Waves Tutorial 7 - Interference
Interference of Waves
When two progressive waves of the same type coincide, they superpose. The results of the superposition is interference.
can demonstrate this with water waves in a ripple tank.
The two dippers act as two different sources which are in phase and
having identical wavelengths and frequencies.
We say that the sources are coherent.
Coherent waves have:
Coherent waves have:
The same wavelength;
The same velocity;
The same frequency;
The same phase relationship.
The waves are NOT necessarily in phase. The phase relationship is constant.
the pattern we can see:
Regions of constructive interference where crests meet
crests and troughs meet troughs. The
amplitude is larger.
Where the waves are in antiphase, there is cancellation.
should note the following:
At any point where there is constructive interference,
the water is still rising and falling at the same frequency, but with
The position of the pattern remains steady, not altering
The total energy of the system remains constant.
While there is no energy in the regions of cancellation, there is
more energy in the regions of reinforcement.
can demonstrate the same thing with sound waves.
We set up two loudspeakers driven by the same signal generator. If we use a microphone connected to a CRO, we can detect
regions of reinforcement and cancellation.
(In reality we don’t get complete cancellation.)
We can explain our observations in terms of path differences. Suppose we go along the centre line between the two sources. At all points we are the same distance from either of the sources. There is zero path difference. Since the waves are in phase and produced at the same frequency and travelling at the same speed, they must still be in phase. So they must reinforce.
We also see regions of constructive interference symmetrically on either side of the centre line. Thus the waves must be in phase. This is because the waves have a path difference of one or more whole wavelengths. We often describe this in terms of half wavelengths, so for there to be constructive interference, there must be a path difference of an even number of half wavelengths.
What is meant by the term path difference? How is it measured?
The reverse side of the argument applies to odd numbers of half wavelengths. If the path difference is ½ a wavelength or 1 ½ and so on, we get regions of cancellation. This is because the waves are in antiphase.
path differences are needed for constructive and destructive interference? Explain your answers.
can demonstrate similar effects with microwaves and sound.
In general, if the separation of the sources is smaller compared to the
wavelength, the pattern of constructive and destructive interference is more
The uses of this are not confined to the laboratory. Freak waves in storms can occur due to this. Attempts at sound deadening using high speed computers to produce sound waves in antiphase have been successful. These are now used by pilots in high quality headphones used in the noisy environment of an aeroplane cockpit.
loudspeakers are set 3.0 m apart in a room.
A microphone connected to a CRO is placed at the apex of a triangle 4 m
from the line separating the two loudspeakers as shown below:
The microphone picks up sound waves of a very large amplitude. If the microphone is moved along the line AB by 10 cm to the right of the central point, a point of minimum amplitude is found. What is the wavelength of the waves? What is the frequency of the signal? Speed of sound is 340 m s-1. (Hint - you will need to do some geometry!)
Getting two coherent light sources is extremely difficult due to the nature of the production of light. Light is produced by the excitation of individual groups of atoms in bursts lasting less than nanoseconds (<1 × 10-9 s). These are random so that there is no constancy in the phase relationships, even from a small region of the light source. Although we need not go into the explanation for this, it has been found that the coherence length for two rays of light rarely exceeds 1 mm.
What is meant by coherent wave sources?
Thomas Young first demonstrated interference in 1801. He didn’t have a laser, just a candle He split the light from a single source into two. In this way he got coherent beams. Using modern laboratory equipment, we can reproduce his set up. A coloured filter prevents dispersion of the light into a spectrum.
Even with a bright ray-lamp, the results are not exactly convincing. How he managed with a candle...
We have lasers now which make it easy to demonstrate.
produces a single high intensity monochromatic
(one wavelength) beam where all the waves are in a constant phase relationship.
If we can split this, we can easily demonstrate interference effects.
You are not expected to know how the LASER works, but a detailed discussion can
be found in
Quantum Physics Tutorial 8.
You are not expected to know how the LASER works, but a detailed discussion can be found in Quantum Physics Tutorial 8.
When the laser shines on the double slit, all the waves are in phase so the slits act as coherent sources. In the diagram the point O is the centre point on the screen and is equidistant from the two sources. Therefore there must be reinforcement, because the waves arrive in phase. A bright fringe is produced. This fringe is made by waves whose path difference is zero.
At P there is a dark fringe,
where there is no light. The waves
must be in antiphase to cancel out so the path difference must be one half
wavelength. At Q, the path difference
is two half wavelengths, so another bright fringe or maximum
is found. Where the path difference
is an odd number of half wavelengths, minima
are found; even numbers of half wavelengths produce maxima.
was found that the wavelength could be found according to the formula:
the wavelength (m)
is the fringe spacing (m)
is the slit spacing (m)
D is the distance from the slits to the screen (m).
Click HERE if you want to see how the equation is derived.
Young's Double Slit experiment using a laser of wavelength 638 nm, the
screen is placed at a distance of 2.5 m from the double slit.
If the slit separation is 0.50 mm, what is the distance between
s = 0.50
D = 2.50 m
w = 2.50 m
0.00319 m = 3.2 mm
0.50 ´10-3 m
Remember always to convert nanometres to metres. 1 nm = 1 × 10-9 m. You will avoid this bear-trap.
In a Young’s Slit experiment, laser light of 630 nm is used to illuminate two slits of separation 0.20 mm. Calculate the fringe separation on a screen 3 m away.
Young's Double Slits is a required practical for the first year of the A-level course. You will use a laser for this experiment.