Interactive practical
Converging-lens focal length and magnification
Focus inverted images on an optical bench, read rulers and investigate focal length and magnification using two graphs.
Open interactive simulationA-Level Physics
Converging-lens focal length and magnification
CCEA 1210 — matched to your course.
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At a glance
Practical focus
- Use real images to determine focal length and test the relationship between magnification and image/object distance.
Practical method
Aim
Determine the focal length of a converging lens and investigate magnification using \(\dfrac{1}{f}\) = \(\dfrac{1}{u}\) + \(\dfrac{1}{v}\).
Variables
Independent variable: object distance u
Dependent variable: image distance v (and image height)
Control variables: the same lens; the same illuminated object.
Apparatus
- Converging lens in a holder
- Illuminated object (cross-wire on a lamp box)
- White screen
- Metre rule
- Ruler for image heights
Method
- Estimate f by focusing a distant window onto the screen.
- Place the object at a measured distance u greater than f and move the screen until the image is sharpest; measure v.
- Repeat for at least five object distances, finding the sharp image from both directions and averaging.
- Measure the object and image heights to find magnification.
- Plot \(\dfrac{1}{v}\) against \(\dfrac{1}{u}\): a straight line with intercepts \(\dfrac{1}{f}\) on both axes.
Sample results
Lens f ≈ 0.15 m; object height 20 mm.
| u / m | v / m | 1/u / m⁻¹ | 1/v / m⁻¹ | Magnification |
|---|---|---|---|---|
| 0.200 | 0.600 | 5.00 | 1.67 | 3.0 |
| 0.250 | 0.375 | 4.00 | 2.67 | 1.5 |
| 0.300 | 0.300 | 3.33 | 3.33 | 1.0 |
| 0.400 | 0.240 | 2.50 | 4.17 | 0.60 |
| 0.600 | 0.200 | 1.67 | 5.00 | 0.33 |
Analysis
\(\dfrac{1}{u}\) + \(\dfrac{1}{v}\) = 6.67 m⁻¹ for every row, so f = 0.150 m. Magnification equals \(\dfrac{v}{u}\) (and image height/object height): 3.0 at u = 0.200 m, 1.0 at u = 2f = 0.300 m.
Uncertainty
Judging the sharpest image is uncertain by about ±5 mm. For v = 0.600 m this is <1%, but for v = 0.200 m it is 2.5%. The \(\dfrac{1}{v}\)–\(\dfrac{1}{u}\) graph averages these.
Errors and improvements
| Problem | Effect | Improvement |
|---|---|---|
| Judging the sharpest image | v uncertain | Approach from both sides and average |
| Measuring to the lens holder edge | Systematic error in u and v | Measure to the optical centre |
| Object inside focal length | No real image | Keep u > f |
Risk assessment
| Hazard | Risk | Control measure |
|---|---|---|
| Hot lamp box | Burns | Allow to cool |
| Focusing sunlight | Eye damage or fire | Never point the lens at the Sun |
Common mistakes
- Using u and v measured from the holder rather than the lens centre.
- Forgetting the image is inverted.
- Using a diverging lens sign convention by mistake.
Worked analysis
u = 0.400 m, v = 0.240 m: \(\dfrac{1}{f}\) = \(\dfrac{1}{0.400}\) + \(\dfrac{1}{0.240}\) = 2.50 + 4.17 = 6.67 m⁻¹, so f = 0.150 m. Magnification = \(\dfrac{v}{u}\) = 0.60.
Common misconception
A distant object focuses at the focal point, but nearer objects form images further from the lens than f.
Assessment coverage
Practice this practical
2 questions available, worth 4 marks in total.
Retrieval cards
What is the lens equation?
\(\dfrac{1}{f}\) = \(\dfrac{1}{u}\) + \(\dfrac{1}{v}\).
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How this practical fits your course
The label below uses the wording from your course specification.
- CCEA 1210 — Specification experiment: Appendix 4 experiments 19–20 – converging-lens focal length and magnification. Appendix 4 includes determining focal length for a converging lens and verifying that real-image magnification equals image distance divided by object distance. Checked 18 September 2026. View official course source