Section 1 of 5Describing an Image
Three pairs of words are used to describe any image formed by a lens.
Real
- The refracted rays actually meet at the image, so it can be caught on a screen.
Virtual
- The rays only appear to come from the image when traced backwards, so it cannot be shown on a screen.
Magnified or diminished
- The image is larger than the object when magnified and smaller than the object when diminished.
Upright or inverted
- An upright image has the same orientation as the object, and an inverted image is upside down.
Section 2 of 5Convex Lens Ray Diagrams
Two rays from the top of the object are enough to locate the top of the image.
1
Draw a ray from the top of the object, parallel to the principal axis, to the lens; it refracts through the principal focus on the far side.
2
Draw a second ray from the top of the object through the centre of the lens; it carries straight on without bending.
3
Where the two rays meet is the top of the image, and the image is the line down to the axis from that point.
4
If the object is beyond the focal length the rays meet, so the image is real and inverted.
5
If the object is closer than the focal length the rays still spread out, so trace them backwards with dotted lines to find a virtual, upright, magnified image.
Section 3 of 5Concave Lens Ray Diagrams
Concave Lens Ray Diagrams
Parallel ray
- The ray parallel to the axis diverges after the lens, as if it came from the principal focus on the object side.
Central ray
- The ray through the centre of the lens passes straight through without any change of direction.
Image
- The emerging rays never meet, so the dotted lines traced back cross to form a virtual image.
Result
- The image is always virtual, upright and diminished, whatever the position of the object.
| Lens and object position | Image |
|---|---|
| Convex, object beyond focal length | Real, inverted (size depends on distance) |
| Convex, object inside focal length | Virtual, upright, magnified |
| Concave, any position | Virtual, upright, diminished |
Section 4 of 5Worked Example
Question
- A convex lens used as a magnifying glass produces a virtual image 5.0 cm tall of an object 2.0 cm tall.
- Calculate the magnification.
- Write the equationmagnification = image height ÷ object height.
- Substitute the valuesmagnification = 5.0 ÷ 2.0.
- Calculatemagnification = 2.5, which has no unit because it is a ratio of two lengths.
- State that the image is 2.5 times as tall as the object.
Answer Magnification = 2.5
Section 5 of 5Exam Checklist & Tips
5.3.3 – Key Points for the Exam
- Real image An image formed where rays actually meet, which can be projected onto a screen.
- Virtual image An image formed where rays only appear to meet and which cannot be projected.
- Magnification Image height divided by object height.
- Inverted Upside down compared with the object.
Common Mistakes
- Drawing the refracted ray bending at the surfaces;
- in a ray diagram change direction once at the centre line of the lens.
- Forgetting to use dotted construction lines for virtual images.
- Calling an image real because it is large; it depends on whether the rays actually meet.
- Describing a concave lens image as magnified; it is always diminished.