Chapter 24Chapter 24Wave OpticsWave Optics

Conceptual questions: 3, 4, 13, 14, 17, 18

Quick Quizzes: 1, 2, 3, 4

Problems: 10, 17, 34

InterferenceInterference

Light waves interfere with each Light waves interfere with each other much like mechanical waves other much like mechanical waves dodo

Two conditions which must be metTwo conditions which must be met1.1. The sources must be The sources must be coherentcoherent. They . They

must maintain a constant phase with must maintain a constant phase with respect to each otherrespect to each other

2.2. The waves must have identical The waves must have identical wavelengthswavelengths

Producing Coherent Producing Coherent SourcesSources

Old method:Old method: Light from a monochromatic source is Light from a monochromatic source is

allowed to pass through a narrow slitallowed to pass through a narrow slit The light from the single slit is allowed The light from the single slit is allowed

to fall on a screen containing two to fall on a screen containing two narrow slitsnarrow slits

The first slit is needed to insure the The first slit is needed to insure the light comes from a tiny region of the light comes from a tiny region of the source which is coherentsource which is coherent

New method: use a laserNew method: use a laser

Young’s Double Slit Young’s Double Slit ExperimentExperiment

The narrow slits, SThe narrow slits, S11 and Sand S2 2 act as act as sources of wavessources of waves

The waves The waves emerging from the emerging from the slits originate from slits originate from the same wave the same wave front and therefore front and therefore are always in phaseare always in phase

Interference PatternsInterference Patterns

Constructive Constructive interference interference occurs at the occurs at the center pointcenter point

The two waves The two waves travel the same travel the same distance and they distance and they arrive in phasearrive in phase

Interference Patterns, 2Interference Patterns, 2

The upper wave The upper wave travels one travels one wavelength fartherwavelength farther Therefore, the Therefore, the

waves arrive in waves arrive in phasephase

A bright fringe A bright fringe occursoccurs

Interference Patterns, 3Interference Patterns, 3 The upper wave The upper wave

travels one-half of travels one-half of a wavelength a wavelength farther than the farther than the lower wavelower wave

The trough of the The trough of the bottom wave bottom wave overlaps the crest overlaps the crest of the upper waveof the upper wave

This is destructive This is destructive interferenceinterference A dark fringe occursA dark fringe occurs

Interference EquationsInterference Equations The path The path

difference, difference, δ, is δ, is found from the found from the tan triangletan triangle

δ = rδ = r22 – r – r11 = d sin = d sin θθ This assumes the This assumes the

paths are parallelpaths are parallel

Interference Equations, 2Interference Equations, 2

For a bright fringe, produced by For a bright fringe, produced by constructive interference, the path constructive interference, the path difference must bedifference must be δ = m λδ = m λ m = 0, ±1, ±2, … m = 0, ±1, ±2, …

δ = d sin θδ = d sin θbrightbright = m λ = m λ m is called the m is called the order numberorder number

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

maximummaximum

Interference Equations, 3Interference Equations, 3

When destructive interference When destructive interference occurs, a dark fringe is observedoccurs, a dark fringe is observed

This needs a path difference of an This needs a path difference of an odd half wavelength; odd half wavelength; δ = (m + ½) δ = (m + ½) λλ

δ = d sin θδ = d sin θdarkdark = (m + ½) λ = (m + ½) λ m = 0, ±1, ±2, … m = 0, ±1, ±2, …

Interference Equations, Interference Equations, 44 The positions of the fringes can be The positions of the fringes can be

measured vertically from the measured vertically from the zeroth order maximumzeroth order maximum

y = L tan y = L tan θ ~ θ ~ L sin L sin θθ ApproximationApproximation

θ is small and therefore θ is small and therefore tantanθ ~ θ ~ sin sin θθ

For bright fringesFor bright fringes

For dark fringesFor dark fringes

2,1,0mmd

Lybright

2,1,0m2

1m

d

Lydark

Quick quiz 24-1Quick quiz 24-1 In a two slit interference pattern In a two slit interference pattern

projected on a screen the fringes projected on a screen the fringes are equally spaced on the screen are equally spaced on the screen

A.A. everywhereeverywhere B. only for large anglesB. only for large angles C. only for small anglesC. only for small angles

Problem 24.10Problem 24.10

A pair of slits, separated by 0.150 mm, is illuminated by light having a wavelength of λ = 643 nm. An interference pattern is observed on a screen 140 cm from the slits. Consider a point on the screen located at y = 1.80 cm from the central maximum of this pattern.

(a) What is the path difference δ for the two slits at the location y?

(b) Express this path difference in terms of the wavelength.

(c) Will the interference correspond to a maximum, a minimum, or an intermediate condition?

Phase Changes Due To Phase Changes Due To ReflectionReflection

An electromagnetic An electromagnetic wave undergoes a wave undergoes a phase change of phase change of 180° upon reflection 180° upon reflection from a medium of from a medium of higher index of higher index of refraction than the refraction than the one in which it was one in which it was travelingtraveling Analogous to a Analogous to a

reflected pulse on a reflected pulse on a stringstring

Phase Changes Due To Phase Changes Due To Reflection, contReflection, cont

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

pulse in a string pulse in a string reflecting from a reflecting from a free supportfree support

Interference in Thin FilmsInterference in Thin Films

Rules to rememberRules to remember1.1. An electromagnetic wave traveling from a An electromagnetic wave traveling from a

medium of index of refraction nmedium of index of refraction n11 toward a toward a medium of index of refraction nmedium of index of refraction n22 undergoes a undergoes a 180° phase change on reflection when n180° phase change on reflection when n22 > n > n11

2.2. There is no phase change in the reflected There is no phase change in the reflected wave if nwave if n22 < n < n11

3.3. The wavelength of light The wavelength of light λλnn in a medium with in a medium with index of refraction n is λindex of refraction n is λnn = λ/n where λ is the = λ/n where λ is the wavelength of light in vacuumwavelength of light in vacuum

Interference in Thin Films, 2Interference in Thin Films, 2 Ray 1 undergoes a Ray 1 undergoes a

phase change of phase change of 180° with respect to 180° with respect to the incident raythe incident ray

Ray 2, which is Ray 2, which is reflected from the reflected from the lower surface, lower surface, undergoes no phase undergoes no phase change with respect change with respect to the incident waveto the incident wave

Ray 2 also travels Ray 2 also travels an additional an additional distance of 2t distance of 2t before the waves before the waves recombinerecombine

Interference in Thin Films, Interference in Thin Films, 33

For constructive interferenceFor constructive interference 2 n t = (m + ½ ) 2 n t = (m + ½ ) λλ m = 0, 1, 2 … m = 0, 1, 2 …

This takes into account both the difference This takes into account both the difference in optical path length (in optical path length (2t2t) for the two rays ) for the two rays and the 180° phase change (and the 180° phase change (1/2 1/2 λλ))

For destruction interferenceFor destruction interference 2 n t = m 2 n t = m λλ m = 0, 1, 2 … m = 0, 1, 2 …

Interference in Thin Films, Interference in Thin Films, 44

An example of An example of different indices of different indices of refractionrefraction

A coating on a solar A coating on a solar cellcell

Quick quiz 24-2Quick quiz 24-2

Supposed Young’s experiment is Supposed Young’s experiment is carried out in air, and then, in a carried out in air, and then, in a second experiment, the apparatus is second experiment, the apparatus is immersed in water. In what way does immersed in water. In what way does the distance between bright fringes the distance between bright fringes change?change? A. they move further apartA. they move further apart B. they move closer togetherB. they move closer together C. no changeC. no change

Problem 24.17Problem 24.17

A coating is applied to a lens to minimize reflections. The index of refraction of the coating is 1.55, and that of the lens is 1.48. If the coating is 177.4 nm thick, what wavelength is minimally reflected for normal incidence in the lowest order?

Newton’s ringsNewton’s rings

Conceptual questionsConceptual questions3. Consider a dark fringe in an interference 3. Consider a dark fringe in an interference

pattern, at which almost no light energy pattern, at which almost no light energy is arriving. Light from both slits is is arriving. Light from both slits is arriving at this point, but the waves are arriving at this point, but the waves are canceling. Where does the energy go?canceling. Where does the energy go?

4. If Young’s double slit experiment were 4. If Young’s double slit experiment were performed under water, how would the performed under water, how would the observed interference pattern be observed interference pattern be affected?affected?

13.Would it be possible to place a 13.Would it be possible to place a nonreflective coating on an airplane to nonreflective coating on an airplane to cancel radar waves of wavelength 3 cm?cancel radar waves of wavelength 3 cm?

Reading a CDReading a CD As the disk rotates, the As the disk rotates, the

laser reflects off the laser reflects off the sequence of lands and sequence of lands and pits into a photodectorpits into a photodector The photodector converts The photodector converts

the fluctuating reflected the fluctuating reflected light intensity into an light intensity into an electrical string of zeros electrical string of zeros and onesand ones

The pit depth is made The pit depth is made equal to one-quarter of equal to one-quarter of the wavelength of the the wavelength of the lightlight

land

DiffractionDiffraction

Huygen’s principle Huygen’s principle requires that the requires that the waves spread out waves spread out after they pass after they pass through slitsthrough slits

This spreading out This spreading out of light from its of light from its initial line of travel is initial line of travel is called called diffractiondiffraction

Fraunhofer DiffractionFraunhofer Diffraction

Fraunhofer Fraunhofer DiffractionDiffraction occurs occurs when the rays leave when the rays leave the diffracting object the diffracting object in parallel directionsin parallel directions

A bright fringe is A bright fringe is seen along the axis seen along the axis ((θ = 0) with θ = 0) with alternating bright alternating bright and dark fringes on and dark fringes on each sideeach side

Single Slit DiffractionSingle Slit Diffraction According to Huygen’s According to Huygen’s

principle, each portion of principle, each portion of the slit acts as a source of the slit acts as a source of waveswaves

The light from one portion The light from one portion of the slit can interfere of the slit can interfere with light from another with light from another portionportion

The resultant intensity on The resultant intensity on the screen depends on the the screen depends on the direction direction θθ

Wave 1 travels farther Wave 1 travels farther than wave 3 by an amount than wave 3 by an amount equal to the path equal to the path difference (a/2) sin difference (a/2) sin θ θ

destructive destructive interferenceinterference occurs occurs when when

sin sin θθdarkdark = mλ / a = mλ / a

Single Slit Diffraction, 2Single Slit Diffraction, 2

A broad central bright A broad central bright fringe is flanked by fringe is flanked by much weaker bright much weaker bright fringes alternating fringes alternating with dark fringeswith dark fringes

The points of The points of constructive constructive interference lie interference lie approximately approximately halfway between the halfway between the dark fringesdark fringes

Problem 34Problem 34

A screen is placed 50.0 cm from a single slit, which is illuminated with light of wavelength 680 nm. If the distance between the first and third minima in the diffraction pattern is 3.00 mm, what is the width of the slit?

In a single-slit diffraction experiment, as the width of the slit is made smaller, the width of the central maximum of the diffraction pattern becomes (a) smaller, (b) larger, (c) remains the same.

Quick quiz 24.3

Diffraction GratingDiffraction Grating The condition for The condition for

maximamaxima is is d sin d sin θθbrightbright = m λ = m λ

m = 0, 1, 2, …m = 0, 1, 2, … The integer m is the The integer m is the

order numberorder number of the of the diffraction patterndiffraction pattern

If the incident radiation If the incident radiation contains several contains several wavelengths, each wavelengths, each wavelength deviates wavelength deviates through a specific through a specific angleangle

If laser light is reflected from a phonograph record or a compact disc, a diffraction pattern appears. This occurs because both devices contain parallel tracks of information that act as a reflection diffraction grating. Which device, record or compact disc, results in diffraction maxima that are farther apart?

QUICK QUIZ 24.4

Diffraction Grating in CD Diffraction Grating in CD TrackingTracking

A diffraction grating can A diffraction grating can be used in a three-beam be used in a three-beam method to keep the method to keep the beam on a CD on trackbeam on a CD on track

The central maximum of The central maximum of the diffraction pattern is the diffraction pattern is used to read the used to read the information on the CDinformation on the CD

The two first-order The two first-order maxima are used for maxima are used for steeringsteering

Polarization of Light Polarization of Light WavesWaves

Each atom Each atom produces a wave produces a wave with its own with its own orientation of Eorientation of E

This is an This is an unpolarized unpolarized wavewave

Polarization of Light, contPolarization of Light, cont A wave is said to be A wave is said to be linearly linearly

polarizedpolarized if the resultant if the resultant electric field vibrates in the electric field vibrates in the same direction at all times same direction at all times at a particular pointat a particular point

Polarization can be obtained Polarization can be obtained from an unpolarized beam from an unpolarized beam by by selective absorptionselective absorption reflectionreflection scatteringscattering

Polarization by Selective Polarization by Selective AbsorptionAbsorption

The most common technique for polarizing lightThe most common technique for polarizing light Uses a material that transmits waves whose electric field Uses a material that transmits waves whose electric field

vectors in the plane parallel to a certain direction and vectors in the plane parallel to a certain direction and absorbs waves whose electric field vectors are perpendicular absorbs waves whose electric field vectors are perpendicular to that directionto that direction

Malus’ law: I = IMalus’ law: I = Ioo cos cos22 θθ

Polarization by Polarization by ReflectionReflection

The angle of incidence for The angle of incidence for which the reflected beam is which the reflected beam is completely polarized is completely polarized is called the called the polarizing anglepolarizing angle, , θθpp

θθpp is also called Brewster’s is also called Brewster’s AngleAngle

Brewster’s Law relates the Brewster’s Law relates the polarizing angle to the index polarizing angle to the index of refraction for the materialof refraction for the material

pp

p tancos

sinn

Polarization by Polarization by ScatteringScattering

The horizontal part of the The horizontal part of the electric field vector in the electric field vector in the incident wave causes the incident wave causes the charges to vibrate charges to vibrate horizontallyhorizontally

The vertical part of the The vertical part of the vector simultaneously vector simultaneously causes them to vibrate causes them to vibrate verticallyvertically

Horizontally and vertically Horizontally and vertically polarized waves are emittedpolarized waves are emitted

Conceptual questionConceptual question

14. Certain sunglasses use a polarizing 14. Certain sunglasses use a polarizing material to reduce intensity of light material to reduce intensity of light reflected from shiny surfaces, such as reflected from shiny surfaces, such as water or a hood of a car. What orientation water or a hood of a car. What orientation of the transmission axis should the of the transmission axis should the material have to be most effective?material have to be most effective?

18. Can a sound wave be polarized?18. Can a sound wave be polarized?17. When you receive a chest x-ray at a 17. When you receive a chest x-ray at a

hospital, the ex-ray passes through a series hospital, the ex-ray passes through a series of parallel ribs in your chest. Do the ribs of parallel ribs in your chest. Do the ribs act as a diffraction grating for x-rays?act as a diffraction grating for x-rays?

Optical ActivityOptical Activity

Certain materials display the Certain materials display the property of property of optical activityoptical activity A substance is optically active if it A substance is optically active if it

rotates the plane of polarization of rotates the plane of polarization of transmitted lighttransmitted light

Liquid Liquid CrystalsCrystals

Rotation of a polarized light beam by a liquid Rotation of a polarized light beam by a liquid crystal when the applied voltage is zerocrystal when the applied voltage is zero

Light passes through the polarizer on the Light passes through the polarizer on the right and is reflected back to the observer, right and is reflected back to the observer, who sees the segment as being brightwho sees the segment as being bright

Liquid Liquid CrystalsCrystals

When a voltage is applied, the liquid crystal does When a voltage is applied, the liquid crystal does not rotate the plane of polarizationnot rotate the plane of polarization

The light is absorbed by the polarizer on the right The light is absorbed by the polarizer on the right and none is reflected back to the observerand none is reflected back to the observer

The segment is darkThe segment is dark

MCADMCAD

Two light sources produce light with Two light sources produce light with wavelength wavelength . The sources are . The sources are placed 22.5 placed 22.5 and 45 and 45 away from away from point P. When both sources are point P. When both sources are turned on and their intensities, I, at turned on and their intensities, I, at point P are equal, the resultant point P are equal, the resultant intensity at point P will beintensity at point P will be A. 0A. 0 B. 0.5 IB. 0.5 I C. IC. I D. 2ID. 2I

The process discussed in the previous question is called

a. Diffraction b. Refraction

c. Interference d. Dispersion

Which of the following would result in greatest diffraction?

a. Small wavelengths moving through a small opening

b. Large wavelengths moving through a small opening

c. Small wavelengths moving through a large opening

d. Large wavelengths moving through a large opening