---
title: "Final image of convex and concave lenses"
subject: "Engineering Physics"
chapter: 5
key: "semester_1/engineering_physics:convex-concave-lens-final-image"
source: https://www.bctengineer.com/q/semester_1/engineering_physics/convex-concave-lens-final-image
site: BCT Engineering
---

# Final image of convex and concave lenses

Engineering Physics · Chapter 5: Geometrical Optics

- Asked in: 2067 Asoj, 5 marks
- 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/convex-concave-lens-final-image
- Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-5

## Question

A thin convex and thin concave lens, each of focal length 50cm, are coaxially situated and separated by 10cm. Find the position and nature of the final image formed of an object placed 20cm from the convex lens.

## Short answer

Lens-by-lens calculation: first convex lens forms virtual image at -33.3 cm; concave lens then produces final virtual image 23.2 cm in front of it (13.2 cm from convex lens), erect, magnification +0.89.

## Answer

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

**Given:** convex lens L1: f₁ = +50 cm; concave lens L2: f₂ = −50 cm; separation d = 10 cm; object u₁ = −20 cm.

**Step 1: first lens (convex)**
```text
1/v₁ = 1/f₁ + 1/u₁ = 1/50 − 1/20 = −0.03
v₁ = −33.33 cm
m₁ = v₁/u₁ = −33.33/−20 = +1.667
```
The object lies inside the focal length of L1, so I₁ is virtual, erect and magnified, 33.33 cm to the left of L1 (on the object side).

**Step 2: second lens (concave).** The distance of I₁ from L2 is 33.33 + 10 = 43.33 cm on its left side, so u₂ = −43.33 cm.
```text
1/v₂ = 1/f₂ + 1/u₂ = −1/50 − 1/43.33 = −0.02 − 0.02308 = −0.04308
v₂ = −23.21 cm
m₂ = v₂/u₂ = −23.21/−43.33 = +0.536
```

**Step 3: net magnification**
```text
m = m₁ m₂ = 1.667 × 0.536 = +0.893
```

```text
 I1      O     I     L1     L2
 |       |     |     |      |
-33.3   -20  -13.2   0     +10     (cm, measured from L1)
```
The final image is at −23.21 cm from L2, i.e. at 10 − 23.21 = −13.21 cm from L1. It lies on the same side as the object, between the object and L1.

**Answer: the final image is virtual and erect, formed 23.2 cm in front of the concave lens (13.2 cm in front of the convex lens, on the object side); its size is 0.89 times the object (m ≈ +0.89).**

## Check yourself

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

1. Object 20 cm from convex lens f = 50 cm gives:
   A. Real image 33 cm right
   B. Virtual, erect image 33.3 cm on the object side
   C. Real inverted image
   D. Image at infinity
   Answer: B. Object inside f: 1/v = 1/50 − 1/20 gives −33.33 cm. (Key point: First image)

2. u2 for the concave lens is:
   A. −33.33 cm
   B. +43.33 cm
   C. −23.33 cm
   D. −43.33 cm
   Answer: D. I1 is 33.33 cm left of L1, plus 10 cm to L2. (Key point: Second object)

3. The final image is:
   A. At infinity
   B. Real, inverted
   C. Virtual, erect, 23.2 cm in front of L2
   D. Virtual, 43 cm behind L2
   Answer: C. 1/v2 = −1/50 − 1/43.33 gives −23.2 cm; m ≈ +0.89. (Key point: Final image)

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