Wave Behaviour

Interference

When two progressive waves of the same type coincide, they superpose.  The results of the superposition is interference.

We 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.

In 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

We 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 greater amplitude.

• The position of the pattern remains steady, not altering with time.

• 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.

We 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.

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.

We 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 spread out.

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.

Interference of Light

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.

However the LASER produces a single high intensity monochromatic (one wavelength) beam where all the waves are in phase.  If we can split this, we can easily demonstrate interference effects.

Thomas Young first demonstrated interference in 1801.  He didn’t have a laser, which makes the original experiment harder to reproduce.  He split the light from a single source into two.  In this way he got coherent beams.  We have lasers now which make it easy to demonstrate.

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.

It was found that the wavelength could be found according to the formula:

where

• l is the wavelength (m)

• w is the fringe spacing (m)

• s 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.

Worked example

Remember always to convert nanometres to metres.  1 nm = 1 × 10^-9 m.  You will avoid this bear-trap.

Diffraction

If we pass waves through a single slit, we observe that the waves spread out due to diffraction.

Notice:

•  If the slit is narrow, the diffraction is more marked .

• The wavelength remains the same.

• Diffraction does not need a slit.  Waves can bend round a barrier by diffraction.  Radio signals can be picked up behind hills for this reason.

• The longer the wavelength, the more the waves will diffract.

• All waves diffract.

We can show the diffraction of light due to a single slit.  (We must be careful not to confuse this with Young's double slits.)  If we have a wide slit, we see just a single bright region with sharp edged shadows.

If we make the slit narrower, we see a pattern emerging with a bright central region, and alternating light and dark bands.  The narrower the slit, the marked the effect.  The central bright region becomes dimmer as well because less light is transmitted

If the light is monochromatic, the bands will be of the same colour.  Red light has a broader pattern than blue light, suggesting that the diffraction effect increases with wave length.  If we use white light, the central band is white, with the fringes being overlapped with the spectrum of colours.  This is called Frauenhoffer diffraction.

We can plot a graph to show the intensity, and we see a bright central maximum, with subsidiary maxima either side.  We can explain the effect of diffraction using the idea of secondary wavelets.  In the middle these form a plane wave-front.  At the edges, circular wave-fronts move into the shadow region.  The maxima and minima are caused respectively by constructive and destructive interference.

We can work out the angle of diffraction using a simple equation:

where q is the angle, l is the wavelength and a is the width of the aperture.

Read the question to see if it's single or double slit.  The key thing is that the pattern from double slits is due to interference.  From a single slit, its due to diffraction.

Example

If the door were less than 1.31 m wide, diffraction could not occur because sin q would be greater than 1, which is impossible.

The single slit diffraction equation  can be used, with modification, to determine the limit to which an optical instrument can resolve.  This is called the resolving power.  (This is sometimes known as Rayleigh's Criterion.)  For a light microscope, the theoretical limit is about 1 mm, so the microscope cannot be used to view atoms.  A beam of electrons is regarded as having wave properties.  So an electron microscope has a much bigger resolution, as the wavelength of the electron beam is much shorter.

Radio waves diffract round hills, which is why we can pick up radio signals behind hills, even though there is no direct line of sight between the transmitter and the receiver.

The Diffraction Grating

A diffraction grating can be used to split light into different wavelengths with a high degree of accuracy, much more so than glass prisms.  A diffraction grating usually consists of a piece of glass with very closely spaced lines ruled on it.  A transmission grating has clear spaces between the lines so that light can pass through it.  A reflection grating has a shiny surface between the lines so that light gets reflected off it.  A compact disc acts as a reflection grating.

The diffraction grating has the advantage over the double slit method of measuring wavelength in that:

• the maxima are more sharply defined;

• the beam passes through more slits than two, so the intensity is brighter;

• the angles are larger so that they can be measured with greater precision.

The derivation of the diffraction grating formula is on the syllabus, but experience has shown that most students struggle with it.  It is more important that they know how to use the formula.  If you want to see the derivation, click HERE.  The formula is:

    d sin q = nl

The term n is called the spectrum order.  If n = 1, we have the first diffraction maximum.  The other physics codes:

• q is the angle,

• l is the wavelength,

• and d is the slit width

Sin q can never be greater than 1, so there is a limit to the number of spectra that can be obtained.

Worked Example

If the answer to the problem had been 6.87, the maximum order would still be 6, even though the nearest whole number was 7.

If we did further calculations we could see that the red light is diffracted more than blue light.  The pattern would be like this:

Note that:

• There is a central undeviated maximum, which would be white light.

• The pattern would be symmetrical with orders either side of the maximum.  We have not showed the ones below here.

• The third order does not have a red light ray. 

The diffraction grating is a very good way of selecting light of a specific wavelength.  Chemists and astronomers use diffraction gratings in spectroscopy, which allows them to see the specific spectra given out by different elements.  Each element has its own individual spectrum.  This allows astronomers to:

• See what elements there are in stars.

•  If the spectrum “fingerprint” is shifted at all, astronomers can tell that a star is moving towards us (blue shift) or away from us (red shift).  This is due to the Doppler effect.