Physics · Simulation and virtual lab Free
Convex lens: find the focal length on an optical bench
Move the lens and the white screen along the optical bench until the image is sharp, then use your u and v readings to find the focal length.
- You will learn to
- Find the position of a sharp real image on a screen.
- Describe images (real or virtual, inverted, magnified or diminished).
- Find the focal length from a graph of 1/v against 1/u.
- You will measure
- Object distance u and image distance v (cm); then the focal length f.
- Time needed
- About 35–45 min
- Equipment
- Nothing extra. Real version: optical bench or metre rule, convex lens in a holder, lamp box with cross-wire object, white screen.
- Price
- Free: no payment, no account.
Virtual lab
Follow the steps with the simulation above (go back to the simulation ↑). Each time you tap Record, a row is added to the table and a point to the graph.
Aim
To investigate the images formed by a convex lens and to determine its focal length from a graph of 1/v against 1/u.
What to do
- Choose Lens A. Set the lens position to u = 20.0 cm.
- Move the screen slowly until the image of the object is sharp. Read v and tap Record. (The “Help me” button moves the screen slowly for you.)
- Repeat for u = 25, 30, 35, 40 and 50 cm.
- Look at the image each time: is it inverted? Magnified or diminished?
- Look at the graph of 1/v against 1/u. Read the intercept on the 1/v axis and calculate f = 1 ÷ intercept.
- Set u smaller than f: can you find an image on the screen? Repeat the whole experiment with Lens B or C.
Results
Questions
State the lens formula.
1/f = 1/u + 1/v, where u is the object distance, v the image distance and f the focal length (real-is-positive convention).
With u = 30 cm a sharp image formed at v = 30 cm. Find f.
1/f = 1/30 + 1/30 = 2/30, so f = 15 cm. (Object at 2F gives an image at 2F, the same size.)
Describe the image when the object is between F and 2F.
Real, inverted and magnified, formed beyond 2F on the other side of the lens.
Why is no image formed on the screen when u is less than f?
The rays leaving the lens spread out (diverge), so they never meet on the other side. The image is virtual, upright and magnified, seen by looking through the lens (magnifying glass).
How do you know that the image on the screen is real?
It can be caught on a screen: the light rays really meet there.
Why do we draw a graph instead of using one pair of u and v?
Each reading has an error in judging the sharp image. A best line through many points averages the errors and gives a more reliable f.
For teachers: expected results and how the model works
Lens A f = 15 cm, lens B f = 10 cm, lens C f = 20 cm. Lens A: u = 20 → v = 60; 25 → 37.5; 30 → 30; 40 → 24; 50 → 21.4 cm. The image is “sharp” when the blur circle is below 0.12 cm (4 cm aperture), so learners are typically within ±0.5 cm of v. The graph has gradient −1 and both intercepts equal 1/f. When u < f no real image forms (a message explains the virtual image).
Exam practice Exam set
National-exam style questions on this simulation (15 marks). Try each one on paper first, then open the marking scheme.
An object is 24 cm from a convex lens of focal length 8 cm. (a) Calculate the image distance. (b) Calculate the magnification. (c) State three properties of the image. (6 marks)
Marking scheme: (a) 1/v = 1/8 − 1/24 = 2/24 (1), v = 12 cm (1). (b) m = v/u = 12/24 = 0.5 (1, 1). (c) Real, inverted, diminished (1 for two, 2 for three).
Draw a ray diagram to show how a convex lens forms the image of an object placed between F and the lens. (4 marks)
Marking scheme: Ray parallel to the axis refracted through F (1); ray through the optical centre undeviated (1); rays diverge, traced back with dashed lines (1); virtual, upright, magnified image on the same side as the object (1).
In an experiment the graph of 1/v against 1/u cut the 1/v axis at 0.067 cm⁻¹ and the 1/u axis at 0.066 cm⁻¹. Find the focal length. (3 marks)
Marking scheme: f = 1 ÷ 0.067 = 14.9 cm (1) and f = 1 ÷ 0.066 = 15.2 cm (1); mean f ≈ 15 cm (1).
Give two precautions to obtain accurate results in this experiment. (2 marks)
Marking scheme: Move the screen backwards and forwards to find the position of the sharpest image (1); read the positions at eye level on the metre rule / keep the object, lens and screen centres on the same axis (1).
Donat Sciences and MathsPhysics · Virtual lab worksheet
School: ........................................ Name: ................................ Class: ........ Date: ............
Convex lens: find the focal length on an optical bench
Aim: To investigate the images formed by a convex lens and to determine its focal length from a graph of 1/v against 1/u.
Apparatus (real laboratory): optical bench or metre rule, convex lens in a holder, lamp box with cross-wire object, white screen.
What to do (with the simulation on www.donatstem.co.rw)
- Choose Lens A. Set the lens position to u = 20.0 cm.
- Move the screen slowly until the image of the object is sharp. Read v and tap Record. (The “Help me” button moves the screen slowly for you.)
- Repeat for u = 25, 30, 35, 40 and 50 cm.
- Look at the image each time: is it inverted? Magnified or diminished?
- Look at the graph of 1/v against 1/u. Read the intercept on the 1/v axis and calculate f = 1 ÷ intercept.
- Set u smaller than f: can you find an image on the screen? Repeat the whole experiment with Lens B or C.
Results
| # | Lens | u (cm) | v (cm) | 1/u (cm⁻¹) | 1/v (cm⁻¹) |
|---|
| 1 | | | | | |
| 2 | | | | | |
| 3 | | | | | |
| 4 | | | | | |
| 5 | | | | | |
| 6 | | | | | |
| 7 | | | | | |
| 8 | | | | | |
Graph
Plot 1/v (vertical axis) against 1/u (horizontal axis). Draw the best straight line and extend it to cut both axes. f = 1 ÷ intercept.
Questions
- State the lens formula. [2]
- With u = 30 cm a sharp image formed at v = 30 cm. Find f. [2]
- Describe the image when the object is between F and 2F. [2]
- Why is no image formed on the screen when u is less than f? [2]
- How do you know that the image on the screen is real? [2]
- Why do we draw a graph instead of using one pair of u and v? [2]
- An object is 24 cm from a convex lens of focal length 8 cm. (a) Calculate the image distance. (b) Calculate the magnification. (c) State three properties of the image. [6]
- Draw a ray diagram to show how a convex lens forms the image of an object placed between F and the lens. [4]
- In an experiment the graph of 1/v against 1/u cut the 1/v axis at 0.067 cm⁻¹ and the 1/u axis at 0.066 cm⁻¹. Find the focal length. [3]
- Give two precautions to obtain accurate results in this experiment. [2]
Donat Sciences and MathsPhysics · Virtual lab worksheet
School: ........................................ Name: ................................ Class: ........ Date: ............
Marking scheme: Convex lens: find the focal length on an optical bench
Expected results
Lens A f = 15 cm, lens B f = 10 cm, lens C f = 20 cm. Lens A: u = 20 → v = 60; 25 → 37.5; 30 → 30; 40 → 24; 50 → 21.4 cm. The image is “sharp” when the blur circle is below 0.12 cm (4 cm aperture), so learners are typically within ±0.5 cm of v. The graph has gradient −1 and both intercepts equal 1/f. When u < f no real image forms (a message explains the virtual image).
Answers
- State the lens formula. [2]
1/f = 1/u + 1/v, where u is the object distance, v the image distance and f the focal length (real-is-positive convention).
- With u = 30 cm a sharp image formed at v = 30 cm. Find f. [2]
1/f = 1/30 + 1/30 = 2/30, so f = 15 cm. (Object at 2F gives an image at 2F, the same size.)
- Describe the image when the object is between F and 2F. [2]
Real, inverted and magnified, formed beyond 2F on the other side of the lens.
- Why is no image formed on the screen when u is less than f? [2]
The rays leaving the lens spread out (diverge), so they never meet on the other side. The image is virtual, upright and magnified, seen by looking through the lens (magnifying glass).
- How do you know that the image on the screen is real? [2]
It can be caught on a screen: the light rays really meet there.
- Why do we draw a graph instead of using one pair of u and v? [2]
Each reading has an error in judging the sharp image. A best line through many points averages the errors and gives a more reliable f.
- An object is 24 cm from a convex lens of focal length 8 cm. (a) Calculate the image distance. (b) Calculate the magnification. (c) State three properties of the image. [6]
(a) 1/v = 1/8 − 1/24 = 2/24 (1), v = 12 cm (1). (b) m = v/u = 12/24 = 0.5 (1, 1). (c) Real, inverted, diminished (1 for two, 2 for three).
- Draw a ray diagram to show how a convex lens forms the image of an object placed between F and the lens. [4]
Ray parallel to the axis refracted through F (1); ray through the optical centre undeviated (1); rays diverge, traced back with dashed lines (1); virtual, upright, magnified image on the same side as the object (1).
- In an experiment the graph of 1/v against 1/u cut the 1/v axis at 0.067 cm⁻¹ and the 1/u axis at 0.066 cm⁻¹. Find the focal length. [3]
f = 1 ÷ 0.067 = 14.9 cm (1) and f = 1 ÷ 0.066 = 15.2 cm (1); mean f ≈ 15 cm (1).
- Give two precautions to obtain accurate results in this experiment. [2]
Move the screen backwards and forwards to find the position of the sharpest image (1); read the positions at eye level on the metre rule / keep the object, lens and screen centres on the same axis (1).