REFLECTION
We can use the idea of wavefronts to see what happens when a wave strikes a barrier. This is demonstrated with a ripple tank.
Figure 433 shows the incident wavefronts and the reflected wavefronts.
incident waves re ected waves
i.e.
i r
i r
Figure 433 Incident and reflected wavefronts
By constructing the associated rays (see Figure 433), we see that the angle at which the waves are reflected from thethe angle at which the waves are reflected from the barrier is equal to the angle at which they are incident on the barrier (the angles are measured to the normal to the barrier). That is ∠i = ∠r . All waves, including light, sound and water obey this rule, the so-called law of reflection.
We can use the idea of reflection to make a very simple measurement of the speed of sound. If you stand about 100 m away from a tall wall and clap your hands once, a short time later you will hear an echo of the clap. The sound pulse produced by your hand clap travels to the
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wall and is reflected back. The trick now is to clap your hands continuously until each clap is synchronous with the echo. You should then get a friend to time the number of claps that you make in say 30 seconds. Let us suppose that this is 50. The number of claps per second is therefore 0.6. This means that it takes the sound 0.6 seconds to travel to the wall and back. The speed of sound can then be approximated as 330 m s–1. If you should get this result
then you have done very well indeed since this is very nearly the value of the speed of sound in air at standard temperature and pressure. However, in the true spirit of experimental physics, you should repeat the experiment several times and at different distances and you should also assess the quantitative error in your result.
Refraction
We now look to see what happens to a wave when it is incident on the boundary between two media and passes from one medium to the other (transmission). As for the single pulse, some energy will be absorbed at the boundary. Also, as well as energy being transmitted by the wave, some energy will be reflected at the boundary. Here we will concentrate on the transmitted energy.
We have discussed previously the idea that the speed of a wave depends only on the nature and properties of the medium through which the wave travels. This gives rise to the phenomenon of refraction. That is the change in direction of travel of a wave resulting from a change in speed of the wave. This is easily demonstrated with a ripple tank by arranging two regions of different depth. To achieve this a piece of flat perspex or glass is placed a short distance from the source of the waves as shown in Figure 434.
source of plane waves
ripple tank perspex sheet water
Figure 434 Ripple tank setup to demonstrate refraction
Figure 435, shows the result of a continuous plane wave going from deep to shallow water.
deep water shallow water deep water
λd λs
L I G H T S O U R C E
NB: λd > λs
perspex sheet
Figure 435 Wavefronts in water of different depth
In this diagram the wave fronts are parallel to the boundary between the two regions. The frequency of the waves does not alter so, as we have mentioned before, the wavelength in the shallow water will be smaller. If the speed of the waves in the deep water is vd and the speed in the shallow water is vs, then, λ λ such that
v
d__
v
s= __
d sIn Figure 436 the wavefronts are now incident at an angle to the boundary between the deep and shallow water.
shallow deep
A B C
barrier
Figure 436 Waves incident at an angle
As well as the wavelength being smaller in the shallow water the direction of travel of the wavefronts also alters. We can understand this by looking at the wavefront drawn in bold. By the time that part A of this wavefront reaches the barrier at B the refracted wave originated from the barrier will have only reached C since it is travelling more slowly.
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4.5.2 SNELL’S
LAW
In 1621 the Dutch physicist Willebrord Snell discovered and published a very important rule in connection with the refraction of light.
Figure 437 shows a light ray travelling from one medium to another. The line AB (called the normal) is a line constructed such that it is at right angles to the surface between the two media. It is used as a reference to enable the measure of the incident angle, i, and the angle of refraction r. Snell discovered that for any two media
θ θ normal medium 1 medium 2
θ
1θ
2 A BFigure 437 Refraction of light rays
This is known as Snell’s law. In fact it enables us to define a property of a given optical medium by measuring the angles when medium 1 is a vacuum. (In the school laboratory air will suffice.). The constant is then a property of medium 2 alone called its refractive index n.
We usually write
When Snell published his law it was essentially an empirical law and the argument as to the nature of light was still debatable.
Although we have described Snell’s law for light rays, we must remember that a ray is a line that is perpendicular to the wavefronts of a wave. In this respect Snell’s law is true for all waves. Example
In Figure 438 the wavefronts in medium 1 travel with speed v1 and in medium 2 with speed v2 .
A θ1 θ2 Y X θ1 θ2 medium 1 medium 2 boundary B
Figure 438 Snell’s law
In the time that it takes point A on the wavefront XA to reach Y, point X will have travelled to point B. If we let this time be Δt then we have
AY = v1Δt and XB = v2Δt
However from the geometry of the situation we have that
AY = XYsinθ1 and XB = XYsinθ2 From which
θ
θ
Δ Δt
t
that is θ θThat is, the constant in Snell’s law is the ratio of the speed of light in medium 1 to that in medium 2. The result will of course be valid for all types of waves.
For light we have that the refractive index of a material is the ratio of the speed of light in vacuo (c) to that in the material (v). We can write this as
As mentioned above this result has been confirmed for light by a direct measurement of the speed of light in water and subsequently in other materials.
Example
The refractive index of a certain type of glass is 1.5. The speed of light in free space is 3.0 × 108 m s-1.
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Solution
From Snell’s law, the speed of light in glass can be found using c = 3 × 108 m s-1.
So, with n = 1.5 and , we have that v = 2 .0 × 108 m s–1.
4.5.3 Explain and discuss qualitatively the diffraction of waves at apertures and obstacles.
4.5.4 Describe examples of diffraction.
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4.5.3 DIFFRACTION
When waves pass through a slit or any aperture, or pass the edge of a barrier, they always spread out to some extent into the region that is not directly in the path of the waves. This phenomenon is called diffraction. This is clearly demonstrated in a ripple tank.
Figure 439 (a) shows plane waves incident on a barrier in which there is a narrow slit, the width of which is similar in size to the wavelength of the incident waves. Figure 439 (b) shows plane waves passing the edge of a barrier.
(a)
(b)
Figure 439
(a) Diffraction at a slit (b) Diffraction at an edge
To understand this, we use an idea put forward by Christiaan Huygens (1629-1695). Huygens suggested that each point on any wavefront acts as a source of a secondary wave that produce waves with a wavelength equal to the wavelength associated with the wavefront. In Figure 440 we can see how this works in the case of plane wavefronts.
new wavefront secondary
wave
wavefront Figure 440 Planar waves
Each point on the wavefront is the source of a secondary wave and the new wavefront is found by linking together the effect of all the secondary waves. Since there are, in this case an infinite number of them, we end up with a wavefront parallel to the first wavefront. In the case of the plane waves travelling through the slit in Figure 438 (b), it is as if the slit becomes a secondary point source.
If we look at the effect of plane waves incident on a slit whose width is much larger than the incident wavelength as shown in Figure 441 then we see that diffraction effects are minimal. We can understand this from the fact that each point on the slit acts as a secondary source and we now have a situation where the waves from the secondary sources results in a wavefront that is nearly planar (see Figure 440).
Figure 441 Diffraction at a large aperture
Diffraction effects at edges can also be understood on the basis of Huygens’ suggestion. By sketching the wavefronts of secondary sources at points on a wavefront close to the edge, it easy to see that diffraction effects become more pronounced as the wavelength of the incident wave is increased.
4.5.4 EXAMPLES
OFDIFFRACTION
We have seen that the ripple tank can be used to demonstrate the phenomenon of diffraction, i.e. the spreading out of a wave has it goes through an aperture or encounters an object.
Diffraction of sound waves can be demonstrated using the set up shown Figure 442.
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The barrier would be expected to prevent the waves reaching points such as X. However, because of diffraction at the slit, the waves spread out in the region beyond the slit and the microphone will detect sound at points such as X. CRO speaker signal generator microphone X barrier
Figure 442 Diffraction of sound waves
You can also demonstrate the diffraction of light by shining laser light through a single narrow slit such that after passing through the slit, the light is incident on a screen. The effect of diffraction at the slit produces a pattern on the screen that consists of areas of illumination (bright fringes) separated by dark areas (dark fringes). In this situation, each point on the slit is acting as a secondary source and the pattern of light and dark fringes is a result of the interference (see next section) of the waves from these sources. (The diffraction of light is considered in more detail in Option G and Topic 11.3).
As mentioned above, diffraction effects at a slit really only becomes noticeable when the slit width is comparable to the wavelength. In this respect, if the laser is replaced by a point source, then as the slit is made wider, the diffraction pattern tends to disappear and the illumination on the screen becomes more like what one would expect if light consisted of rays rather than waves. Historically, the diffraction of light was strong evidence for believing that light did indeed consist of waves.
The diffraction of light can also be demonstrated by looking directly at a point source through a narrow slit. Unless the point source is monochromatic – a monochromatic source is one that emits light of a single colour (i.e. wavelength), you will see a series of different coloured fringes interspersed by dark fringes.
Exercise 4.5 (a)
1. (a) Calculate the wavelengths of
(i) FM radio waves of frequency 96 MHz
(ii) long wave radio waves of frequency 200 kHz.(Speed of light in free space
c = 3.0 × 108 m s-1)
(b) Use your answers to (i) and (ii) to explain why if your car is tuned to FM, it cuts out when you enter a tunnel but doesn’t if you are tuned to long wave reception. (Hint: consider diffraction)
2. Suggest one reason why ships at sea use a very low frequency sound for their foghorn.
4.5.5 State the principle of superposition and
explain what is meant by constructive
interference and by destructive interference.
4.5.6 State and apply the conditions for constructive and for destructive
interference in terms of path difference and phase difference.
4.5.7 Apply the principle of superposition to determine the resultant of two waves
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4.5.5 THE
PRINCIPLEOF
SUPERPOSITION
We now look at what happens when two waves overlap. For example, in Figure 443 what will happen when the two pulses on a string travelling in opposite directions cross each other?
pulse 1 pulse 2
Figure 443 Two pulses approaching each other
There is a very important principle in physics that applies not only to waves but to other situations as well. This is the principle of superposition. What it effectively tells us is that if you want to find out the effect of two separate
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causes then you need to add the effects of each separate cause. In the example of the two pulses, to find the net displacement at any point we add the displacement of pulse 1 at that point to the displacement of pulse 2 at the point. We show two examples of this in Figure 444 below as the pulses move across each other.
pulse 1 pulse 2
pulse 1 pulse 2 a
2a
Figure 444 (a) Constructive interference
1. In the Figure 444(a) the pulses do not fully overlap and in the second diagram they do. In the second diagram we have we what call full constructive
interference. The two pulses add to give a single
pulse of twice the amplitude of each separate pulse.
Figure 444 pulse 1 pulse 2 pulse 1 pulse 2 pulse 1 pulse 2 (b) Destructive interference
2. We now consider two pulses as shown in Figure 444 (b).
When these two pulses completely overlap the net displacement of the string will be zero. We now have what we call complete destructive interference.
4.5.6 PATHDIFFERENCE
ANDPHASE
DIFFERENCE
The schematic diagram in Figure 445 shows part of the interference pattern that we get from two point sources, S1 and S2. The left hand side is a reflection of the pattern on the right hand side. The dashed line shows where the crests from S1 meet the crests from S2, creating a double
crest, i.e., constructive interference.
S1 S2
The dashed line shows where the crests from S meet the crests from S creating a double crest constructive interference.
constructive interference destructive interference
(crest meets trough) (crest meets crest)
(antinodes)
(nodes) 1 2,
, i.e.,
Figure 445 Two source interference
The sources are two dippers connected to a bar that is vibrated by an electric motor. The dippers just touch the surface of the water in a ripple tank. The sources therefore have the same frequency. They also are in phase. By this we mean that when a crest is created by one dipper a crest is created by the other dipper. Sources that have the same frequency and that are in phase are called coherent
sources.
The bold lines in the diagram show the places where a trough of one wave meets the crest of the other wave producing complete destructive interference. The water at these points will not be displaced and such points are called nodal points or nodes. Points of maximum constructive interference are called antinodes. The bold lines are therefore called nodal lines.
The overall pattern produced by the interfering waves is called an interference pattern.
To emphasise the idea of phase the diagrams in Figure 446 show snapshots of two waves in phase and two waves that are π out of phase.
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Two waves in phase Two waves π out of phase. wave A
wave B wave A
wave B
Diagram 1 Diagram 2
Figure 446 Phase difference
The idea of π out of phase comes from the idea that if the space displacement of wave A in diagram 2 is represented by y = Asinθ , then the space displacement of wave B is
y = Asin (θ + π). At the instant shown the waves in diagram 1
will reinforce and produce constructive interference whereas the waves in diagram 2 will produce destructive interference.
Let us now look at the interference between wave sources from two points in a little more detail.
In Figure 447 we want to know what will be the condition for there to be a point of maximum or minimum interference at P.
S1 S2 X
P
Figure 447 Interference
S1 and S2 are two coherent point sources. For there to be a maximum at P, a trough must meet a trough or a crest must meet a crest. The waves from the sources will have travelled a different distance to P and therefore will not be necessarily in phase when they reach P. However, if a crest meets a crest (or trough a trough) at P then they will be in phase. They will be in phase only if the difference in the distances travelled by the two waves is an integral number of wavelengths. The interference of waves is discussed in further detail in Option G (Chapter 18).
This means that path difference
S2P – S1P = S2X = nλ, n = 0, 1, 2, ...
The waves will be out of phase if the difference in the distance travelled, is an odd number of half-wavelengths.
So for a minimum we have path difference = S2P – S1P = S2X =
(
n + 1 __ 2)
λ, n = 0, 1, 2, ...4.5.7 APPLYING
THE
PRINCIPLEOF
SUPERPOSITION
It will help you understand the principle of superposition to do the following exercise.
Exercise 4.5 (b)
Sophia sounds two tuning forks A and B together and places them close to her ear.
The graph in Figure 448 shows the variation with time t of the air pressure close to her ear over a short period of time.
Figur -1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1 t pr es su re /a rb it ar y uni ts tuning fork A tuning fork B
e 448 Tuning fork graph
On a separate piece of graph paper, use the principle of superposition to construct a graph that shows the variation with time t of the resultant pressure close to Sophia’s ear.
Use your graph to suggest the nature of the sound that Sophia will hear up until the time that the vibrations of the tuning forks die out.