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

# Principal points, object for infinity image

Engineering Physics · Chapter 5: Geometrical Optics

- Asked in: 2070 Asar, 5 marks
- Video: [Coaxial lens system: Cardinal  points (object  at infinity)](https://www.youtube.com/watch?v=Bi0giRVpdcU) — Dr.Shailendra Kolhe
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/engineering_physics/two-identical-lenses-infinity-object
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-5

## Question

Two thin identical convex lenses of focal length 8 cm each are coaxial and 4 cm apart. Find the principal points and the position of object for which image is formed at infinity.

## Short answer

Equivalent focal length F = 5.33 cm. H1 is 2.67 cm right of the first lens, H2 is 2.67 cm left of the second lens. For an image at infinity, the object must be placed at the first focal point, 2.67 cm in front of the first lens.

## Answer

**Given:** f₁ = f₂ = 8 cm, d = 4 cm.

#### Equivalent focal length
```text
1/F = 1/8 + 1/8 − 4/(8×8) = 1/4 − 1/16 = 3/16
F   = 16/3 = 5.33 cm
```

#### Principal points
```text
α = F d / f₂ = 5.33 × 4 / 8 = 2.67 cm  (H₁: 2.67 cm right of L1)
β = F d / f₁ = 5.33 × 4 / 8 = 2.67 cm  (H₂: 2.67 cm left of L2)
```
H₂ is 4 − 2.67 = 1.33 cm to the right of L1, so H₂ is nearer to L1 than H₁ is.

```text
 L1        H2  H1          L2
 |---------.---.-----------|
 0        1.33  2.67       4.0   (cm from L1)
```

#### Object for an image at infinity
An image at infinity is formed when the object is at the **first focal point** F₁ of the system.
```text
F₁ from L1 = α − F = 2.67 − 5.33 = −2.67 cm
```
So the object must be **2.67 cm in front of the first lens**.

**Check:** for L1, u = −2.67 cm: 1/v₁ = 1/8 − 1/2.67 = −1/4, v₁ = −4 cm. This image is 4 + 4 = 8 cm to the left of L2, which is exactly f₂, so the final rays leave L2 parallel (image at infinity).

**Answer:** F = 5.33 cm; H₁ is 2.67 cm to the right of L1; H₂ is 2.67 cm to the left of L2; the object must be 2.67 cm in front of L1.

## Check yourself

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

1. What is F for two 8 cm lenses 4 cm apart?
   A. 4 cm
   B. 5.33 cm
   C. 6 cm
   D. 8 cm
   Answer: B. 1/F = 1/4 − 1/16 = 3/16, so F = 16/3 cm. (Key point: Equivalent focal length)

2. Where are H1 and H2?
   A. Both at the lens centres
   B. H1 1.33 cm right of L1
   C. H1 4 cm right of L1
   D. H1 2.67 cm right of L1; H2 2.67 cm left of L2
   Answer: D. α = β = F d / f = 2.67 cm. (Key point: Principal points)

3. For image at infinity, the object must be at:
   A. The principal point H1
   B. The second focal point
   C. The first focal point F1 of the system
   D. Twice F
   Answer: C. Rays from F1 emerge parallel. (Key point: Object position)

4. How far in front of L1 is the object?
   A. 8 cm
   B. 5.33 cm
   C. 2.67 cm
   D. 1.33 cm
   Answer: C. F1 = α − F = 2.67 − 5.33 = −2.67 cm from L1. (Key point: Result)

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