Ray diagrams for converging lenses
Key idea: Two rays from the top of an object locate its image: a real image where they cross, or a virtual image where their backward extensions meet. With the object at F, no image forms.
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Guided steps
- Follow rays from one point and their backward extensions.
- Follow the rays and describe the image.
- Draw to scale, then find the image distance and height.
- Compare object positions on each side of the focus.
- Predict the three ray patterns before using the model.
- Change object distance and compare the three ray patterns.
- Explain how parallel rays differ from a virtual-image construction.
- Use the 2F case to test the smaller-image claim.
- Use the image position to find the object region and properties.
- Explain why the screen search can miss a real image.
- Follow diverging rays and their dashed backward extensions.
- Sketch both stamp positions and explain the screen outcome.
- Explain why parallel outgoing rays have no finite image point.
- Sketch the rays and explain why no screen position works.
- Trace the rays backwards to find object distance and height.
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The same converging lens can make a small upside-down picture of a distant tree, a large upside-down picture on a projector screen, or an enlarged upright view of a stamp. The image type and relative size depend on the object’s distance compared with the focal length. A ray diagram, built from two simple rays, shows where the image forms and what it looks like.
Drawing a ray diagram
An illuminated object point can send light towards different parts of the lens. To find where the lens forms the image of the top of the object, you only need two of those rays, because their paths through the lens are known:
- a ray parallel to the principal axis bends through the principal focus F on the far side;
- a ray through the optical centre carries straight on.
When the outgoing rays actually cross, their intersection locates a real image of the top. If they diverge, their backward extensions can locate a virtual image. At the focus they emerge parallel with no finite intersection. Other rays from the same object point form the same image point. These light paths are reversible: you can trace the same principal rays backwards from an image to locate its object.
- Draw the principal axis and the lens. Mark F and 2F on both sides, at distances f and 2f from the lens. Use a scale if distances are given.
- Draw the object as an upright arrow standing on the axis.
- From the top of the object, draw the parallel ray and the central ray, with arrows.
- For a real image, use the actual outgoing intersection; for a virtual image, use the crossing of dashed backward extensions. Draw the image arrow from the axis to that point. Parallel outgoing rays have no finite image point.
- Describe the image: where it is, whether it is real or virtual, inverted or upright, and magnified, the same size or diminished.
A real image is formed where light rays actually meet, so it can be caught on a screen. A virtual image is formed where rays only appear to come from, found by extending them backwards as dashed lines; it cannot be caught on a screen.
Positions and image heights follow the ideal thin-lens model, with rays changing direction at the lens’s centre plane. The principal axis and backward extensions carry no travel arrows.
Worked example 1
An object between F and 2F
Problem
An object stands between F and 2F in front of a converging lens, as in the diagram above. Describe the image.
Show full solution
Find where the rays meet
Method
Follow the parallel ray through F on the far side and the central ray straight through O.
Reason
Where they cross is the image of the top of the object.Working
Answer
The rays cross below the axis, beyond 2F on the far side.
Describe the image
Method
The image is beyond 2F, real, inverted and magnified.
Reason
The rays actually meet, so the image is real. It is below the axis, so it is upside down. It is taller than the object, so it is magnified.
Working
Answer
Beyond 2F; real; inverted; magnified.
Quick check
Check your understanding 1 Draw it to scale
A converging lens has a focal length of 10 cm. An object 4.0 cm tall stands 15 cm from the lens. Draw a ray diagram to scale, using 1 cm on paper for 5 cm. Find where the image forms and how tall it is.
Show answer
On the scale drawing, the parallel ray and the central ray cross 30 cm from the lens on the far side (6.0 cm on paper), and the image is 8.0 cm tall (1.6 cm on paper), below the axis.
The image is real, inverted and magnified, twice the height of the object. The object is between F and 2F, so the image lies beyond 2F (20 cm), as expected. Check that your drawing shows a principal axis, O, focus points 2 cm from O on paper, the object 3 cm from O and 0.8 cm high, two real rays from the same top point, and image 6 cm from O and 1.6 cm below the axis. Draw light arrows towards and away from the lens. Allow for ruler precision when comparing your ray intersection with the model.
How the image changes as the object moves
For object positions beyond F with the lens unchanged, moving the object towards F moves the real image farther away and increases magnification. Crossing inside F changes the case to a virtual image; the real-image trend must not be extended blindly through the focus.
| Object distance, u | Image position | Real or virtual | Orientation | Size |
|---|---|---|---|---|
| very far away | approximately at F | real | inverted | very diminished |
| beyond 2F | between F and 2F on the far side | real | inverted | diminished |
| at 2F | at 2F on the far side | real | inverted | same size |
| between F and 2F | beyond 2F on the far side | real | inverted | magnified |
| at F | rays leave parallel; no finite image | — | — | — |
| between F and the lens | on the same side as the object | virtual | upright | magnified |
Check your understanding 2 Predict each object-distance case
A lens has focal length 10 cm. Before using the model, predict the outgoing rays for an object at 15 cm, 10 cm and 6 cm. Keep the lens unchanged.
Show answer
At 15 cm the rays converge to a real, inverted, magnified image beyond 2F. At 10 cm they leave parallel, with no finite sharp image. At 6 cm they diverge; dashed backward extensions meet on the object side to locate a virtual, upright, magnified image.
Compare the outgoing rays
Keep Focal length, f at 10 cm. Set Object distance, u to 15, 10 and 6 cm in turn. Follow the outgoing rays and compare the Image readout with your predictions. Use labelled distances rather than screen pixels; the model can change its display scale.
A converging lens of focal length 10 centimetres with an object 30 centimetres away. The image is real, inverted, diminished, 15.0 centimetres from the lens.
- Object distance, u
- 30.0 cm
- Image distance, v
- 15.0 cm
- Magnification
- 0.500
- Image
- Real, inverted, diminished
- Angle of incidence, i
- 35.0 °
- Angle of reflection
- 35.0 °
- Angle of refraction, r
- 22.5 °
- sin i / sin r
- 1.500
- Critical angle, c
- — °
Try this
0 of 4 doneMake a real image the same size as the object. (not done yet)
The object is at 2F, so the image is at 2F on the other side: real, inverted and the same size.
Make a virtual image three times the height of the object. (not done yet)
With the object nearer than F the rays leave diverging. Traced back, they meet at a virtual, upright, magnified image on the object's side, as in a magnifying glass.
From glass into air, make a refracted ray at least 75° from the normal while light still escapes. (not done yet)
The refracted ray approaches the boundary as incidence approaches the critical angle, c ≈ 41.8° for this glass–air pair. It is still escaping below c; only at exactly c is the limiting refraction angle 90°. The half-degree control brackets the transition between 41.5° and 42°.
Make total internal reflection happen. (not done yet)
It needs light in the denser medium hitting the boundary at more than the critical angle. Optical fibres trap light this way.
You’ve tried every task.
Check your understanding 3 Parallel or diverging?
At object distances 10 cm and 6 cm, neither case gives a sharp screen image. Use the model’s ray paths to explain the difference.
Show answer
At 10 cm, the object is at F and the outgoing rays are parallel, so they have no finite intersection, even when extended backwards. At 6 cm, the rays diverge and their backward extensions meet on the object side: there is a virtual image. Compare these two distinct ray shapes in your sketch.
Spot the mistake 1
Are real images always smaller?
Learner response
A student says every real image made by a converging lens is diminished. An object is placed at 2F. Use this case to show the claim is wrong, and describe the image fully.
Diagnose before viewing the correction
Show solution step by step
Draw the 2F case
Method
The rays cross at 2F on the far side.Reason
The object and image are the same distance from the lens, so the diagram is symmetrical and the image is the same height as the object.
Working
Answer
Image at 2F, same size.Describe the image
Method
The image is real, inverted and the same size.Reason
It is a real image that is not diminished, so the student’s claim is wrong.
Working
Answer
At 2F; real; inverted; same size.
Common mistake
Exam-style question 1
Working back from the image
Examination question
A converging lens forms a real image between F and 2F on the far side of the lens. State where the object must be, and describe the orientation and size of the image. [3 marks]
Infer the object position and image properties
Show solution step by step
Find the object position
1 markMethod
The object is beyond 2F.Reason
That is the only object position that puts a real image between F and 2F.
Working
Answer
u > 2fDescribe the image
2 marksMethod
The image is inverted and diminished.Reason
The rays cross below the axis, closer to the lens than the object, so the image is upside down and smaller.
Working
Answer
Inverted; diminished.
Self-mark with the mark scheme
Tick each mark point that your answer includes.
Self-mark each point separately.
Common mistake
Calling the image magnified loses the last mark: an object beyond 2F gives a diminished image.
Check your understanding 4 Decide what a missing screen image means
A lens has focal length 10 cm. Sketch the outgoing rays for an object at 11 cm, then at 9 cm. A screen can move only up to 50 cm from the lens. Neither case gives a sharp image there. Does that mean both images are virtual?
Show answer
No. At 11 cm the object is outside F: outgoing rays converge to a real, inverted image, beyond the screen’s 50 cm range. At 9 cm the object is inside F: outgoing rays diverge, and their backward extensions locate a virtual, upright image on the object side.
Compare your two sketches with converging and diverging rays. A real image outside the search range is still real. A missing screen image must be interpreted using the object position and ray paths.
Object closer than the focus
When the object is between F and the lens, the lens cannot bend the rays enough to make them meet. They still spread out after the lens. But if you look back through the lens, your eye traces them back in straight lines, and they seem to come from a point behind the object. That point is a virtual image. This is how a magnifying glass works.
To find it, extend the two rays backwards as dashed lines until they cross.
Guided practice 1
An object inside the focal length
Problem
A lens has focal length 12 cm. A stamp moves from 6 cm to 9 cm from it. Sketch both arrangements using two rays from the stamp’s top. Can either image be caught on a screen? Describe the image after the move.
Decide whether the rays really meet
Hints
Hint 1
follow the real rays
After the lens, do the two rays come together or spread apart?
Hint 2
extend backwards
Extend them backwards with dashed lines and see where they cross.
Show solution step by step
Follow the rays
Method
At both 6.0 cm and 9.0 cm the stamp is inside the 12 cm focal length. The two rays still spread apart after the lens, so they never meet.
Reason
Moving farther from the lens is insufficient if the object still remains inside the focus.
Working
Answer
The move does not produce a sharp screen image. Moving the screen cannot make the diverging real rays meet.
Find the virtual image
Method
Extended backwards, the rays cross on the same side as the object, further from the lens.
Reason
That is where the light appears to come from when you look through the lens.
Working
Answer
At 9.0 cm the image is virtual, upright and magnified, on the same side as the stamp. For a real, magnified image, place the stamp more than 12 cm but less than 24 cm from the lens.
Quick check
Object at the focus
When the object is exactly at F, the rays from any one point on it leave the lens parallel to each other. Parallel rays never meet, and they cannot be traced back to a meeting point either, so no image forms at any finite distance.
Mind stretcher 1 Why no image at the focus?Extension
Construct rays from the top of an object placed at the focal distance, then explain why no sharp image can be caught on a screen at any finite position. Compare the outgoing rays with those for an object inside F.
Show answer
Rays from each point on the object leave the lens parallel to one another. They never come together, so there is no place where a screen could catch them as a single sharp point. You could say the image is “at infinity”. This is different from the case inside F, where the rays spread out but their backward extensions do meet.
Check your understanding 5 Locate the object from a projected image
A lens has focal length 8 cm. Its real image is 12 cm from the lens and 2 cm high below the axis. Draw the image and focal points to scale on paper. Trace principal-ray paths backwards to find the object’s distance and height, then its image magnification.
Show the steps Optional working
Show answer
Mark F at x = −8 cm and +8 cm and the image top at (12, −2) cm. Reverse the optical-centre path through O; it reaches the object side above the axis. Reverse the image-to-far-focus path back to the lens, then continue it parallel to the axis on the left. The two reverse construction paths meet at (−24, 4) cm. Thus the object is 24 cm away and 4 cm high; the image-height ratio is 2/4 = 0.50.
This image is real, inverted and diminished, consistent with an object beyond 2F and an image between F and 2F.
Compare your scaled diagram: both principal rays connect the same object top to the image top. Show physical light arrows travelling from the object to the lens and then to the real image. The reverse tracing is a construction method, not light emitted by the screen.
Common mistakes
- Drawing the two rays from different points on the object.
- Forgetting to mark F and 2F before drawing rays.
- Drawing the parallel ray through the focus on the near side, instead of the far side.
- Drawing real rays dashed, or putting arrows on backward extensions.
- Describing an image without all three properties: real or virtual, inverted or upright, magnified or diminished.
- Saying a virtual image can be projected onto a screen.
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