Wave Optics

Chapter 25

Wave OpticsThe field of wave optics studies the properties of

light that depend on its wave natureOriginally light was thought to be a particle and that

model successfully explained the phenomena discussed in geometric options

Other experiments revealed properties of light that could only be explained with a wave theory

Maxwell’s theory of electromagnetism convinced physicists that light was a wave

Wave vs. Geometric OpticsThe wavelength of light plays a key role in determining

when geometric optics can or cannot be usedWhen discussing image characteristics over distances

much greater than the wavelength, geometric optics is extremely accurate

When dealing with sizes comparable to or smaller than the wavelength, wave optics is requiredExamples include interference effects and propagation

through small openingsEven more experiments led to the quantum theory of

lightLight has properties of both waves and particles

InterferenceOne property unique to

waves is interferenceInterference of sound

waves can be produced by two speakers

When the waves are in phase, their maxima occur at the same time at a given point in space

Section 25.1

Interference, cont.The total wave displacement at the listener’s

location is the sum of the displacements of the two individual waves

If two waves are in phase, the sum of their displacements is largeThe wave interfere constructivelyThis produces a large amplitude and a large intensity

Section 25.1

Interference, finalThe maximum of one wave can coincide with the

minimum of the other waveThese waves are out of phase

The interference is destructive when the waves are out of phaseIf the waves are 180° out of phase, the sum of the

displacements of the two waves is zero

Section 25.1

Conditions for InterferenceTwo or more interfering waves travel through

different regions of space over at least part of their propagation from source to destination

The waves are brought together at a common pointThe waves must have the same frequency and must

also have a fixed phase relationshipThis means that over a given distance or time interval

the phase difference between the waves remains constant

Such waves are called coherent

Section 25.1

CoherenceThe eye cannot follow variations of every cycle of

the wave, so it averages the light intensityFor waves to interfere constructively, they must stay

in phase during the time the eye is averaging the intensity

For waves to interfere destructively, they must stay out of phase during the averaging time

Both of these possibilities involve the wave having precisely the same frequency

Section 25.1

Coherence, cont.

With slightly different frequencies, the interference changes from constructive to destructive and back

Over a large number of cycles, the waves average no interference

Section 25.1

Michelson InterferometerThe Michelson

interferometer is based on the interference of reflected waves

Two reflecting mirrors are mounted at right angles

A third mirror is partially reflectingCalled a beam splitter

Section 25.2

Michelson Interferometer, cont.The incident light hits the beam splitter and is

divided into two wavesThe waves reflect from the mirrors at the top and

right and recombine at the beam splitterThe only difference between the two waves is that

they travel different distances between their respective mirrors and the beam splitter

The path length difference is ΔL = 2L2 – 2L1

Section 25.2

Michelson Interferometer, finalThe path length

difference is related to the wavelength of the light

If N is an integer, the two waves are in phase and produce constructive interference

If N is a half-integer the waves will produce destructive interference

LN

Section 25.2

Interference ConditionsFor constructive interference, ΔL = m λFor destructive interference, ΔL = (m + ½) λ

m is an integer in both casesIf the interference is constructive, the light intensity

at the detector is largeCalled a bright fringe

If the interference is destructive, the light intensity at the detector is zeroCalled a dark fringe

Section 25.2

Measuring Length with a Michelson InterferometerUse the light from a laser and adjust the mirror to

give constructive interferenceThis corresponds to one of the bright fringes

The mirror is then moved, changing the path lengthThe intensity changes from high to zero and back to

high every time the path length changes by one wavelength

If the mirror moves through N bright fringes, the distance d traveled by the mirror is

2N

d Section 25.2

Measuring Length, cont.The accuracy of the measurement depends on the

accuracy with which the wavelength is knownMany laboratories use helium-neon lasers to make

very precise length measurementsA similar type of interference effect is used to read

information from CDs and DVDs

Section 25.2

LIGOLIGO – Laser Interferometer Gravitational Wave

ObservatoryDesigned to detect very small vibrations associated

with gravitational waves that arrive at the Earth from distant galaxies

By using a long distance between the beam splitter and the mirrors, the LIGO interferometers are sensitive to very small percentage changes in that distance

Section 25.2

Thin-Film Interference

Assume a thin soap film rests on a flat glass surfaceThe upper surface of the soap film is similar to the

beam splitter in the interferometerIt reflects part of the incoming light and allows the

rest to be transmitted into the soap layer after refraction at the air-soap interface

Section 25.3

Thin-Film Interference, cont.The transmitted ray is partially reflected at the

bottom surfaceThe two outgoing rays meet the conditions for

interferenceThey travel through different regions

One travels the extra distance through the soap film

They recombine when they leave the filmThey are coherent because they originated from the

same source and initial ray

Section 25.3

Thin-Film Interference, finalThe index of refraction of the film also needs to be

accounted forFrom the speed of the wave inside the film

The wavelength changes as the light wave travels from a vacuum into the filmThe frequency does not change

The number of extra wavelengths is

ƒfilm filmfilm

cn

Section 25.3

2

film

dN

Frequency of a Wave at an InterfaceWhen a light wave

passes from one medium to another, the waves must stay in phase at the interface

The frequency must be the same on both sides of the interface

Section 25.3

Phase Change and ReflectionWhen a light wave reflects from a surface it may be

invertedInversion corresponds to a phase change of 180°

There is a phase change whenever the index of refraction on the incident side is less than the index of refraction of the opposite side

If the index of refraction is larger on the incident side the reflected ray in not inverted and there is no phase change

Section 25.3

Phase Change and Reflection, Diagram

Section 25.3

Phase Changes in a Thin FilmThe total phase change in a thin film must be

accounted forThe phase difference due to the extra distance

traveled by the rayAny phase change due to reflection

For a soap film on glass, nair < nfilm < nglass

There are phase changes for both reflections at the soap film interfacesThe reflections at both the top and bottom surfaces

undergo a 180° phase changeSection 25.3

Thin-Film Interference, Case 1Both waves reflected by a thin film undergo a phase

changeThe number of extra cycles traveled by the ray

inside the film completely determines the nature of the interference

If the number of extra cycles, N, is an integer, there is constructive interference

If the number of extra cycles is a half-integer, there is destructive interference

Section 25.3

Case 1, cont.Equations are

These equations apply whenever nair < nfilm < n(substance

below the film)

2

12

2

film

film

md constructive interference

n

md destructive interference

n

Section 25.3

Thin-Film Interference, Case 2Assume the soap bubble is surrounded by airThere is a phase change at the top of the bubbleThere is no phase change at the bottom of the

bubbleSince only one wave undergoes a phase change,

the interference conditions are12

2

2

film

film

md constructive interference

n

md destructive interference

n

Section 25.3

Thin-Film Interference: White LightEach color can interfere

constructively, but at different anglesBlue will interfere

constructively at a different angle than red

When you look at the soap film the white light illuminates the film over a range of angles

Section 25.3

Antireflection CoatingsNearly any flat piece of

glass may act like a partially reflecting mirror

To avoid reductions in intensity due to this reflection, antireflective coatings may be used

The coating makes a lens appear slightly dark in color when viewed in reflected light

Section 25.3

Antireflective Coatings, cont.Many coatings are

made from MgF2

nMgF2 = 1.38

There is a 180° phase change at both interfaces

Destructive interference occurs when

2

12

2MgF

md

n

Section 25.3

Antireflective Coatings, finalThe smallest possible value of d that gives

destructive interference corresponds to m = 0MgF2 is a popular material for antireflective coatings

because it can be made into very uniform films with small thicknesses

Although an antireflective coating will work best only at one wavelength, it will give partially destructive interference at nearby wavelengths

To function over the entire range of visible wavelengths, the coatings are made using multiple layers that give perfect destructive interference at different wavelengths

Section 25.3

Thin-Film Interference

Assume a thin soap film rests on a flat glass surfaceThe upper surface of the soap film is similar to the

beam splitter in the interferometerIt reflects part of the incoming light and allows the

rest to be transmitted into the soap layer after refraction at the air-soap interface

Section 25.3

Phase Change and Reflection, Diagram

Section 25.3

Case 1, cont.Equations are

These equations apply whenever nair < nfilm < n(substance

below the film)

2

12

2

film

film

md constructive interference

n

md destructive interference

n

Section 25.3

Thin-Film Interference, Case 2Assume the soap bubble is surrounded by airThere is a phase change at the top of the bubbleThere is no phase change at the bottom of the

bubbleSince only one wave undergoes a phase change,

the interference conditions are12

2

2

film

film

md constructive interference

n

md destructive interference

n

Section 25.3

Thin-Film Interference

Section 25.3

2

12

2

film

film

md constructive interference

n

md destructive interference

n

12

2

2

film

film

md constructive interference

n

md destructive interference

n

m=0,1,2..

Coherent Interference IntensityI = utotal x c = ½ εo c

Eo2

Electric Fields add linearly

Intensity units W/m2

21max III

21min III

Light Through a Single Slit

Light passes through a slit or opening and then illuminates a screen

As the width of the slit becomes closer to the wavelength of the light, the intensity pattern on the screen and additional maxima become noticeable

Section 25.4

Single-Slit DiffractionWater wave example

of single-slit diffractionAll types of waves

undergo single-slit diffraction

Water waves have a wavelength easily visible

Diffraction is the bending or spreading of a wave when it passes through an opening

Section 25.4

Huygens’ PrincipleIt is useful to draw the

wave fronts and rays for the incident and diffracting waves

Huygen’s Principle can be stated as all points on a wave front can be thought of as new sources of spherical waves

Section 25.4

Double-Slit InterferenceLight passes through two very narrow slitsWhen the two slits are both very narrow, each slit

acts as a simple point source of new wavesThe outgoing waves from each slit are like simple

spherical wavesThe double slit experiment showed conclusively that

light is a waveExperiment was first carried out by Thomas Young

around 1800

Section 25.5

Young’s Double-Slit ExperimentLight is incident onto

two slits and after passing through them strikes a screen

The incident light shown is a plane waveThis is easy to achieve

with a laserCould also be achieved

with a lens

Section 25.5

Young’s Experiment, cont.The experiment satisfies the

general requirements for interference The interfering waves travel through

different regions of space as they travel through different slits

The waves come together at a common point on the screen where they interfere

The waves are coherent because they come from the same source

Interference will determine how the intensity of light on the screen varies with position

Section 25.5

Young’s Experiment, finalAssume the slits are very narrowAccording the Huygen’s principle, each slit acts as a

simple source with circular wave fronts as viewed from above

The light intensity on the screen alternates between bright and dark as you move along the screenThese areas correspond to regions of constructive

interference and destructive interference

Section 25.5

Double Slit AnalysisDetermine the path

length between each slit and the screen

Assume W is very largeIf the slits are separated

by a distance d, then the difference in length between the paths of the two rays is

ΔL = d sin θ

Section 25.5

Double Slit Analysis, cont.If ΔL is equal to an integral number of complete

wavelengths, then the waves will be in phase when they strike the screenThe interference will be constructiveThe light intensity will be large

If ΔL is equal to a half number of complete wavelengths, then the waves will not be in phase when they strike the screenThe interference will be destructiveThe light intensity will be zero

Section 25.5

Conditions for InterferenceFor constructive interference, d sin θ = m λ

m = 0, ±1, ±2, … Will observe a bright fringe

For destructive interference, d sin θ = (m + ½) λm = 0, ±1, ±2, … Will observe a dark fringe

Section 25.5

Double-Slit Intensity PatternThe angle θ varies as

you move along the screenEach bright fringe

corresponds to a different value of m

Negative values of m indicate that the path to those points on the screen from the lower slit is shorter than the path from the upper slit

Section 25.5

Spacing Between SlitsNotation:

d is the distance between the slitsW is the distance between the slits and the screenh is the separation between the slits

For m = 1,

Since the angle is very small, sin θ ~ θ and θ ~ λ/dBetween m = 0 and m = 1, h = W tan θ

sind

Section 25.5

ApproximationsSince the wavelength of light is small, the angles

involved in the double-slit analysis are also smallFor small angles, sin θ ~ θ and tan θ ~ θUsing the approximations, h = W θ = W λ / d

Section 25.5

Interference with Monochromatic LightThe conditions for interference state the interfering

waves must have the same frequencyThis means they must have the same wavelength

Light with a single frequency is called monochromaticConsists of one color

Light sources with a variety of wavelengths are generally not useful for double-slit interference experimentsThe bright and dark fringes may overlap or the total

pattern may be a “washed out” sum of bright and dark regions

No bright or dark fringes will be visible

Section 25.5

Single-Slit Interference

Slits may be narrow enough to exhibit diffraction but not so narrow that they can be treated as a single point source of waves

Assume the single slit has a width, wLight is diffracted as it passes through the slit

and then propagates to the screen

Section 25.6

Single-Slit AnalysisThe key to the

calculation of where the fringes occur is Huygen’s principle

All points across the slit act as wave sources

These different waves interfere at the screen

For analysis, divide the slit into two parts

Section 25.6

Single-Slit Fringe LocationsIf one point in each part of the slit satisfies the

conditions for destructive interference, the waves from all similar sets of points will also interfere destructively

Destructive interference will produce a dark fringeConditions for destructive interference are

w sin θ = ±m λm = 1, 2, 3, … The negative sign will correspond to a fringe below the

center of the screenSection 25.6

Single-Slit Analysis – Bright FringesThere is no simple formula for the angles at which

the bright fringes occurThe intensity on the screen can be calculated by

adding up all the Huygens wavesThere is a central bright fringe with other bright

fringes that are lower in intensityThe central fringe is called the central maximumThe central fringe is about 20 times more intense than

the bright fringes on either side

Section 25.6

Single-Slit – Central MaximumThe width of the

central bright fringe is approximately the angular separation of the first dark fringes on either side

The full angular width of the central bright fringe = 2 λ / w

If the slit is much wider than the wavelength, the light beam essentially passes straight through the slit with almost no effect from diffraction

Section 25.6

Double-Slit Interference with Wide SlitsWhen the slits of a double-slit experiment are not

extremely narrow, the single-slit diffraction pattern produced by each sit is combined with the sinusoidal double-slit interference pattern

A full calculation of the intensity pattern is very complicated

Section 25.6

Diffraction GratingAn arrangement of

many slits is called a diffraction grating

AssumptionsThe slits are narrow

Each one produces a single outgoing wave

The screen is very far away

Section 25.7

Diffraction Grating, cont.Since the screen is far away, the rays striking

the screen are approximately parallelAll make an angle θ with the horizontal axis

If the slit-to-slit spacing is d, then the path length difference for the rays from two adjacent slits isΔL = d sin θ

If ΔL is equal to an integral number of wavelengths, constructive interference occurs

For a bright fringe, ΔL = d sin θ = m λm = 0, ±1, ±2, …

Section 25.7

Diffraction Grating, finalThe condition for bright

fringes from a diffraction grating is identical to the condition for constructive interference from a double slit

The overall intensity pattern depends on the number of slits

The larger the number of slits, the narrower the peaks

Section 25.7

Grating and Color Separation

A diffraction grating will produce an intensity pattern on the screen for each color

The different colors will have different angles and different places on the screen

Diffraction gratings are widely used to analyze the colors in a beam of light

Section 25.7

Diffraction and CDsLight reflected from

the arcs in a CD acts as sources of Huygens waves

The reflected waves exhibit constructive interference at certain angles

Light reflected from a CD has the colors “separated”

Section 25.7

Optical ResolutionFor a circular opening

of diameter D, the angle between the central bright maximum and the first minimum is

The circular geometry leads to the additional numerical factor of 1.22

D1.22

Section 25.8

Telescope Example

Assume you are looking at a star through a telescopeDiffraction at the opening produces a circular

diffraction spotAssume there are actually two starsThe two waves are incoherent and do not interfereEach source produces its own different pattern

Section 25.8

Rayleigh Criterion

If the two sources are sufficiently far apart, they can be seen as two separate diffraction spots (A)

If the sources are too close together, their diffraction spots will overlap so much that they appear as a single spot (C)

Section 25.8

Rayleigh Criterion, cont.Two sources will be resolved as two distinct sources

of light if their angular separation is greater than the angular spread of a single diffraction spot

This result is called the Rayleigh criterionFor a circular opening, the Rayleigh criterion for

the angular resolution is

Two objects will be resolved when viewed through an opening of diameter D if the light rays from the two objects are separated by an angle at least as large as θmin

Dmin

1.22

Section 25.8

ScatteringWhen the wavelength is

larger than the reflecting object, the reflected waves travel away in all direction and are called scattered waves

The amplitude of the scattered wave depends on the size of the scattering object compared to the wavelength

Blue light is scattered more than redCalled Rayleigh

scatteringSection 25.9

Blue SkyThe light we see from the

sky is sunlight scattered by the molecules in the atmosphere

The molecules are much smaller than the wavelength of visible lightThey scatter blue light

more strongly than redThis gives the

atmosphere its blue color

Section 25.9

Scattering, Sky, and SunBlue sky

Although violet scatters more than blue, the sky appears blue

The Sun emits more strongly in blue than violetOur eyes are more sensitive to blueThe sky appears blue even though the violet light is

scattered moreSun near horizon

There are more molecules to scatter the lightMost of the blue is scattered away, leaving the red

Section 25.9

Nature of LightInterference and diffraction show convincingly that

light has wave propertiesCertain properties of light can only be explained with

a particle theory of lightColor vision is one effect that can be correctly

explained by the particle theoryHave strong evidence that light is both a particle and

a waveCalled wave-particle dualityQuantum theory tries to reconcile these ideas

Section 25.10

Quantum Mechanics?!Single electrons fired

through double slitInterference? With…?Quantum Computing