---
title: "Image by two lenses 30 cm and 40 cm"
subject: "Engineering Physics"
chapter: 5
key: "semester_1/engineering_physics:two-lenses-thirty-forty-image"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/two-lenses-thirty-forty-image
site: BCT Engineering
---

# Image by two lenses 30 cm and 40 cm

Engineering Physics · Chapter 5: Geometrical Optics

- Asked in: 2074 Chaitra, 5 marks; 2072 Chaitra, 5 marks (asked 2 times)
- Video: [Physics - Optics: Lenses (1 of 5) Lens Combinations - Two Converging Lenses](https://www.youtube.com/watch?v=aHHa0cK_3as) — Michel van Biezen
- 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-thirty-forty-image
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-5

## Question

Two thin converging lenses of focal lengths 30 cm and 40 cm respectively are placed co-axially in air separated by a distance of 20 cm. An object is placed 40 cm in front of the first lens. Find the position and nature of the image.

## Short answer

First lens: v₁=120 cm, m₁=-3. For second lens, u₂=+100 cm (virtual), v₂=28.6 cm, m₂=0.286. Total magnification -0.86. Final image is 28.6 cm right of second lens, real, inverted, diminished.

## Answer

Use the lens formula 1/v − 1/u = 1/f with the Cartesian sign convention (distances measured from the lens, positive in the direction of light).

**Given:** f₁ = +30 cm, f₂ = +40 cm, separation d = 20 cm, object at u₁ = −40 cm.

```text
 O                L1        L2
 |------40cm-----|---20cm---|
```

**Step 1: image by the first lens**
```text
1/v₁ = 1/f₁ + 1/u₁ = 1/30 − 1/40 = 1/120
v₁ = +120 cm   (image I₁, 120 cm right of L1)
m₁ = v₁/u₁ = 120/(−40) = −3
```

**Step 2: I₁ is the object for the second lens.** L2 is 20 cm from L1, so I₁ lies 120 − 20 = 100 cm beyond L2. It is a **virtual object**, u₂ = +100 cm.
```text
1/v₂ = 1/f₂ + 1/u₂ = 1/40 + 1/100 = 0.035
v₂ = +28.57 cm
m₂ = v₂/u₂ = 28.57/100 = 0.2857
```

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

**Check by the equivalent-lens method:** 1/F = 1/30 + 1/40 − 20/(30×40) = 1/24, so F = 24 cm. The principal points are 12 cm right of L1 and 16 cm left of L2. The object is 52 cm from H₁, so the image is at 44.57 cm from H₂, which is 28.57 cm from L2. This agrees.

**Answer: the final image is 28.6 cm to the right of the second lens; it is real, inverted and diminished (m ≈ −0.86).**

## Check yourself

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

1. What image does the first lens (f = 30 cm, u = −40 cm) form?
   A. Virtual at 120 cm left of L1
   B. Real at 120 cm right of L1
   C. Real at 17.1 cm right of L1
   D. Virtual at 17.1 cm left
   Answer: B. 1/v = 1/30 − 1/40 = 1/120, so v = +120 cm. (Key point: First image)

2. How is I1 treated by the second lens, 20 cm from L1?
   A. Real object, u2 = −100 cm
   B. Virtual object, u2 = +100 cm
   C. Real object, u2 = −140 cm
   D. Virtual object, u2 = +140 cm
   Answer: B. I1 is 120 − 20 = 100 cm beyond L2, so it acts as a virtual object, u2 = +100 cm. (Key point: Virtual object)

3. What is the final image?
   A. 40 cm right of L2, enlarged
   B. 28.6 cm right of L2, real, inverted, diminished
   C. 100 cm right of L2, real
   D. 28.6 cm left of L2, virtual, erect
   Answer: B. 1/v2 = 1/40 + 1/100 gives 28.57 cm; m = −3 × 0.2857 = −0.86. (Key point: Final image)

## Close variants set in other exams

- [Image and magnification, 3 cm and 4 cm lenses](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Atwo-lenses-three-four-lateral-magnification) — Engineering Physics, chapter 5, asked 3×

## Related in Engineering Physics

- [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
- [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
- [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
- [Final image of convex and concave lenses](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Aconvex-concave-lens-final-image) — Engineering Physics, chapter 5
- [Principal points, object for infinity image](https://www.bctengineer.com/question.md?key=semester_1%2Fengineering_physics%3Atwo-identical-lenses-infinity-object) — 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