### Position of the image
**Lens by lens:**
```text
Lens 1: 1/v₁ = 1/0.2 − 1/0.6 = 3.333  ->  v₁ = +0.30 m
Lens 2: object I₁ is 0.30 − 0.10 = 0.20 m beyond L2, u₂ = +0.20 m
        1/v₂ = 1/0.3 + 1/0.2 = 8.333  ->  v₂ = +0.12 m
m = (0.30/−0.6)(0.12/0.20) = (−0.5)(0.6) = −0.30
```
**Check by principal points:** U = −(0.6 + 0.05) = −0.65 m; 1/V = 1/0.15 − 1/0.65 gives V = 0.195 m from H₂, i.e. 0.195 − 0.075 = 0.12 m from L2. This agrees.

**Answer:** F = 0.15 m; H₁ is 0.05 m behind the first lens; H₂ is 0.075 m in front of the second lens (0.025 m behind the first lens); the final image is 0.12 m (12 cm) beyond the second lens, real and inverted (m = −0.3).

## Check yourself

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

1. What is F for 0.2 m and 0.3 m lenses 0.1 m apart?
   A. 0.5 m
   B. 0.12 m
   C. 0.15 m
   D. 0.25 m
   Answer: C. 1/F = 5 + 3.333 − 1.667 = 6.667. (Key point: F)

2. Where are the principal points?
   A. H1 10 cm
   B. H1 7.5 cm right; H2 5 cm left
   C. Both at L1
   D. H1 5 cm right of L1; H2 7.5 cm left of L2
   Answer: D. α = Fd/f2 = 0.05 m; β = Fd/f1 = 0.075 m. (Key point: H1 and H2)

3. For the first lens u = −0.6 m, v1 is:
   A. +0.60 m
   B. +0.15 m
   C. −0.30 m
   D. +0.30 m
   Answer: D. 1/v1 = 5 − 1.667 = 3.333. (Key point: Intermediate image)

4. The final image is:
   A. 0.3 m beyond L2
   B. 0.2 m beyond L2
   C. 0.12 m beyond L2, real, inverted
   D. 0.12 m before L2
   Answer: C. I1 is 0.2 m beyond L2; 1/v2 = 1/0.3 + 1/0.2 gives 0.12 m; m = −0.3. (Key point: Final image)

## Related in Engineering Physics

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