5.3 Lenses and ray diagrams — Learn

Prerequisites: Section 5.2, especially refraction and the normal; measuring lengths in millimetres, centimetres and metres.

By the end of this section you can: describe the action of a convex lens and a concave lens; draw a ray diagram to find the image formed by a convex lens; and calculate the magnification of a lens.

1. Convex and concave lenses

A lens is a piece of transparent material that refracts light. The refraction happens at the two curved surfaces, so a lens can bring light to a point or spread it out.

A convex lens is thicker in the middle than at the edges. It brings parallel rays of light together. It is a converging lens.
A concave lens is thinner in the middle than at the edges. It spreads parallel rays of light apart. It is a diverging lens.

The point where a convex lens brings parallel rays together is the principal focus. The distance from the centre of the lens to the principal focus is the focal length. A lens with a short focal length bends the light more strongly.

2. Ray diagrams

A ray diagram shows how light from an object forms an image. For a convex lens you draw three things.

The principal axis, a straight line through the centre of the lens.
The lens, as a vertical line across the axis, with a principal focus marked on each side.
Two rays from the top of the object.

There are two rules for the rays.

A ray that travels parallel to the principal axis is refracted through the principal focus on the other side.
A ray that passes through the centre of the lens continues in a straight line.

The two rays meet at the top of the image. The image of a distant object is real and inverted. A real image can be put on a screen, because the rays really meet there.

If the object is closer to the lens than the principal focus, the rays do not meet on the far side. They only appear to come from a point on the same side as the object. This image is virtual. It is upright, and it is what a magnifying glass produces. A virtual image cannot be put on a screen.

3. Magnification

The size of an image is described by the magnification:

\text{magnification} = \frac{\text{image height}}{\text{object height}}

Magnification is a ratio of two lengths, so it has no unit. A magnification greater than 1 means the image is larger than the object. A magnification between 0 and 1 means the image is smaller.

Worked example 1. A lens projects a picture 2.0 cm tall onto a wall. The image is 0.30 m tall. Calculate the magnification.

First change the units so that both heights are the same:

0.30\,\text{m} = 30\,\text{cm}

Then

\begin{aligned} \text{magnification} &= \frac{30\,\text{cm}}{2.0\,\text{cm}} \\ &= 15 \end{aligned}

The image is 15 times taller than the object.

Worked example 2. A lens forms an image of a coin. The coin is 3.0 mm across, and the image is 15 mm across. Calculate the magnification.

\begin{aligned} \text{magnification} &= \frac{15\,\text{mm}}{3.0\,\text{mm}} \\ &= 5.0 \end{aligned}

Worked example 3. A lens has a magnification of 3.0. The object is 8.0 mm tall. Calculate the height of the image.

\begin{aligned} \text{image height} &= \text{magnification} \times \text{object height} \\ &= 3.0 \times 8.0 \\ &= 24\,\text{mm} \end{aligned}

4. Changing the object distance

The figure shows a convex lens with an object in front of it. You can change how far the object is from the lens.

Predict first. Do not move the slider yet. Write down your answers.

As the object moves closer to the lens, does the image get bigger or smaller?
Does the image stay the same way up, or turn upside down?

Test. Move the slider and watch the image. Stop with the object twice as far from the lens as the principal focus, and read the magnification.

Explain. The rays still obey the two ray rules at every setting. Why does moving the object change the size of the image? Write your answer in two or three sentences.

Watch out. A real image from a convex lens is inverted, so it is upside down compared with the object. A virtual image from a magnifying glass is upright. Also check the units before you divide: a height in metres must be changed to centimetres or millimetres so that it matches the other height. Magnification has no unit, so an answer in centimetres is wrong.

5. Progress check

Return to the chapter problem. The visitor centre lens projects a picture 2.0 cm tall so that its image is 0.30 m tall. Work out the magnification. Then decide whether the image is real or virtual, and explain your answer.

You should now be able to

describe the action of a convex lens and a concave lens;
explain what the principal focus and the focal length of a lens are;
draw a ray diagram for a convex lens, using the two ray rules;
describe a real image and a virtual image, and say how they differ;
calculate the magnification of a lens;
find an image height or an object height from the magnification.