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14.3Diffraction and interference

The spreading of a wave as it goes past an obstacle or through an aperture is called diffraction the spreading of a wave through an aperture or past an obstacle. Let us consider a plane wave of wavelength λ moving towards an aperture (Figure 14.28). What will the wavefronts look like after the wave has gone through the aperture? The answer depends on the size of the wavelength compared to the size of the aperture.

In Figure 14.28a the wavelength is small compared to the aperture, so there is little diffraction. In Figure 14.28b the wavelength is similar to the aperture, so there is a lot of diffraction.

Figure 14.28: Waves in a ripple tank passing through an aperture, demonstrating the principle of diffraction. a When these waves pass through a large aperture they change shape and form ﬂattened concentric waves centred on the aperture. The amount by which the waves change shape depends on the size of the aperture. Diffraction is greatest when the aperture size is similar to the wavelength. This is seen in b, where waves of the same wavelength are passing through a smaller aperture.

Exam tip

When you draw diffraction diagrams, make sure that you do not change the distance between the wavefronts.

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Figure 14.29 shows diffraction around obstacles.

Figure 14.29: Diffraction around obstacles.

Diffraction explains why we can hear, but not see, around corners. For example, a person talking in a corridor outside a room can be heard through the open door because sound diffracts through the opening of the door—the wavelength of sound for speech is roughly the same as the door size. On the other hand, light does not diffract through the door since its wavelength is much smaller than the opening of the door. Hence we can hear through the open door, even though we cannot see the speaker.

Double-source interference

When two identical waves meet at the same point in space, the principle of superposition states that the resulting wave has a displacement that is the sum of the individual displacements. This results in a phenomenon called interference when two or more waves meet to form a resultant wave of greater, lower or the same amplitude: the resulting amplitude can be anything from zero to twice the amplitude of one of the waves. All waves show interference.

Consider two identical sources S1 and S2 (Figure 14.30). Wavefronts from the two sources meet at various points. The waves from both sources have the same speed, wavelength, frequency and amplitude. Let us focus on point P. Point P is a distance from source S1 equal to 2λ and a distance of 3λ from S2. The path difference the difference in distance of a point from each of the two sources is the difference in distance of the point from the two sources, Δr = |S1P − S2P|. For P, the path difference is equal to λ. All the points marked red in Figure 14.15 have a path difference that is one wavelength, λ. At point P the waves from both sources arrive as crests. At point Q (the path difference is still λ) the waves arrive as troughs.

Figure 14.30: Wavefronts from two sources meet and interfere.

In either case, when the individual waves are added (as the principle of superposition says) the resulting wave will have the same wavelength, frequency and speed as the individual waves but double the amplitude. This is true for all the points in red.

Now look at the points marked in black on Figure 14.30. The path difference for all of these points is Δ r=λ 2. At point R, the wave from source S2 arrives as a crest but the wave from S1 arrives as a trough. When the two waves are added the resulting wave has amplitude zero—the waves cancel each other out and vanish! This is true for all the points in black. We say there is constructive interference superposition of waves leading to maximum amplitude for the points in red and destructive interference superposition of waves leading to zero (or minimum) amplitude for those in black.

We can imagine joining points with the same path difference with a smooth curve. Figure 14.31 shows this. Red curves go through points whose path difference is 0, λ, 2λ. Black curves go through points whose path difference is λ 2 and 3 λ 2. (If the diagram showed more wavefronts from the two sources, we would be able to find points with larger path differences.)

Figure 14.31: Curves in red join points where the path difference is an integral multiple of the wavelength. Curves in black join points with a path difference that is a half integral multiple of the wavelength. (The curves are hyperbolas.)

From this we can make the general observation that points in red have a path difference that is an integer times the wavelength, whereas points in black have a path difference that is a half-integer times the wavelength. So we conclude that:

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Constructive interference occurs when the path difference:

|S 1 P−S 2 P|=n λ with n=0,1,2,3,…

Destructive interference occurs when the path difference:

|S 1 P−S 2 P|=(n+1 2)λ with n=0,1,2,3,…

So, constructive interference occurs when the waves are in phase (meaning the waves meet crest to crest and trough to trough), and destructive interference occurs when they are exactly out of phase (meaning the waves meet crest to trough)—see Figure 14.32.

Figure 14.32: Displacement–time curves for waves that interfere a constructively and b destructively.

Worked Example 14.4

Identical waves leaving two sources arrive at point P. Point P is 12 m from the first source and 16.5 m from the second. The waves from both sources have a wavelength of 3.0 m.

State and explain what is observed at P.

Answer

The path difference is 16.5 − 12 = 4.5 m.

Dividing the path difference by the wavelength 4.5 3.0=1.5; in other words, the path difference is a half-integral multiple of the wavelength so there is destructive interference.

If the path difference is anything other than an integral or half-integral multiple of the wavelength, then the resultant amplitude of the wave at P will be some value between zero and 2A, where A is the amplitude of one of the waves.

When sound waves from two sources interfere, points of constructive interference are points of high intensity of sound. Points of destructive interference are points of no sound at all. If the waves involved are light waves, constructive interference produces points of bright light, and destructive interference results in points of darkness. Complete destructive interference takes place only when the two waves have equal amplitudes.

Young’s two-slit experiment

Interference for light was first demonstrated in 1801 by Thomas Young, an English polymath who also came very close to deciphering hieroglyphics. This experiment definitively established the wave nature of light. Figure 14.33 shows plane wavefronts of light approaching two extremely thin, parallel, vertical slits. Because of diffraction, the wavefronts spread out from each slit. Wavefronts from the slits arrive on a screen and interfere. At those points on the screen where the path difference is an integral multiple of the wavelength of light constructive interference takes place. The screen looks bright at those points, marked on the diagram as n = 0, ± 1,….The value of n indicates that the path difference is n λ. At other points where the path difference is a half-integral multiple of λ, the screen looks dark: we have destructive interference.

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Figure 14.33: Double-slit interference for light. The fringe spacing s is also marked.

Figure 14.34a shows two rays which have diffracted through each slit and meet at point P on the screen. The diagram is not to scale: the distance between the slits, d, is only a fraction of a millimetre and the distance to the screen is about a metre. This means that these rays are almost parallel as shown in Figure 14.19b. (The two angles labelled θ are equal because they have their sides mutually perpendicular). Using this diagram we see that the path difference is the distance S2Z.

Figure 14.34: The geometry of Young’s double-slit experiment. a The nth maximum at P is a distance s n from the centre of the screen. b Using geometry to find the path difference.

Using trigonometry, the distance S2Z, which is the path difference, is equal to d sin θ. Here, d is the separation of the two slits. This gives the conditions necessary for constructive and destructive interference at P.

For constructive interference: | d sin θ=n λ | n=0,±1,±2,…
For destructive interference: | d sin θ=(n+1 2)λ

Because the distance d is very small in comparison to the distance to the screen, the angle θ is quite small, so sin θ is small. This means we can approximate sin θ by tan θ. You can check on your calculator that, for small angles θ in radians, it is an excellent approximation that sin θ ≈ tan θ ≈ θ.

The derivation that follows is not important—you should know the result. The angle θ is zero for the position directly opposite the slits, where we find the central maximum. Here n = 0 (Figure 14.34). The angle θ increases for each successive maximum moving away from the central maximum.

From Figure 14.34 we deduce that:

tan θ=s n D

where D is the distance of the slits from the screen and s n is the distance of the point P from the middle point of the screen.

Using the small angle approximation, we have:

tan θ≈sin θ

⇒s n=D sin θ

The condition for constructive interference is d sin θ=n λ, so:

sin θ=n λ d

Substituting in the equation for s n:

s n=n λ D d

The linear separation on a screen of two consecutive maxima, s, is thus:

s=s n+1−s n=(n+1)λ D d−n λ D d=λ D d

s=λ D d

This last formula shows that the maxima of the interference pattern are equally separated. (Additional work shows that these maxima are also equally bright if the slit width is negligibly small.)

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Notice that the waves arriving at the two slits originate from the same source and so the phase difference between them does not change with time. We say that the waves are coherent sources with the same frequency and a constant phase difference between them. If the light originated from two separate sources, such as two flashlights, the phase difference between the two sources would be very rapidly changing. There would be an interference pattern, but the maxima and minima would be changing places so fast that we would see the average of many different interference patterns, i.e. uniform illumination on a screen and no fringes.

CHECK YOURSELF 4

Red light and blue light are incident on two narrow slits. Bright and dark fringes are formed on a screen beyond the slits. In which case is the fringe separation greater?

Worked Example 14.5

In a Young’s two-slit experiment, a source of light of unknown wavelength is used to illuminate two very narrow slits a distance of 0.150 mm apart. On a screen at a distance of 1.30 m from the slits, bright spots are observed separated by a distance of 4.95 mm. What is the wavelength of light being used?

Answer

Use the equation s=λ D d.

Rearranging to make λ the subject, and substituting the values from the question, we get:

λ=s d D=4.95×10−3×1.50×10−4 1.30=5.71×10−7 m

The wavelength of the light is 571 nm.

In Figure 14.35, the units on the vertical axis are arbitrary. An intensity of one unit corresponds to the intensity from only one slit. At points of constructive interference the amplitude is double that of just one wave. Since intensity is proportional to the square of the amplitude, the intensity is four times as large.

Figure 14.35: The intensity pattern for two slits of negligible width. The horizontal axis label y refers to the distance from the centre of the screen.

Worked Example 14.6

Use the graph in Figure 14.35 for this question. In a double-slit interference experiment the two slits are separated by a distance of 4.2 × 10−4 m and the screen is 3.8 m from the slits.

aDetermine the wavelength of light used in this experiment.

bSuggest the effect on the separation of the fringes of decreasing the wavelength of light.

cState the feature of the graph that enables you to deduce that the slit width is negligible.

Answer

aReading from the graph, the separation of the bright fringes is 0.50 cm.

Applying s=λ D d gives:

λ=d s D=4.2×10−4×0.50×10−2 3.8=5.5×10−7 m

bFrom the separation formula we see that if we decrease the wavelength the separation decreases.

cThe intensity of the side fringes is equal to the intensity of the central fringe.

Exam tip

‘Check the units!’

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nature of science

Competing theories and progress in science

In 1817 Augustin-Jean Fresnel published a new wave theory of light. The mathematician Siméon Poisson favoured the particle theory of light and worked out that Fresnel’s theory predicted the presence of a bright spot in the shadow of a circular object, which he believed was impossible. François Arago, a supporter of Fresnel, was able to show by experiment that there was indeed a bright spot in the centre of the shadow. In further support of his theory, Fresnel was able to show that the polarisation of light could only be explained if light was a transverse wave. The wave theory then took precedence until new evidence showed that light could behave as both a wave and a particle.

TEST YOUR UNDERSTANDING

18Planar waves of wavelength 1.0 cm approach an aperture whose opening is also 1.0 cm. Draw the wavefronts of this wave as they emerge through the aperture.

19Repeat question 18 for waves of wavelength 1.0 mm approaching an aperture of size 10 cm.

20Two speakers 16 m apart emit sound waves in phase of wavelength 0.80 m. Point P is a distance 9.0 m from the left speaker. (See Figure 14.36.)

Figure 14.36: For question 20.

What is observed at P?

21Two speakers emit sound waves in phase of wavelength 2.0 m, as shown in Figure 14.37.

Figure 14.37: For question 21.

aExplain what will be heard at point P.

bThe wavelength of the waves is changed to 1.6 m. Explain what will be observed at P now.

22A sound wave of wavelength 24 cm enters a tube as shown in Figure 14.38. The wave splits at A and recombines at B.

Figure 14.38: For question 22.

No sound is heard from the other end of the tube. What is the smallest possible distance x?

23Figure 14.39 shows wavefronts emitted from two identical sources.

Figure 14.39: For question 23.

State and explain what will be observed at points P, Q, R and S.

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24Figure 14.40 shows wavefronts emitted from two identical sources. An observer walks along the dotted line from P to Q.

Figure 14.40: For question 24.

On the axes, draw a sketch graph to show how the intensity of sound heard by the observer varies with distance from P.

25A radio station antenna, R, emits radio waves of wavelength 1600 m which reach a house, H, directly and after reflecting from a mountain, M, behind the house (see Figure 14.41). The reception at the house is very poor. Estimate the shortest possible distance between the house and the mountain. (The wave reflected from the mountain will undergo a phase change by π.)

Figure 14.41: For question 25.

26In a Young’s two-slit experiment, we have destructive interference at a point P on a screen. Where did the energy each of the waves carried to P go?

27In a Young’s two-slit experiment, a coherent source of light of wavelength 680 nm is used to illuminate two very narrow slits a distance of 0.12 mm apart. A screen is placed at a distance of 1.50 m from the slits. Calculate the separation of two successive bright spots.

28Explain why two identical flashlights pointing light at the same spot on a screen will never produce an observable interference pattern.

29In a Young’s two-slit experiment it is found that an nth-order maximum for a wavelength of 680.0 nm coincides with the (n + 1)th maximum of light of wavelength 510.0 nm. Determine n.

30The graph (Figure 14.42) shows the intensity pattern from a two-slit interference experiment.

Figure 14.42: For question 30.

aDetermine the separation of the slits in terms of the wavelength of light used.

bSuggest how the pattern in a changes if the slit separation is halved.

31Light is incident normally on two narrow parallel slits a distance of 1.00 mm apart. A screen is placed a distance of 1.2 m from the slits. The distance on the screen between the central maximum and the centre of the n = 4 bright spot is measured to be 3.1 mm.

aDetermine the wavelength of light.

bThis experiment is repeated in water. Suggest how the distance of 3.1 mm would change, if at all.

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32Figure 14.43 shows two radio sources 3.0 m apart that emit identical waves, in phase, of wavelength 0.60 m.

Figure 14.43: For question 32.

aA sensor is moved along the line PQ. How many maxima will the sensor record?

bHow many maxima would be recorded if the sensor moved in a full circle around the sources?

33In a two-slit experiment the angular separation θ of the central maximum and the first minimum is 0.012 rad. The wavelength of light used is 656 nm. (See Figure 14.44.)

Figure 14.44: For question 33.

aWhat is the slit separation d?

bWhat would be the effect on the angular separation θ, if any, if d is made smaller?

34In a two-slit experiment the screen is 1.20 m from the slits and the slit separation is 0.150 mm. The distance on the screen between the first minimum and the fourth minimum is 12.6 mm. What is the wavelength of light used in the experiment?