Physics
Cambridge IGCSE
Cambridge IGCSE
TOPIC 3: WAVES
Refraction is the bending of the direction of travel of a wave as it moves from one substance, or 'medium', into another. Many of you will be familiar with the refraction of light, but refraction works with all types of waves.
In this section, we will be learning about the refraction of light. To understand how refraction works, we will use the example of a rectangular glass block. Have a look at figure 1 to see how the path of the light ray is affected by the glass:
Figure 1. The path of a light ray through a glass block
As you can see from the diagram. a ray that hits the side of the glass at 90° - perpendicular to the surface - is not refracted. However at angles of incidence between 0 and 90°, the light ray bends towards the normal line as it enters the glass, and away from the normal line as it leaves. This happens with water, clear plastic, or any other transparent medium denser than air. Note that the ray does not travel along the normal line, just closer to it inside the glass.
Why does the light ray change direction like this? The answer lies in the fact that glass is denser than air, and that this typically makes light travel more slowly. The change in speed causes a change in direction.
The best way to try to understand why refraction happens is to draw a wavefront diagram. (See section 3.1 for more on wavefronts). In figure 2, we look at a horizontal block of glass, and compare it to a line of people walking holding hands.
Figure 2. Refraction of wavefronts
As you can see from the diagram, the wavefronts change direction as they slow down, and also become closer together. (This means the wavelength is shorter). The diagram on the right is a good way of trying to picture why this happens - if a line of people walk into mud as shown, the first to enter the mud slow down, causing the line to bend, just a shown in the first diagram.
Remember that wavefronts are always drawn perpendicular to the ray which shows the direction the wave is travelling.
A standard refraction experiment that you need to learn involves semi-circular 'D-shaped' blocks. These blocks work well in this context because - as shown in figure 3 - the light ray is not refracted as it leaves the glass, making it easy to measure angles and draw the emerging ray. Note that, as for reflection, the angles of incidence (i) and refraction (r) are measured from the normal line as shown here:
Figure 3. Refraction through a semi-circular block
A triangular glass or plastic block is called a prism. In the arrangement shown in figure 4, the white light is refracted 'downwards' at both surfaces. This magnifies the effect called dispersion - the fact that different colours actually refract by slightly different angles. The refraction of light depends on the frequency of the light. As white light is made of many frequencies of light, the prism splits the light. We see the familiar traditional 7 colours of the spectrum: Red, orange, yellow, green, blue, indigo and violet. Note that the violet coloured light is refracted more than the red colour.
Figure 4. Dispersion in a prism
For this topic, a standard practical you will need to describe involves investigating refraction using rectangular glass blocks, triangular prisms and semi-circular glass blocks. The theory of these is covered above, but you should know how to carry out this investigation. Typically, you will need:
Learn the outcome of these experiments as described in the section above. A typical question will ask you to measure angles from a diagram, or sketch where the light ray is expected to emerge, as shown in figure 1, shown again here.
Questions:
1. A fish is swimming in a shallow lake. The sun shines on the fish, and then light reflected from the fish moves upwards. One ray is shown in the following diagram:
a) The normal line should be perpendicular to the water surface. It is usually drawn as a dotted line, but as long as it is labelled N, that is fine.
b) The refracted line labelled R bends away from the normal line, so it is shifted towards the horizontal.
c) The reflected line labelled L should be (approximately) the same angle to the normal line as the incident ray already given.
Example solution:
The diagrams above show how light rays are bent by glass and other dense, transparent materials like plastic and water. However it becomes immediately apparent that glass bends light more than water. Scientists say it is 'optically' dense, and measure how much the light is refracted using a measurement called the refractive index.
The diagram below shows some results from a refraction experiment using a glass block. The angles of incidence and refraction are - as always - measured from the normal line.
Figure 5. Measuring refraction angles
It is hard to spot a pattern in these results. The angle of refraction is always smaller than the angle of incidence, but there does not seem to be an obvious pattern.
The 'law' that governs refraction is called 'Snell's Law', and shows that it is trigonometry that helps us find the pattern - by taking the sine of both angles:
| sin i | = constant for any substance |
| sin r |
For the first example above, using this formula we get:
| sin 30° | = 1.5 |
| sin 19° |
This calculated constant is called the refractive index (n) of the substance. From above, the glass in figure 5 has a refractive index of 1.5. There are no units for 'n', as it is just a ratio.
The final formula for Snell's Law is usually given in the following format:
| n = | sin i |
| sin r |
The refractive index for a substance depends on the speed of the wave in that substance. Typically, the denser the substance, the slower the speed of light is and the higher the refractive index for light. In fact, the refractive index is just the ratio of speeds before and after the boundary between two different substances. The value of n= 1.5 for glass means that light travels 1.5 times slower in glass than in air (or a vacuum).
Here are a few standard values for the refractive index:
| Substance | refractive index (n) |
|---|---|
| Glass | 1.5 to 1.6 |
| Water | 1.3 |
| Perspex | 1.5 |
Table 1. Refractive index of some common substances (to 2 sig. figs.)
Here's an example of a typical exam question on this topic:
Example:
A student is investigating the refractive index of ethanol. When a light ray is incident on the liquid with an angle of incidence of 60.0o, the angle of refraction is 39.6o.
a) State the formula used to calculate refractive index.
b) Calculate the refractive index of ethanol. State your answer to 2 significant figures.
Answer:
a) The formula for refractive index is n = sin i / sin r (as shown above).
b) Using this formula and substituting in the angles from above gives:
| n = | sin i | = | sin 60.0 |
| sin r | sin 39.6 |
It is easy to make mistakes with this formula on a calculator, so always check your answer again, and make sure it makes sense. (For example, if the calculated angle is 0.045°, far to small to be measured, then something has probably gone wrong!).
Have a go at these practice questions to gain confidence with Snell's Law. Try to do the questions yourself before checking the solution. If you are not sure how to use your calculator to find the sine of an angle, or to use the sin-1 function, ask for help before starting:
Questions:
2. A light ray is incident on the surface of a glass of water with an angle of incidence of 25°. If the refractive index of water is 1.33, calculate the angle of refraction of the light ray.
We know that n = 1.33, and i = 25°. Using Snell's Law we get:
| sin 25° | = 1.33 |
| sin r |
rearranging this and finding sin 25° gives:
| sin r = | 0.423 |
| 1.33 |
So sin r = 0.318,
r = sin -1 (0.318)
r = 18.50 , or 19° (to 2 sig. figs.)
3. The diagram below shows a perspex block with a refractive index of 1.5.
Calculate the angle of incidence i that produces this refracted ray with an angle of refraction of 38°.
We know that n = 1.5, and r = 38°. Using Snell's Law we get:
| sin i | = 1.5 |
| sin 38° |
Therefore:
sin i = 1.5 x sin 38°
sin i = 0.93
i = sin-1 0.93
i = 67° (to 2 sig.figs.)
Strange things happen when a light ray moves from a substance with a high refractive index back into the air. The ray will bend away from the normal line, as described earlier in this section. However, there is a point at which this no longer happens, as shown in figure 6:
Figure 6. The path of a light ray through a glass block
As you can see from figure 6 diagram (a), the ray leaving the glass is refracted, but some of the light is reflected inside the glass. This happens with all glass - even though glass is transparent, some light is actually reflected, and this is why you can see your reflection in a window on a dark night.
As we increase the angle of incidence, the refracted ray gets closer to the surface of the glass, as shown in diagram (b). At about i = 42°, the light is refracted at nearly 90° to the normal line, and is almost parallel to the surface.
This means that, at angles of incidence greater than about 42° for this glass, no light can escape from the glass, and all of the light is reflected, as shown in diagram (c). This effect is called total internal reflection. The glass is perfectly reflecting the light, and is actually better at it than a mirror where some light is absorbed at the surface.
For the glass shown, at any angle greater than about 42°, total internal reflection occurs. This angle is called the critical angle (c). Water has a critical angle of 48°, as it is not as optically dense as glass.
| sin c = | 1 |
| n |
Example:
Diamond has a refractive index of 2.42. Calculate the critical angle for diamond.
Answer:
We know that n = 2.42 and the formula for the critical angle is:
| sin c = | 1 |
| n |
| sin c = | 1 |
| 2.42 |
Questions:
4. Water has a refractive index of 1.33.
a) Using the formula above, and n= 1.33 gives:
| sin C = | 1 |
| 1.33 |
sin C = 0.75
C = sin-1 (0.75)
C = 48.°
b) At an angle of incidence of 50°, the ray will be reflected, because this angle is greater than the critical angle calculated in part (a).
Extension
You do not need to know where the total internal reflection formula comes from, but if you are interested in a challenge, can you derive the equation from Snell's Law? The solution is given here:
We know from Snell's Law that:
| n = | sin i |
| sin r |
To find the critical angle, we use the fact that if light travels inside glass it is refracted away from the normal as shown in figure 6. If we reverse the direction of the ray from figure 6 diagram (b), and assume it is incident on the glass at 90°, then the refracted ray will be at the critical angle. Therefore we use i = 90° in the Snell's Law formula, and r = C :
| n = | sin 90° |
| sin C |
sin 90° = 1, so:
| n = | 1 |
| sin C |
The reflection inside a pure glass block is very efficient as no light can be refracted. For this reason, it is an excellent effect to use in optical devices such as binoculars, telescopes and microscopes. The use of total internal reflection usually involves a 45° prism as shown in figure 7:
Figure 7. Total internal reflection using a 45° glass prism
As you can see, the light ray travels down through the glass, and is not refracted at the top as it hits the glass at 90° to the surface. At the bottom surface, the angle of incidence is 45°. This is greater than the critical angle of the glass, which as explained above, is typically about 42° for glass of refractive index of 1.5.
A periscope can be made with two 45° prisms as shown in question 5 below. Telescopes and binoculars use prisms instead of mirrors as they give a clearer image with less loss of light.
An optical fibre is a very thin tube of glass or a similar transparent substance. When it is this thin, it is reasonably flexible and can be bent round corners. Any ray of light travelling down the fibre will hit the inside surface at an angle of incidence much larger than the critical angle. That means it will undergo total internal reflection, and bounce down the fibre without any light emerging from the sides of the glass. All of the light is channelled down the fibre, and this can continue over many kilometres.
Figure 8. Light being channelled by an optical fibre
Optical fibres can be used for a range of applications:
1. Fibres can be used to send signals at high speeds, for the internet or other communication links. Long under-sea fibre optic cables link countries together so that we can communicate using the internet, and for telephone communications.
2. Bundles of thin fibres can be used in cameras called endoscopes, used to see inside the human body. Some fibres shine light into the body, and other fibres are used to produce the image. An endoscope is a tremendous development for medical examinations, as it means we can look inside the body - for example the throat or stomach - without the need for surgery.
Questions:
5. The diagram below shows a standard periscope used in submarines and in other applications. Two mirrors are used in this device.
Draw a diagram to show how two 45° prisms can be used to replace the mirrors in this device. Show the path of the light ray through the prisms.
The mirrors should be removed from the above diagram, and replaced with 45° prisms as shown below. It is important that the prism orientation is exactly as shown, so that the light ray hits the back of the 45° sloping edge.
Now test your understanding using these quick, 10 minute questions on this topic from Grade Gorilla:
