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14.1Reflection and refraction
This section deals with ways to describe waves and the phenomena of reflection and refraction.
Wavefronts and rays
Imagine waves on the surface of water approaching the shore (Figure 14.1). These waves are propagating in a horizontal direction. Imagine planes going through the crests of the waves; the planes will be normal to the direction of the wave. These planes are called wavefronts. Lines at right angles to the wavefronts show the direction of wave propagationâthese are called rays.

Figure 14.1: Surfaces through crests and normal to the direction of energy propagation of the wave are called wavefronts. Rays are mathematical lines perpendicular to the wavefronts and give the direction of energy transfer.

Now imagine the waves on the surface of water caused by a stone dropped in a pool of water. These waves radiate out across the water surface from the point of impact. In this case the wavefronts are cylindrical surfaces (Figure 14.2a).
A point source emits waves in all directions. The wavefronts from a point source are spherical (Figure 14.2b).
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Figure 14.2: Example of cylindrical and spherical wavefronts. a The cylinders go through the crests and are normal to the plane of the paper. b The wavefronts from a point source radiate in all directions. For clarity, only half of each spherical wavefront is shown.

Reflection
This is an everyday phenomenon: you can see your reflection in a mirror, and the reflection of the blue sky in sea water makes the water look blue.
The law of reflection states that the angle of incidence i (angle between the ray and the normal to the reflecting surface at the point of incidence) is equal to the angle of reflection r (angle between the normal and the reflected ray).
The reflected and incident rays and the normal to the surface lie on the same plane, called the plane of incidence.
The ray diagrams in Figure 14.3 illustrate reflection from a plane surface.

Figure 14.3: a Reï¬ection at a plane (ï¬at) surface. b The position of an image seen in a plane mirror.

Reflection takes place when the reflecting surface is sufficiently smooth. This means that the wavelength of the incident wave has to be larger than the size of any irregularities of the surface. The wavelength of the reflected waves is the same as that of the incident wave.
Refraction
Light travels with a velocity of 3.0 Ã 108 m sâ1 in a vacuum. In all other media, the velocity of light is smaller. Refraction is the phenomenon of the passage of light or any other wave from one medium into another where it has a different speed. Refraction changes the direction of the incident ray (unless the incident ray is normal to the boundary of the two media).
When a ray strikes an interface between two media, there is both reflection and refraction. The top three diagrams in Figure 14.4 show the case when a wave enters a medium where the speed of the wave is lower. In this case the ray bends towards the normal. The parts of the wavefronts that have reached the interface are shown in blue. The lower three diagram show the reverse case where the wave enters a medium where the speed is higher. The ray bends away from the normal.

Figure 14.4: A ray of light incident on the interface of two media partly reï¬ects and partly refracts. The top three diagrams show the case of a wave that enters a medium where the speed is lower. In the lower three diagrams the wave enters a medium where the wave speed is higher. a Ray diagram. b Wavefront diagram for refraction. c Wavefront diagram for reflection. The angle between a ray and the normal is the same as that between a wavefront and the surface.

The frequency cannot change as the wave moves into the second medium: imagine an observer right at the boundary of the two media. The frequency can be found from the number of wavefronts that cross the interface per second. This number is the same for both media. So, since the frequency does not change but the speed does, it follows that in refraction the wavelength changes.
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CHECK YOURSELF 1
Light of wavelength 656 nm in air enters glass where the speed of light is 2.0 Ã 108 m sâ1. What is the wavelength of the light in glass?
WORKED EXAMPLE 14.1
Light of wavelength 686 nm in air enters water where the speed of light is 2.3 Ã 108 m sâ1, making an angle of 40Â° with the normal. Determine:
aif the angle of refraction will be greater or smaller than 40Â°
bthe wavelength of light in water.
Explain your working.
Answer
aThe wave has a lower speed in water so the ray will bend towards the normal so the angle of refraction will be less than 40Â°.
bThe frequency in air is f=cÎ» where Î» is the wavelength in air. The frequency in water is the same. Therefore cÎ»=cwÎ»w and so Î»w=cwcÎ»=2.3Ã1083.0Ã108Ã686=526Â nm.
WORKED EXAMPLE 14.2
The paths of rays of red and violet light passing through a glass prism are as shown in Figure 14.5.
Discuss what can be deduced about the speed of light in glass for red and violet light.

Figure 14.5: For worked example 14.2.

Answer
Considering the first refraction when the rays first enter the glass, we see that violet makes a smaller angle of refraction (draw the normal at the point of incidence to see that this is so). Hence the speed of light for violet in glass must be less than that for red.
Snellâs law
There is a relation between the angle of incidence and the angle of refraction known as Snellâs law. It states that (see Figure 14.4)
sinÎ¸1c1=sinÎ¸2c2
where c1 is the speed of the wave in medium 1 and c2 that in medium 2. So revisiting worked example 14.1, the angle of refraction is found from:
sin40Â°3.0Ã108=sinÎ¸22.3Ã108âsinÎ¸2=2.3Ã1083.0Ã108Ãsin40Â°=0.493âÎ¸2â30Â°
If we multiply both sides of Snellâs equation by c, the speed of light in vacuum, we get
cc1sinÎ¸1=cc2sinÎ¸2
We then define the ratio cc1 to be the refractive index n1 of medium 1 and cc2 to be the refractive index n2 of medium 2. So, in terms of the refractive indices Snellâs law for light can be written as
n1sinÎ¸1=n2sinÎ¸2
Knowing the refractive index of a medium allows us to find the speed of light in that medium. For example, glass with refractive index 1.5 has a light speed of
1.5=3.0Ã108cglasÂ âcglassÂ =3.0Ã1081.5=2.0Ã108Â mÂ sâ1.
Consider light of wavelength Î» = 600 nm in air that enters glass of refractive index 1.5. The frequency will stay the same but the speed will change and so the wavelength will change as well:
f=cÎ»=cglassÂ Î»glassÂ 
This can be rewritten as
ccglassÂ =Î»Î»glassÂ , i.e. nglassÂ =Î»Î»glassÂ  and so Î»glassÂ =Î»nglassÂ =6001.5=400Â nm.
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SCIENCE IN CONTEXT
Total internal reflection
An interesting phenomenon occurs when a wave moving in an optically dense medium arrives at the interface with a less dense medium, for example light in water reaching the boundary with air. Some light is reflected at the boundary and some light is refracted. As shown in Figure 14.6a, the angle of refraction is greater than the angle of incidence. As the angle of incidence increases the angle of refraction eventually reaches 90Â°, as shown in Figure 14.6b. The angle of incidence for which the angle of refraction is 90Â° is called the critical angle.

Figure 14.6: Total internal reï¬ection occurs when the angle of incidence is greater than the critical angle.

As shown in Figure 14.6c there is no refracted ray when the angle of incidence is greater than the critical angle; there is just the reflected ray, and so we call this phenomenon total internal reflection.
One of the great modern applications of total internal reflection is the propagation of digital signals carrying coded information securely in optical fibres. The signal stays within the core, as shown in Figure 14.7. This means that the signal can travel large distances without much loss of energy and so does not require frequent amplification. The signal suffers little distortion which means the transmitted signal is of high quality.

Figure 14.7: Laser light carrying coded information travels down the length of the optical fibre in a sequence of total internal reï¬ections.

THEORY OF KNOWLEDGE
It can be shown that the path followed by light as it moves from one medium into another is such so as to minimize the time taken. In other words, light follows a path of least time. This principle was formulated in 1662 by Pierre Fermat. The principle was developed further by Joseph Louis Lagrange in the eighteenth century and has led to a new
formulation of classical mechanics that could be used to describe quantum theory two centuries later. The principle of least time immediately raised the question: how does light âknowâ what is the path of least time without sampling the time taken for all other paths? The resolution of this conundrum came later when it was firmly established that light is a wave.
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TEST YOUR UNDERSTANDING
1In the context of wave motion, explain, with the aid of a diagram, the terms:
awavefront
bray.
2Figure 14.8 shows planar wavefronts incident on a surface.

Figure 14.8: For question 2.

Complete the wavefronts in blue to show reflected wavefronts. (Keep the length of the wavefront the same as the wavefronts in black.)
3Figure 14.9 shows planar wavefronts incident on the boundary of two media. The wave speed in each medium is shown.

Figure 14.9: For question 3.

Complete the wavefronts in blue to show refracted wavefronts. (Keep the length of the wavefront the same as the wavefronts in black.)
4Figure 14.10 shows planar wavefronts incident on the boundary of two media. The wave speed in each medium is shown.

Figure 14.10: For question 4.

Complete the wavefronts in blue to show refracted wavefronts. (Keep the length of the wavefront the same as the wavefronts in black.)
Figure 14.11 shows rays of light going through a converging lens.

Figure 14.11: For question 5.

5Draw a wavefront in each region A, B and C.
6Figure 14.12 shows wavefronts after they have gone through a diverging lens.

Figure 14.12: For question 6.

Continue the three incident rays shown to the right of the lens.
7Plane wavefronts of sound are approaching land from the sea, as shown in Figure 14.13.

Figure 14.13: For question 7.

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On a hot summer day, the temperature over land is greater than the temperature over the water. Suggest how the wavefronts in blue move over land. Explain your reasoning.
8Light of frequency 6.0 Ã 1014 Hz is emitted from point A and is directed towards point B a distance of 3.0 m away. The light will travel 1.5 m in a vacuum and 1.5 m in glass where the speed of light is 2.0 Ã 108 m sâ1. (See Figure 14.14.)

Figure 14.14: For question 8.

aDetermine how long it will take light to get to B.
bCalculate how many full waves fit in the space between A and B.
cDraw three wavefronts in the vacuum and three in the glass.
9Figure 14.15 shows an electromagnetic wave travelling in a vacuum until x = 2.

Figure 14.15: For question 9.

At x = 2 the wave enters a new medium. What is the speed of light in the new medium? (Units on both axes are arbitrary.)
10Describe what is meant by total internal reflection. Describe one application of total internal reflection.
11A ray of light of wavelength 680 nm is travelling in a medium X where the refractive index is 1.20 and enters a medium Y of refractive index 1.80. The ray makes an angle of 35Â° with the normal at the boundary. Calculate:
athe angle of refraction
bthe wavelength of the light in Y.
cA ray moving in Y enters X making an angle of 32Â° with the normal. What is the angle of refraction?
dWhat is the critical angle for a ray moving from Y to X?
12A ray of light is incident on a rectangular block of glass of refractive index 1.45 at an angle of 40Â°, as shown in Figure 14.16.

Figure 14.16: For question 12.

The thickness of the block is 4.00 cm.
aExplain why the ray that exits the glass is parallel to the initial ray incident on the glass.
bCalculate the distance d by which the ray gets deviated.
