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E2.5 · Predict and test converging-lens images with rays and equations

Learn to predict and test converging-lens images with rays and equations through clear examples and targeted practice.

Ontario Grade 10 Science

Physics: Light and Geometric Optics

Use ray diagrams and a Grade 10 lens equation to predict and test images

A magnifying glass can make nearby print look larger. The same lens can form an image on a screen when it is aimed at an object farther away. The image changes because the object’s position relative to the lens’s focal point changes. This lesson uses a ray diagram and a course-level lens equation to predict what happens. A ray diagram is a model that traces selected light paths. It is not a photograph of every ray.

What you will learn

  • Describe how a converging lens changes the direction of light.
  • Use principal rays to predict an image’s location and appearance.
  • Use a lens equation and a simple size comparison to make predictions.
  • Compare a prediction with a safe observation and explain possible differences.

1. Bridge from light to lenses

Light travels in straight lines through a uniform material. A ray is a simple line that represents the direction light travels. When light passes from one material into another, it can change direction. This bending is called refraction.
A converging lens is thicker in the middle than at its edges. It refracts parallel incoming rays toward one another. The focal point is where those parallel rays meet. The focal length, written as ff, is the distance from the centre of the lens to the focal point.
An object is the item whose light enters the lens. An image is formed where light rays meet, or where they appear to meet. Object distance, written as dod_o, is measured from the object to the centre of the lens. Image distance, written as did_i, is measured from the lens to the image. Object height and image height describe the sizes of the object and image.
f=distance from lens centre to focal point
  • A converging lens bends parallel rays toward a focal point.
  • Focal length is measured from the centre of the lens to its focal point.
  • The object’s position affects the image.

2. Predict with a ray diagram

Draw the lens as a vertical line and the principal axis as a horizontal line through its centre. A principal axis is the reference line through the middle of the lens. Mark one focal point on each side, each one focal length from the lens. Draw the object as an upright arrow.
Use these ray rules to draw rays from the top of the object. A ray parallel to the principal axis bends through the focal point on the far side. A ray through the centre of the lens continues in a straight line in this simple model. A ray directed through the focal point on the object side emerges parallel to the principal axis.
Where refracted rays actually meet, the image forms. This is a real image, and it can be caught on a screen. If the refracted rays spread apart, extend their paths backward with dotted lines. Where the extensions appear to meet is a virtual image. A virtual image cannot be caught on a screen.
When the object is farther from the lens than the focal point, the refracted rays meet on the far side. The image is real and inverted, meaning upside down compared with the object. When the object is between the lens and the focal point, the rays spread apart. Their backward extensions meet on the object’s side, forming an upright virtual image. A careful drawing should show at least two rays from the object top meeting at the same image point.
  • Draw at least two principal rays from the object’s top.
  • Actual rays meeting indicate a real image; backward extensions meeting indicate a virtual image.
  • An object inside the focal length forms an upright virtual image.

3. Use an equation and test a prediction

A Grade 10 lens equation can connect focal length, object distance, and image distance. In the cases used here, the focal length and object distance are positive. The image distance is positive for a real image and negative for a virtual image. Keep all distances in the same unit.
The image-size equation relates image height to object height. A negative image height means the image is inverted; a positive height means it is upright. The image’s height magnitude gives its size. Use the ray diagram first to predict the image type, then check whether the equation gives a matching result.
To test a prediction, a classroom setup can use a suitable converging lens, an object, and a screen. Keep the object and lens steady while moving the screen to find a sharp image. Measure from the lens centre. Record the screen position, whether the image is sharp, its orientation, and its approximate size. A screen test detects a real image, not a virtual one.
Treat the numerical data in the examples as hypothetical, not as measured laboratory results. Real measurements may differ from predictions because distances and image edges are not perfectly precise. Use only a teacher-approved light source and classroom equipment. Never point a lens at the Sun or focus sunlight; concentrated sunlight can harm eyes or start a fire.
1f=1do+1di,hiho=−dido\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i},\quad \frac{h_i}{h_o}=-\frac{d_i}{d_o}
  • Use the ray diagram to predict image type before calculating.
  • Use the equations to predict image distance and size.
  • Compare a prediction with an observation and note measurement limits.

4. Check whether the results agree

A prediction is strongest when the ray diagram, equation, and observation are consistent. For an object beyond the focal point, expect a real, inverted image. For an object inside the focal point, expect a virtual, upright image. If the equation’s signs or the observed image do not match the ray diagram, recheck the object position, distance measurements, arithmetic, and ray paths.
Equations give a model-based prediction. An observation is evidence from a test. Differences between them can point to measurement uncertainty or an unclear image. They do not mean that the image type should be chosen without checking the evidence.
  • Use the ray model and equation together.
  • Check the image type and signs against the diagram and observation.

Worked example

A real image on a screen

Hypothetical values: a converging lens has a focal length of 12 cm. An object is 36 cm from the lens and is 3.0 cm tall. Predict the image distance, height, and orientation.
  1. Predict the image type
    The object is farther from the lens than the focal point because 36 cm is greater than 12 cm. The ray diagram predicts a real, inverted image.
  2. Find image distance
    Substitute the distances into the lens equation. The positive result places the real image on the far side of the lens.
    112=136+1di,di=18 cm\frac{1}{12}=\frac{1}{36}+\frac{1}{d_i},\quad d_i=18\text{ cm}
  3. Find image height
    The image is half as tall as the object. The negative height means it is inverted.
    hi3.0 cm=−1836,hi=−1.5 cm\frac{h_i}{3.0\text{ cm}}=-\frac{18}{36},\quad h_i=-1.5\text{ cm}
Answer: The image is real, inverted, 18 cm from the lens, and 1.5 cm tall.
Check: The positive image distance agrees with a real image, and the negative image height agrees with an inverted image.

Worked example

An object inside the focal length

Hypothetical values: a converging lens has a focal length of 10 cm. An object is 6.0 cm from the lens and is 2.0 cm tall. Predict the image distance, height, and type.
  1. Predict the image type
    The object is closer to the lens than the focal point. The refracted rays spread apart, and their backward extensions meet on the object’s side. The image is virtual and upright.
  2. Find image distance
    Use the lens equation. A negative image distance means the image is virtual and on the object’s side of the lens.
    110=16.0+1di,di=−15 cm\frac{1}{10}=\frac{1}{6.0}+\frac{1}{d_i},\quad d_i=-15\text{ cm}
  3. Find image height
    The negative image distance in the size equation gives a positive image height. The image is upright and 2.5 times as tall as the object.
    hi2.0 cm=−−156.0,hi=5.0 cm\frac{h_i}{2.0\text{ cm}}=-\frac{-15}{6.0},\quad h_i=5.0\text{ cm}
Answer: The image is virtual, upright, 15 cm from the lens on the object’s side, and 5.0 cm tall.
Check: The image cannot be caught on a screen. Its positive height agrees with the upright ray-diagram prediction.

Worked example

Use a prediction to plan a screen search

Hypothetical values: a converging lens has a focal length of 8.0 cm. An object is 12 cm from the lens and is 4.0 cm tall. Predict the image distance and height, then state where to look for a screen image.
  1. Predict the image type
    The object is beyond the focal point, so the rays meet on the far side of the lens. Expect a real, inverted image that can be found with a screen.
  2. Find image distance
    Use the lens equation. The positive result predicts that the screen image is 24 cm from the lens on the far side.
    18.0=112+1di,di=24 cm\frac{1}{8.0}=\frac{1}{12}+\frac{1}{d_i},\quad d_i=24\text{ cm}
  3. Find image height
    The image is twice as tall as the object. The negative height indicates that it is inverted.
    hi4.0 cm=−2412,hi=−8.0 cm\frac{h_i}{4.0\text{ cm}}=-\frac{24}{12},\quad h_i=-8.0\text{ cm}
Answer: Look for a real, inverted image 24 cm from the lens on the far side. Its predicted height is 8.0 cm.
Check: The object is beyond the focal point, so a real inverted image and a positive image distance match the ray model.

Common mistakes and how to avoid them

Calling every image real because light passes through the lens.
Correction: A real image forms where refracted rays actually meet. A virtual image forms where backward extensions appear to meet.
Assuming a virtual image can be caught on a screen.
Correction: Only a real image forms where rays actually meet on the screen side.
Ignoring the sign of image height.
Correction: In this model, a negative image height means inverted and a positive image height means upright.
Measuring distances from the lens edge.
Correction: Measure object and image distances from the centre of the lens.

Lesson summary

  • A converging lens bends parallel light rays toward a focal point.
  • Principal rays predict where an image forms and whether it is real or virtual.
  • The lens equation predicts image distance, and the image-size equation predicts height and orientation.
  • Use consistent units and compare predictions with safe observations.

Check your understanding

Question 1

An object is inside the focal length of a converging lens. What does the ray model predict?
  1. A real, inverted image on the far side
  2. A virtual, upright image on the object’s side
  3. A real, upright image on the object’s side
  4. A virtual, inverted image on the far side
Show answer and explanation
A virtual, upright image on the object’s side
Inside the focal length, refracted rays spread apart. Their backward extensions meet on the object’s side, making an upright virtual image.

Question 2

A lens has focal length 15 cm and an object is 30 cm away. What image distance is predicted?
  1. 10 cm
  2. 15 cm
  3. 30 cm
  4. 45 cm
Show answer and explanation
30 cm
The equation gives 1di=115−130=130\frac{1}{d_i}=\frac{1}{15}-\frac{1}{30}=\frac{1}{30}, so the image distance is 30 cm.

Question 3

In the image-size equation used here, what does a negative image height indicate?
  1. The image is virtual
  2. The image is inverted
  3. The image is larger
  4. The focal length is negative
Show answer and explanation
The image is inverted
The sign of image height describes orientation. A negative image height means the image is inverted.

Key terms

Converging lens
A lens that bends parallel incoming light rays toward one another.
Focal point
The point where parallel rays meet after passing through a converging lens.
Focal length
The distance from the centre of a lens to its focal point.
Principal axis
A reference line through the centre of a lens.
Principal ray
A selected ray with a predictable path, used to draw an image.
Real image
An image formed where light rays actually meet; it can be caught on a screen.
Virtual image
An image formed where light rays appear to meet when extended backward; it cannot be caught on a screen.

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About this lesson and its review

Published by DoAssignment. This reviewed lesson follows Ontario Grade 10 Science (SNC2D), expectation E2.5. It is a study resource, not an official curriculum publication.

Before publication, content is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability. Errors can still occur, so corrections are welcomed.

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