---
title: "Image and magnification, 3 cm and 4 cm lenses"
subject: "Engineering Physics"
chapter: 5
key: "semester_1/engineering_physics:two-lenses-three-four-lateral-magnification"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/two-lenses-three-four-lateral-magnification
site: BCT Engineering
---

# Image and magnification, 3 cm and 4 cm lenses

Engineering Physics · Chapter 5: Geometrical Optics

- Asked in: 2071 Shrawan, 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-three-four-lateral-magnification
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-5

## Question

Two thin converging lenses of focal length 3cm and 4cm respectively are placed coaxially in air and separated by distance of 2cm. An object is placed 4cm in front of the first lens. Find the position and nature of the image and its lateral magnification.

## Short answer

First lens produces v1 = 12 cm (magnification m1 = −3). The image serves as a virtual object for the second lens at u2 = +10 cm, giving v2 = 2.86 cm (m2 = 0.2857). Net lateral magnification m = −0.857; the final image is real, inverted, and 2.86 cm right of the second lens.

## Answer

Use the lens formula 1/v − 1/u = 1/f with the Cartesian sign convention.

**Given:** f₁ = +3 cm, f₂ = +4 cm, d = 2 cm, u₁ = −4 cm.

**Step 1: first lens**
```text
1/v₁ = 1/3 − 1/4 = 1/12
v₁ = +12 cm            (real image I₁, 12 cm right of L1)
m₁ = v₁/u₁ = 12/(−4) = −3
```

**Step 2: second lens.** I₁ is 12 − 2 = 10 cm beyond L2, a virtual object: u₂ = +10 cm.
```text
1/v₂ = 1/f₂ + 1/u₂ = 1/4 + 1/10 = 7/20
v₂ = +2.857 cm
m₂ = v₂/u₂ = 2.857/10 = 0.2857
```

**Step 3: net lateral magnification**
```text
m = m₁ × m₂ = (−3)(0.2857) = −0.857
```

```text
 O --4cm-- L1 ---2cm--- L2 -2.86cm- I
   (object)  I1 at 12 cm from L1 (virtual object for L2)
```

**Answer: the final image is 2.86 cm to the right of the second lens; it is real and inverted; the lateral magnification is −0.857 (the image is 0.857 times the size of the object).**

## Check yourself

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

1. What is v1 and m1 for the first lens?
   A. v1 = +1.7 cm, m1 = −0.4
   B. v1 = −12 cm, m1 = +3
   C. v1 = +12 cm, m1 = −3
   D. v1 = +12 cm, m1 = +3
   Answer: C. 1/v1 = 1/3 − 1/4 = 1/12 and m1 = 12/(−4) = −3. (Key point: First lens)

2. What is u2?
   A. −10 cm
   B. +10 cm
   C. −12 cm
   D. +14 cm
   Answer: B. I1 is 12 − 2 = 10 cm beyond L2 and is a virtual object. (Key point: Second object)

3. What is the net lateral magnification and nature of the image?
   A. +0.857, erect
   B. +2.86, virtual
   C. −3, real
   D. −0.857, real, inverted
   Answer: D. m = m1 × m2 = (−3)(0.2857) = −0.857. (Key point: Net magnification)

## Close variants set in other exams

- [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×

## Related in Engineering Physics

- [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
- [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
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