Chapter 21-22

Waves Optics: Interference of Light Waves

The Principle Superposition

...= + + =∑1 2 ii

D D D D

Di is displacement of the medium caused by wave i.The waves are

Added (constructive interference)Subtracted (destructive interference)

The Principle SuperpositionStanding waves

Standing wavesTwo waves have the same frequency, same amplitude and wavelength (one incident one reflected)The standing wave is a superposition of the two waves in opposite directions:

antinodesnodesIntensity proportional to the square of amplitude,

, ,= 2 ( )I cA where A is amplitude

Standing waves

Two waves traveling to the left and rightUse superposition principle:

SHMAmplitude function A(x)

( ), ,= + ( )R LD x t D D net dispacement of the medium

Mathematics of Standing Waves

Net displacement

The net displacement due to two counter-propagating sinusoidal wavesThese phenomena include:

NodesAntinodes

Wave OpticsWave optics is a study concerned with phenomena that cannot be adequately explained by geometric (ray) opticsThese phenomena include:

InterferenceDiffractionPolarization

The Path Difference and Phase Difference

( )Φ0avg

Path difference, ΔxPhase difference, ΔΦ

Average phase, Average distance to the two sources, avgx

The Phase Difference and wave number

ΔΔΦ = + ΔΦ02 xπ

λ

Mathematics of Interference

( )( )cos sinΔΦ⎡ ⎤= − + Φ⎢ ⎥⎣ ⎦0

22 av av

aD kr wtr

+= 1 2

2avr rr

Interference in One Dimension

Interference in Two and Three Dimensions

( )( )cos sinΔΦ⎡ ⎤= − + Φ⎢ ⎥⎣ ⎦0

22 av av

aD kr wtr

( ) ( )

= + =

− + + − +

1 2

1 10 1 20sin sin

D D Da akr ω t φ kr ω t φr r

Examples: Problem 2

Principle superposition comes into play whenever the pulses overlap

The overlapping parts of the two waves are shown by the dotted lines

Examples: Problem 4

The overlapping parts of the two pulses are shown by the dotted lines

Examples: Prob. 6

At t = 8 s both the reflected pulse waves are inverted and they are both moving to the leftA wave pulse reflected

from the string-wall boundary is inverted and its amplitude is unchanged. Phase shift 1800.

Examples: Prob. 6

At t = 8 s both the reflected pulse waves are inverted and they are both moving to the leftA wave pulse reflected

from the string-wall boundary is inverted and its amplitude is unchanged. Phase shift 1800.

Examples: Prob. 32

Because the wave pulses travel along the string at a speed of 100 m/s, they move a distance d = vt = (100 m/s)(0.05 s) = 5 m in 0.05 s

the snapshot graph of the wave pulse moving left at t= 0.05 s is the same as at t = 0 s (shown as a dotted line) except that it is shifted to the left by 5 m

Wave OpticsWave optics is a study concerned with phenomena that cannot be adequately explained by geometric (ray) opticsThese phenomena include:

InterferenceDiffractionPolarization

InterferenceIn constructive interference the amplitude of the resultant wave is greater than that of either individual waveIn destructive interference the amplitude of the resultant wave is less than that of either individual waveAll interference associated with light waves arises when the electromagnetic fields that constitute the individual waves combine

Conditions for InterferenceTo observe interference in light waves, the following two conditions must be met:1) The sources must be coherent

They must maintain a constant phase with respect to each other

2) The sources should be monochromaticMonochromatic means they have a single wavelength

Producing Coherent SourcesLight from a monochromatic source is used to illuminate a barrierThe barrier contains two narrow slits

The slits are small openingsThe light emerging from the two slits is coherent since a single source produces the original light beamThis is a commonly used method

DiffractionFrom Huygens’s principle we know the waves spread out from the slitsThis divergence of light from its initial line of travel is called diffraction

Young’s Double-Slit Experiment: SchematicThomas Young first demonstrated interference in light waves from two sources in 1801The narrow slits S1 and S2 act as sources of wavesThe waves emerging from the slits originate from the same wave front and therefore are always in phase

Resulting Interference PatternThe light from the two slits forms a visible pattern on a screenThe pattern consists of a series of bright and dark parallel bands called fringesConstructive interferenceoccurs where a bright fringe occursDestructive interferenceresults in a dark fringe

Interference PatternsConstructive interference occurs at point PThe two waves travel the same distance

Therefore, they arrive in phase

As a result, constructive interference occurs at this point and a bright fringe is observed

Interference Patterns, 2The upper wave has to travel farther than the lower wave to reach point QThe upper wave travels one wavelength farther

Therefore, the waves arrive in phase

A second bright fringe occurs at this position

Interference Patterns, 3The upper wave travels one-half of a wavelength farther than the lower wave to reach point RThe trough of the bottom wave overlaps the crest of the upper waveThis is destructive interference

A dark fringe occurs

Young’s Double-Slit Experiment: GeometryThe path difference, δ, is found from the tan triangleΔr = r2 – r1 = d sin θ

This assumes the paths are parallelNot exactly true, but a very good approximation if L is much greater than d

Interference EquationsFor a bright fringe produced by constructive interference, the path difference must be either zero or some integral multiple of the wavelengthΔr = d sin θ bright = mλ

m = 0, ±1, ±2, … m is called the order number

When m = 0, it is the zeroth-order maximumWhen m = ±1, it is called the first-order maximum

Interference Equations, 2When destructive interference occurs, a dark fringe is observedThis needs a path difference of an odd half wavelengthΔr = d sin θdark = (m + ½)λ

m = 0, ±1, ±2, …

Interference Equations, 4The positions of the fringes can be measured vertically from the zeroth-order maximumAssumptions

L >> dd >> λ

Approximation:θ is small and therefore the small angle approximation tan θ ~ sin θ can be used

y = L tan θ ≈ L sin θ

Interference Equations, finalFor bright fringes

For dark fringes

bright ( 0 1 2 ), ,λLy m md

= = ± ± K

dark1 ( 0 1 2 )2

, ,λLy m md⎛ ⎞= + = ± ±⎜ ⎟⎝ ⎠

K

Uses for Young’s Double-Slit Experiment

Young’s double-slit experiment provides a method for measuring wavelength of the lightThis experiment gave the wave model of light a great deal of credibility

It was inconceivable that particles of light could cancel each other in a way that would explain the dark fringes

Intensity Distribution: Double-Slit Interference Pattern

The bright fringes in the interference pattern do not have sharp edges

The equations developed give the location of only the centers of the bright and dark fringes

We can calculate the distribution of light intensity associated with the double-slit interference pattern

Intensity Distribution, Assumptions

Assumptions:The two slits represent coherent sources of sinusoidal wavesThe waves from the slits have the same angular frequency, ωThe waves have a constant phase difference, φ

The total magnitude of the electric field at any point on the screen is the superposition of the two waves

Intensity Distribution, Electric Fields

The magnitude of each wave at point P can be found

E1 = Eo sin ωtE2 = Eo sin (ωt + φ)Both waves have the same amplitude, Eo

Intensity Distribution, Phase Relationships

The phase difference between the two waves at P depends on their path difference

Δr = r2 – r1 = d sin θA path difference of λ corresponds to a phase difference of 2π radA path difference of δ is the same fraction of λas the phase difference φ is of 2πThis gives = Δ =

2 2 sin π πφ d θλ λ

Intensity Distribution, Resultant Field

The magnitude of the resultant electric field comes from the superposition principle

EP = E1+ E2 = Eo[sin ωt + sin (ωt + φ)]

This can also be expressed as

EP has the same frequency as the light at the slitsThe magnitude of the field is multiplied by the factor 2 cos (φ / 2)

2 cos sin2 2P oφ φE E ωt⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

Intensity Distribution, EquationThe expression for the intensity comes from the fact that the intensity of a wave is proportional to the square of the resultant electric field magnitude at that pointThe intensity therefore is

2 2max max

sin cos cosI I Iπd θ πd yλ λL

⎛ ⎞ ⎛ ⎞= ≈⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Light Intensity, GraphThe interference pattern consists of equally spaced fringes of equal intensityThis result is valid only if L >> d and for small values of θ

Phase Changes Due To ReflectionAn electromagnetic wave undergoes a phase change of 180° upon reflection from a medium of higher index of refraction than the one in which it was traveling

Analogous to a pulse on a string reflected from a rigid support

Phase Changes Due To Reflection, cont.

There is no phase change when the wave is reflected from a boundary leading to a medium of lower index of refraction

Analogous to a pulse on a string reflecting from a free support

Interference in Thin FilmsInterference effects are commonly observed in thin films

Examples include soap bubbles and oil on water

The varied colors observed when white light is incident on such films result from the interference of waves reflected from the two surfaces of the film

Interference in Thin Films, 2Facts to note

An electromagnetic wave traveling from a medium of index of refraction n1 toward a medium of index of refraction n2 undergoes a 180° phase change on reflection when n2 > n1

There is no phase change in the reflected wave if n2 < n1

The wavelength of light λn in a medium with index of refraction n is λn = λ/n where λ is the wavelength of light in vacuum

Interference in Thin Films, 3Assume the light rays are traveling in air nearly normal to the two surfaces of the filmRay 1 undergoes a phase change of 180° with respect to the incident rayRay 2, which is reflected from the lower surface, undergoes no phase change with respect to the incident wave

Interference in Thin Films, 4Ray 2 also travels an additional distance of 2tbefore the waves recombineFor constructive interference

2nt = (m + ½)λ (m = 0, 1, 2 …)This takes into account both the difference in optical path length for the two rays and the 180° phase change

For destructive interference2nt = mλ (m = 0, 1, 2 …)

Interference in Thin Films, 5Two factors influence interference

Possible phase reversals on reflectionDifferences in travel distance

The conditions are valid if the medium above the top surface is the same as the medium below the bottom surface

If there are different media, these conditions are valid as long as the index of refraction for both is less than n

Interference in Thin Films, 6If the thin film is between two different media, one of lower index than the film and one of higher index, the conditions for constructive and destructive interference are reversedWith different materials on either side of the film, you may have a situation in which there is a 180o phase change at both surfaces or at neither surface

Be sure to check both the path length and the phase change

Interference in Thin Film, Soap Bubble Example

Newton’s RingsAnother method for viewing interference is to place a plano-convex lens on top of a flat glass surfaceThe air film between the glass surfaces varies in thickness from zero at the point of contact to some thickness tA pattern of light and dark rings is observed

These rings are called Newton’s ringsThe particle model of light could not explain the origin of the rings

Newton’s rings can be used to test optical lenses

Newton’s Rings, Set-Up and Pattern

Problem Solving Strategy with Thin Films, 1

Identify the thin film causing the interferenceThe type of interference – constructive or destructive – that occurs is determined by the phase relationship between the upper and lower surfaces

Problem Solving with Thin Films, 2

Phase differences have two causesdifferences in the distances traveledphase changes occurring on reflection

Both causes must be considered when determining constructive or destructive interferenceThe interference is constructive if the path difference is an integral multiple of λ and destructive if the path difference is an odd half multiple of λ

Two pulses on a string approach each other at speeds of 1 m/s. What is the shape of the string at t = 6 s?

(1) (2) (3) (4)

Two pulses on a string approach each other at speeds of 1 m/s. What is the shape of the string at t = 6 s?

(1) (2) (4)(3)

A standing wave on a string vibrates as shown at the top. Suppose the tension is quadrupled while the frequency and the length of the string are held constant. Which standing wave pattern is produced?

(1) (2) (3) (4) (5)

A standing wave on a string vibrates as shown at the top. Suppose the tension is quadrupled while the frequency and the length of the string are held constant. Which standing wave pattern is produced?

(1) (2) (3) (4) (5)

Two loudspeakers emit waves with λ = 2.0 m. Speaker 2 is 1.0 m in front of speaker 1. What, if anything, must be done to cause constructive interference between the two waves?

1. Move speaker 1 backward (to the left) 0.5 m.2. Move speaker 1 backward (to the left) 1.0 m.3. Move speaker 1 forward (to the right) 1.0 m.4. Move speaker 1 forward (to the right) 0.5 m.5. Nothing. The situation shown already causes

constructive interference.

The interference at point C in the figure at the right is

1. maximum constructive.2. destructive, but not

perfect.3. constructive, but less than

maximum.4. there is no interference at

point C.5. perfect destructive.

The interference at point C in the figure at the right is

1. maximum constructive.2. destructive, but not

perfect.3. constructive, but less than

maximum.4. there is no interference at

point C.5. perfect destructive.

These two loudspeakers are in phase. They emit equal-amplitude sound waves with a wavelength of 1.0 m. At the point indicated, is the interference maximum constructive, perfect destructive or something in between?

1. maximum constructive 2. perfect destructive3. something in between

These two loudspeakers are in phase. They emit equal-amplitude sound waves with a wavelength of 1.0 m. At the point indicated, is the interference maximum constructive, perfect destructive or something in between?

1. maximum constructive 2. perfect destructive3. something in between

Chapter 21Reading Quiz

When a wave pulse on a string reflects from a boundary, how is the reflected pulse related to the incident pulse?

1. Shape unchanged, amplitude unchanged2. Shape inverted, amplitude unchanged3. Shape unchanged, amplitude reduced4. Shape inverted, amplitude reduced5. Amplitude unchanged, speed reduced

When a wave pulse on a string reflects from a boundary, how is the reflected pulse related to the incident pulse?

1. Shape unchanged, amplitude unchanged2. Shape inverted, amplitude unchanged3. Shape unchanged, amplitude reduced4. Shape inverted, amplitude reduced5. Amplitude unchanged, speed reduced

There are some points on a standing wave that never move. What are these points called?

1. Harmonics2. Normal Modes3. Nodes4. Anti-nodes5. Interference

There are some points on a standing wave that never move. What are these points called?

1. Harmonics2. Normal Modes3. Nodes4. Anti-nodes5. Interference

Two sound waves of nearly equal frequencies are played simultaneously. What is the name of the acoustic phenomena you hear if you listen to these two waves?

1. Beats2. Diffraction3. Harmonics4. Chords5. Interference

Two sound waves of nearly equal frequencies are played simultaneously. What is the name of the acoustic phenomena you hear if you listen to these two waves?

1. Beats2. Diffraction3. Harmonics4. Chords5. Interference

The various possible standing waves on a string are called the

1. antinodes.2. resonant nodes.3. normal modes.4. incident waves.

The various possible standing waves on a string are called the

1. antinodes.2. resonant nodes.3. normal modes.4. incident waves.

The frequency of the third harmonic of a string is

1. one-third the frequency of the fundamental.2. equal to the frequency of the fundamental.3. three times the frequency of the fundamental.4. nine times the frequency of the fundamental.

The frequency of the third harmonic of a string is

1. one-third the frequency of the fundamental.2. equal to the frequency of the fundamental.3. three times the frequency of the fundamental.4. nine times the frequency of the fundamental.

Chapter 22

A. They get brighter but otherwise do not change.B. They get brighter and closer together.C. They get brighter and farther apart.D. They get out of focus.E. They fade out and disappear.

Suppose the viewing screen in the figure is moved closer to the double slit. What happens to the interference fringes?

A. They get brighter but otherwise do not change.B. They get brighter and closer together.C. They get brighter and farther apart.D. They get out of focus.E. They fade out and disappear.

Suppose the viewing screen in the figure is moved closer to the double slit. What happens to the interference fringes?

Light of wavelength λ1 illuminates a double slit, and interference fringes are observed on a screen behind the slits. When the wavelength is changed to λ2, the fringes get closer together. How large is λ2 relative to λ1?

Α. λ2 is larger than λ1.B. λ2 is smaller than λ1.C. Cannot be determined from this information.

Light of wavelength λ1 illuminates a double slit, and interference fringes are observed on a screen behind the slits. When the wavelength is changed to λ2, the fringes get closer together. How large is λ2 relative to λ1?

Α. λ2 is larger than λ1.B. λ2 is smaller than λ1.C. Cannot be determined from this information.

White light passes through a diffraction grating and forms rainbow patterns on a screen behind the grating. For each rainbow,

A. the red side is on the right, the violet side on the left.B. the red side is on the left, the violet side on the right.C. the red side is closest to the center of the screen, the

violet side is farthest from the center.D. the red side is farthest from the center of the screen,

the violet side is closest to the center.

White light passes through a diffraction grating and forms rainbow patterns on a screen behind the grating. For each rainbow,

A. the red side is on the right, the violet side on the left.B. the red side is on the left, the violet side on the right.C. the red side is closest to the center of the screen, the

violet side is farthest from the center.D. the red side is farthest from the center of the

screen, the violet side is closest to the center.

A Michelson interferometer using light of wavelength λhas been adjusted to produce a bright spot at the center of the interference pattern. Mirror M1 is then moved distance λ toward the beam splitter while M2 is moved distance λaway from the beam splitter. How many bright-dark-bright fringe shifts are seen?

A. 0B. 1C. 2D. 3 E. 4

A Michelson interferometer using light of wavelength λhas been adjusted to produce a bright spot at the center of the interference pattern. Mirror M1 is then moved distance λ toward the beam splitter while M2 is moved distance λaway from the beam splitter. How many bright-dark-bright fringe shifts are seen?

A. 0B. 1C. 2D. 3 E. 4

Chapter 22Reading Quiz

What was the first experiment to show that light is a wave?

A. Young’s double slit experimentB. Galileo’s observation of Jupiter’s moonsC. The Michelson-Morley interferometerD. The Pound-Rebka experimentE. Millikan’s oil drop experiment

What was the first experiment to show that light is a wave?

A. Young’s double slit experimentB. Galileo’s observation of Jupiter’s moonsC. The Michelson-Morley interferometerD. The Pound-Rebka experimentE. Millikan’s oil drop experiment

What is a diffraction grating?

A. A device used to grate cheese and other materialsB. A musical instrument used to direct soundC. A plaque with a tiny circular apertureD. An opaque objects with many closely spaced slitsE. Diffraction gratings are not covered in Chapter 22.

What is a diffraction grating?

A. A device used to grate cheese and other materialsB. A musical instrument used to direct soundC. A plaque with a tiny circular apertureD. An opaque objects with many closely spaced slitsE. Diffraction gratings are not covered in Chapter 22.

When laser light shines on a screen after passing through two closely spaced slits, you see

A. a diffraction pattern.B. interference fringes.C. two dim, closely spaced points of light.D. constructive interference.

When laser light shines on a screen after passing through two closely spaced slits, you see

A. a diffraction pattern.B. interference fringes.C. two dim, closely spaced points of light.D. constructive interference.

This chapter discussed the

A. acoustical interferometer.B. Michelson interferometer. C. Fabry-Perot interferometer.D. Both A and B. E. Both B and C.

This chapter discussed the

A. acoustical interferometer.B. Michelson interferometer.C. Fabry-Perot interferometer.D. Both A and B. E. Both B and C.

The spreading of waves behind an aperture is

A. more for long wavelengths, less for short wavelengths.

B. less for long wavelengths, more for short wavelengths.

C. the same for long and short wavelengths.

D. not discussed in this chapter.

The spreading of waves behind an aperture is

A. more for long wavelengths, less for short wavelengths.

B. less for long wavelengths, more for short wavelengths.

C. the same for long and short wavelengths.

D. not discussed in this chapter.

Apertures for which diffraction is studied in this chapter are

A. a single slit.B. a circle. C. a square. D. both A and B.E. both A and C.

Apertures for which diffraction is studied in this chapter are

A. a single slit.B. a circle. C. a square. D. both A and B.E. both A and C.