1

SuperpositionSuperposition Superposition

• 16. Superposition• Content• 16.1 Stationary waves• 16.2 Diffraction• 16.3 Interference• 16.4 Two-source interference patterns• 16.5 Diffraction grating• Learning Outcomes• * (a) explain and use the principle of superposition in simple applications.• * (b) show an understanding of experiments which demonstrate stationary waves

using microwaves, stretched strings and air columns.• * (c) explain the formation of a stationary wave using a graphical method, and

identify nodes and antinodes.• (d) explain the meaning of the term diffraction.• (e) show an understanding of experiments which demonstrate diffraction including

the diffraction of water waves in a ripple tank with both a wide gap and a narrow gap.

• (f) show an understanding of the terms interference and coherence.

2

• (g) show an understanding of experiments which demonstrate two-source interference using water, light and microwaves.

• (h) show an understanding of the conditions required if two-source interference fringes are to be observed.

• (i) recall and solve problems using the equation λ = ax/D for double-slit interference using light.

• (j) recall and solve problems using the formula d sin θ = nλ and describe the use of a diffraction grating to determine the wavelength of light. (The structure and use of the spectrometer is not included.)

3

Wave interference Wave interference is the phenomenon which occurs when two

waves from 2 coherent sources meet while travelling along the same medium.

2 waves are said to be coherent if They produce waves of the same frequency They produce waves of the same phase (or with

constant phase difference) The interference of waves causes the medium to take on a shape

which results from the net effect of the two individual waves upon the particles of the medium

When waves are produced on the surface of water, the wave crests will act like a convex lens while the troughs will act like a concave lens causing bright and dark fringes

Waves interference can be constructive or destructive A wave-front is a line that joins all the points vibrating in-phase and

is represented by the bright and dark fringes or maxima and minima respectively, collectively called the interference pattern

4

Principle of Superposition

The task of determining the shape of the resultant demands that the principle of superposition is applied.

The principle of superposition is generally stated as follows:

When two or more waves interfere i.e. meet at the same point, the resulting displacement is the algebraic sum of the displacements of the individual waves at that same point

• The principle applies to all types of waves

5

Constructive Interference This is the superposition of 2 waves which are in phase to produce a

resultant wave of maximum amplitude of the same original frequency

In this example, amplitude A + amplitude A = 2A

6

Destructive Interference

This is the superposition of 2 waves which are in anti-phase to produce a resultant wave of zero amplitude

In this example, amplitude A – amplitude A = 0

7

Two waves traveling in the same direction:

Constructive and Destructive Interference

8

Two waves traveling in opposite directions: Constructive and

Destructive Interference

9

Path or phase difference

• Constructive interference occurs when the wave amplitudes reinforce each other, building a wave of even greater amplitude.

• This happens when the waves are in phase i.e. path difference is a whole number of wavelengths, nλ

• Destructive interference occurs when the wave amplitudes oppose each other, resulting in waves of reduced amplitude i.e. path difference is an odd number of wavelengths, (n + ½)λ

10

=

WAVE IN PHASE Constructive interference

Waves in anti-phase

=

Destructive interference

11

Demonstration of interference pattern with a) sound waves b) light waves

• By using a signal generator and 2 speakers (pg 216 Chris Mee fig 8.25)

• For light waves, must have 2 light sources of the same single frequency and not a mixture of frequencies. Also must have a constant phase relationship i.e. coherent

12

Diffraction

• Normally light is thought of as traveling in straight lines, but when light travels through an aperture, passes the edge of an object or passes round a small obstacle, it deviates from the straight-on direction and spreads out like waves

• Greatest effect occurs when the wavelength is about the same size as the aperture, or the gap through which the wave is passing is narrower

• The smaller the size of the obstacle, the greater is the effect of spreading

• This effect is known as diffraction and is powerful evidence that light has wave properties leading to the wave theory of light

• The amplitude of the diffracted wave is smaller than that of the incident wave

• Although it was Sir Isaac Newton who attempted to explain diffraction (but not based on the wave theory of light), it was Christian Huygens and later Augustin Fresnel who developed the wave theory of light

13

Diffraction

• When diffraction occurs:– Wavelength is unchanged– Frequency is unchanged– Speed is unchanged– Wave direction changes hence– Wave velocity changes

14

Huygen’s wavelet explanation

• Huygen’s suggested that at any instant, all points on a wavefront could be regarded as secondary disturbances giving rise to their own outspreading circular wavelets (pg 230 Chris Mee fig 8.49)

• This causes interference patterns • The central region of the pattern is a broad bright area with narrow

dark fringes on either side, beyond which are further successions of bright and dark areas becoming less and less intense

• Path difference and wavelength explanation and equation (pg 231 Chris Mee fig 8.51)

• Path difference between 2 rays:

½ d sin θ = ½ λ where, d is the width of slit, λ is wavelength, or generally,

d sin θ = nλ where n is an integer

Huygen’s principle

• The propagation of light wave can be predicted by assuming that each point on the wave-front acts as a source for secondary wavelets which spread out at all directions.

• The envelope of these 2nd wavelets after a small period of time is the new wave-front.

Wave-front – the large wave Wavelets – small waves

16

Thomas Young’s double-slit experiment – 1801

• (pg 217 Chris Mee fig 8.29)• Monochromatic light > single slit > double slit > dark and bright

interference patterns on screen• Light from monochromatic source; since monochromatic, is of the

same frequency and phase, is diffracted at the slit, producing 2 light sources at the double slit

• This creates an interference pattern of dark and bright fringes on a screen

• The distance x between successive bright fringes on the screen is called the fringe width

• For small angles since sin θ ≈ tan θ the fringe width is related to the wavelength λ of the light source by the equation:

x = λL/d or λ = dx/Lwhere x is the fringe width, L is the distance between the double

slits to the screen, d is the distance between the centre of the slits (distance between 2 coherent sources)

Interference – Young's experiment

sind

Originally performed by Young (1801) to demonstrate the wave-nature of light. Has now been done with electrons, neutrons, He atoms among others.

D

θd

Detecting screen

Incoming coherent beam of particles (or light)

y

Alternative method of detection: scan a detector across the plane and record number of arrivals at each point

THE DOUBLE-SLIT EXPERIMENT

19

Exercises

• Calculate the observed fringe width for a Young’s double slit experiment using light of wavelength 600 nm and slits 0.50 mm apart. The distance from the slits to the screen is 0.80 m. (ans: 0.96 mm)

• 2 loudspeakers connected to the same signal generator produce sounds of frequency 1000 Hz. The separation between the 2 speakers is 2.0 m. An observer who stands 5.0 m in front hears 3 consecutive loud sounds when he moves through a distance of 1.65 m in the direction parallel to the speakers.

Determine a) the wavelength of the sound (ans: 0.33 m)

b) the speed of the sound waves in air (ans: 330 ms-1)

20

Example• A Young’s double slit interferometer with green light of wavelength

546 nm. The double slits are 0.100 mm apart and the screen is 20 cm away from the double slits. Find

i) the angular position of the 1st minimum

ii) the angular position of the 5th maximum

iii) the distance on the screen between the 1st maximum and the 5th maximum

Solution

i) at the 1st maximum the path difference is ½λ;

using d sin θ = λ/2, θ = 0.156°

ii) at the 5th maximum, path difference is 5λ

using d sin θ = 5λ, θ = 1.56°

iii) the fringe spacing is given by x = λL/d = 1.09 mm

hence distance between 1st and 5th maxima is 4x the fringe separation = 4.37 mm

21

Example

• 2 narrow parallel slits are illuminated by light of 2 wavelengths, λ1 = 600nm and λ2 is unknown. On the interference patterns produced on the screen, the 4th bright fringe for λ1 coincides with the 6th bright fringe for λ2 . Find λ2

Solution

Since the 4th fringe and the 6th fringe coincide, this means that the fringes lie at the same angular position.

i.e d sin θ = 4λ1 and d sin θ = 6λ2

therefore 4λ1 = 6λ2 and hence λ2 = 400 nm

22

The diffraction grating

• This is a plate on which there is a large number of parallel, identical, very closely spaced slits

• When monochromatic light is incident on this plate, a pattern of narrow bright fringes is produced due to interference and the principle of superposition

• To obtain constructive interference, the path difference should be an integral number of wavelengths

• (see page 233/234 of A/AS Physics by Chris Mee for explanation and diagram and different orders of diffraction)

• d sin θ = nλ

23

Example

• Monochromatic light is incident on a diffraction grating with 7.00 x 105 lines per meter. A second-order maximum is observed at an angle of diffraction of 40.0°. Calculate the wavelength of the incident light.

Solution

The relationship between the slit spacing and the number N of lines per metre is d = 1/N

Using d sin θ = nλ, λ = 460 nm

24

Example

• A diffraction grating is ruled with 3500 lines per cm. parallel light of wavelength 600 nm is viewed through the grating. Find

i) the maximum number of orders of reinforcement that can be seen through the grating

ii) the number of reinforcement maxima that will be observed

Solution

i) using d sin θ = nλ, n = d sin θ/λ

but n will be maximum when sin θ = 1

so n = d/λ = 4.76 i.e max order is 4

ii) the number of reinforcement maxima will be 9 (1 central beam, plus 4 on each side)

25

The diffraction grating with white light

• When white light is incident on a diffraction grating, each wavelength making up the white light is diffracted by a different amount as described by nλ = d sin θ

• In the visible spectrum(VIBGYOR), red light ,because it has the longest wavelength is diffracted through the largest angle, whereas violet is the least diffracted

• Hence a continuous spectrum is produced• This phenomenon is an important use of diffraction grating in a

spectrometer, used to investigate spectra• By determining the angle of the diffracted image, the wavelength of

light producing that image can be determined

26

Stationary or standing waves

• This is the phenomenon when 2 progressive waves of equal amplitude and frequency travel along the same line with the same speed but in opposite directions [recap: Waves which move energy from place to place are called progressive waves. Waves that do not are called stationary waves]

• The waves interfere producing a wave pattern in which the crests and troughs do not move, unlike progressive waves

• They travel along with the same speed e.g. strings in musical instruments

• Within a stationary/standing wave, regions of constructive interference are called antinodes and regions of destructive interference are called nodes. 

• Nodes and antinodes do not move along the string

27

28

Experiment:- Stretched StringExperiment:- Stretched String

video

29

Stationary waves on strings

• If a stretched string is plucked and allowed to vibrate freely, there are certain frequencies at which it will vibrate with a large amplitude

• This is known as the resonance frequency• The simplest way a stretched string vibrates, has a wave pattern

which has a single loop called the fundamental mode of vibration or the 1st harmonic

30

31

N       -       A       -       Nfundamental frequency/1st harmonic, fo

L = 1 loop = ½ o

N    -  A  -   N   -  A   -   N

1st overtone/2nd harmonic, f1 = 2fo

L = 2 loops = 1

N - A - N - A - N - A- N

2nd overtone/3rd harmonic, f2 = 3fo

L = 3 loops = 1.5 2

32

General string expression

• v = fλ , where v is the speed of the progressive waves which have interfered to produce the stationary wave

• λ of fundamental or 1st harmonic is 2L, hence f0 is v/2L

• λ of 1st overtone/2nd harmonic is L, hence f1 is v/L

• Hence general expression is fn = vn/2L where n is an integer

33

Key features distinguishing stationary waves from progressive waves

• In stationary waves, the nodes & antinodes do not move along the string, whereas in progressive waves, the crests and troughs do move along it

• In stationary waves, the amplitude of vibration varies with position along the string: zero at the node & maximum at an antinode. In a progressive wave, all points have the same amplitude

• Between adjacent nodes, all points of the stationary wave vibrate in phase. i.e. all particles of the string are at their max displacement at the same instant. In a progressive wave, phase varies continuously along the wave

34

Stationary waves explained by interference

• Pg 224 Chris Mee fig 8.41

35

Experiment:- stationary waves in a closed Experiment:- stationary waves in a closed tube or pipe - air columnstube or pipe - air columns

36

Air columns

• For stationary waves in a closed pipe, the air cannot move at the closed end, so it is always a node, N

• The open end is a position of maximum disturbance and hence is an antinode, A

• The particular frequencies at which stationary waves are obtained in a pipe are the resonant frequencies of the pipe

37

Standing waves in open ended pipe on both sides

Standing wave in one closed end and one open ended pipe

38

39

40

Stationary waves using microwaves

• (pg 226 Chris Mee)• Using a microwave source and a metal reflector, and moving a

probe detector (connected to a meter) in between slowly towards and away from the source, the metered signal fluctuates from a minimum to a maximum

• The minima are nodes whereas the maxima are antinodes• The distance between 2 successive nodes or antinodes is half

the wavelength of the microwaves

41

Measuring speed of sound using stationary waves using resonance method

• (pg 226 Chris Mee)• Uses the principle of resonance in a tube closed at one end• A glass tube is placed in a cylinder of water and by raising the tube,

the air column can be increased• A tuning fork of known frequency is used to cause resonance• 1st loud sound is at λ/4, next loud sound at 3λ/4

• Difference is λ/2 = L2 - L1

since c = fλ , c = 2f(L2 - L1)

42

Speed of sound using stationary waves in free air

• (pg 227 Chris Mee)• A signal generator and loudspeaker is used to produce a note of

known frequency • A reflector is used• A microphone and oscilloscope are used to find 2 nodes (i.e.

minimum amplitude points)• Distance between 2 nodes d = λ/2 • Using c = fλ , c = 2fd

43

ExampleThe string at the right is 1.5 meters long and is vibrating as the first harmonic. The string vibrates up and down with 33 complete vibrational cycles in 10 seconds. Determine the frequency, period, wavelength and speed for this wave.

Solution:-

Given: L = 1.5 m33 cycles in 10 secondsSo,f = (33 cycles) / (10 seconds) = 3.3 Hz

T = 1 / (3.3 Hz) = 0.303 seconds

= 2 • L = 2 • (1.5 m) = 3.0 m The speed of a wave can be calculated from its wavelength and frequency using the wave equation:v = f • = (3.3 Hz) • (3. 0 m) = 9.9 m/s

44

Example:A stretched wire of length 60.0 cm vibrates transversely. Waves travel along the wire at a speed 210 m/s. Three antinodes can be found in the stationary waves formed in between the two ends of the wire. Determine

a. the wavelength of the waves which move along the wireb. the frequency

Solution:

L

Antinodes½