college physics b

CollegePhysics B

Review:Thin-FilmInterference

Single-SlitExperiment

Double-SlitExperimentYoung Experiment

Double-Slit Analysis

Single-SlitInterference

DiffractionGrating

College Physics B - PHY2054C

Wave Optics: Diffraction

10/29/2014

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CollegePhysics B






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Outline

1 Review: Thin-Film Interference

2 Single-Slit Experiment

3 Double-Slit ExperimentYoung ExperimentDouble-Slit Analysis

4 Single-Slit Interference

5 Diffraction Grating

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Thin-Film Interference

Assume a thin soap film rests on a flat glass surface.

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The upper surface of the soap film is similar to the beam splitterin the interferometer:

• It reflects part of the incoming light and allows the rest tobe transmitted into the soap layer after refraction at theair-soap interface.

• The transmitted ray ispartially reflected at thebottom surface.

• The two outgoing rays meetthe conditions for interference:

1 Travel through different regions

2 Recombination

3 Coherence

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Index of refraction of the film also needs to be accounted for:

• The wavelength changes as thelight wave travels from a vacuuminto the film:

λ film =vf

=c/n film

f

=λ vac

n film≈

λ air

n film

• Number of extra wavelengths:

N =2d

λ film=

2dλ/n film

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Frequency of Wave at Interface

When a light wave passes from one medium to another, thewaves must stay in phase at the interface. The frequency mustbe the same on both sides of the interface.

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Frequency of Wave at Interface

1 There is a phase change whenever the index of refractionon the incident side is less than the index of refraction ofthe opposite side (wave is inverted).

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

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Phase Changes in a Thin Film

The total phase change in a thin film must be accounted for:

• The phase difference due to the extra distance traveledby the ray.

• Any phase change due to reflection.

• For a soap film on glass: n air < n film < n glass

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There are phase changes for both reflections at the soap filminterfaces:

B The reflections at both the top and bottom surfacesundergo a 180◦ phase change:

1 If the number of extra cycles, N, is an integer, there isconstructive interference: 2d = mλ/n film

2 If the number of extra cycles is a half-integer, there isdestructive interference: 2d = (m + 1/2)λ/n film

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Assume the soap bubble is surrounded by air.C There is a phase change at the top of the bubble. There

is no phase change at the bottom of the bubble:1 If the number of extra cycles, N, is a half-integer, there is

constructive interference: 2d = (m + 1/2)λ/n film

2 If the number of extra cycles is an integer, there isdestructive interference: 2d = mλ/n film

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Outline






CollegePhysics B






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Light Through a Single Slit

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

• As the width of the slit becomes closer to the wavelength ofthe light, the intensity pattern on the screen and additionalmaxima become noticeable.

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Single-Slit Diffraction

All types of waves undergo single-slit diffraction:

• Water waves have a wavelength easily visible.

• The outgoing wave front isapproximately spherical.

Diffraction is the bending orspreading of a wave when itpasses through an opening.

Wave Interference:http://phet.colorado.edu/en/simulation/wave-interference

http://phet.colorado.edu/en/simulation/wave-interference

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Huygens’s Principle

Christiaan Huygens(14 April 1629 - 8 July 1695)

Huygens’s Principle:All points on a wave frontcan be thought of as newsources of spherical waves.

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Outline






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Double-Slit Experiment

Light passes through two very narrow slits:

• When the two slits are both very narrow, each slit actsas a simple point source of new waves.

• The outgoing waves from each slit are like simplespherical waves.

• The double slit experiment showed conclusively that lightis a wave.

Why?XXXXXXX

?

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Question 1What will you observe on the screen in an experimentusing light incident on two slits?

A Nothing, the screen will be dark.

B A complicated pattern with numerous bright lines.

C Two bright lines corresponding to the number of slits.

D No changes compared to a single slit.

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?

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Question 1What will you observe on the screen in an experimentusing light incident on two slits?





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Double-Slit Experiment

Light passes through two very narrow slits:

• When the two slits are both very narrow, each slit actsas a simple point source of new waves.

• The outgoing waves from each slit are like simplespherical waves.

• The double slit experiment showed conclusively that lightis a wave (observation of interference).

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Question 2What will you observe on the screen in an experimentusing a beam of particles incident on two slits?





XXXXXXX

?

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Young’s Double-Slit Experiment

Experiment was first carried out by T. Young around 1800:

• Light is incident onto two slits and after passing throughthem strikes a screen.

Experiment satisfies the general requirements for interference:

1 The interfering waves travel through different regions ofspace as they travel through different slits.

2 The waves come together at a common point on thescreen where they interfere.

3 The waves are coherent because they come from thesame source.

➜ Interference will determine how the intensity of light on thescreen varies with position.

Quantum Wave Interference:phet.colorado.edu/en/simulation/quantum-wave-interference

phet.colorado.edu/en/simulation/quantum-wave-interference

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Determine the path length between each slit and the screen.

Assume W is very large. If the slits are separated by a distanced , then the difference in length between the paths of the tworays is:

∆L = d sin θ

Bright fringe:

d sin θ = m λ

m = 0, ± 1, ± 2, ...

Dark fringe:

d sin θ = (m +12)λ

m = 0, ± 1, ± 2, ...

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Double-Slit Intensity Pattern

The angle θ varies as you move along the screen:

• Each bright fringe corresponds to a different value of m.

• Negative values of m indicate that the path to those pointson the screen from the lower slit is shorter than the pathfrom the upper slit.

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ExampleFor m = 1: sin θ = λ/d

• Since the angle is very small:

tanθ = θ and sin θ = θ and θ = λ/d

• Between m = 0 and m = 1: h = W tanθ = W λ/d

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ExampleFor m = 1: sin θ = λ/d

• Since the angle is very small:

tanθ = θ and sin θ = θ and θ = λ/d

• Between m = 0 and m = 1: h = W tanθ = W λ/d

Suppose the slits are d = 0.1 mm apart and the screen islocated W = 50 cm from the slits. For light with a wavelengthof 630 nm:

h = Wλ

d= (0.5 m)

6.3 × 10−7 m1 × 10−4 m

= 3.2 mm

This fringe spacing is large enough to be seen easily by thenaked eye. This gives also a way to measure the wavelengthof light.

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Question 3

A Young’s double-slit experiment is performed in air andthen the apparatus is submerged in water. What happensto the fringe separation (d sin θ = m λ), and what can beused to explain the change, if any?

A The separation stays the as it is the sameexperiment independent of the medium.

B The separation decreases because the frequency ofthe light decreases in the water.

C The separation increases because the wavelength ofthe light increases in the water.

D The separation decreases because the wavelengthof the light decreases in the water.

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Outline






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Single-Slit Interference

Slits may be narrow enough to exhibit diffraction but not sonarrow that they can be treated as a single point source ofwaves:

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Assume the single slit has a width, w . Light is diffracted as itpasses through the slit and then propagates to the screen:

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Assume the single slit has a width, w . Light is diffracted as itpasses through the slit and then propagates to the screen.

All points across the slit act as wave sources andinterfere at the screen:

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Single-Slit Fringe Locations

If one point in each part of the slit satisfies the conditionsfor destructive interference (dark fringes), the waves from allsimilar sets of points will also interfere destructively:

w sin θ = ±m λ, with m = 1, 2, 3, ...

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If one point in each part of the slit satisfies the conditionsfor destructive interference (dark fringes), the waves from allsimilar sets of points will also interfere destructively:

w sin θ = ±m λ, with m = 1, 2, 3, ...

Different from doubleslits!

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There is no simple formula for the angles at which the brightfringes occur:

• There is a central bright fringe with other bright fringesthat are lower in intensity ➜ Central Maximum

About 20 times moreintense.

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Single-Slit: Central Maximum

The width of the central bright fringe isapproximately the angular separation ofthe first dark fringes on either side:

• The full angular width of thecentral bright fringe is 2λ/w .

• If the slit is much wider than thelight’s wavelength, the light beamessentially passes straight throughthe slit with almost no diffraction.

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Summary

The complete interference pattern that isproduced by two slits is combination of

A the double-slit pattern and

B the single-slit pattern.

C Intensity pattern for slits that arenot extremely narrow.

A full calculation of the intensity patternis very complicated.

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Outline






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Diffraction Grating

An arrangement of many slits is called a diffraction grating.

Assumptions:

1 The slits are narrow.

2 The screen is very faraway.

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Diffraction Grating

An arrangement of many slits is called a diffraction grating.

Assumptions:

1 The slits are narrow.

2 The screen is very faraway.

Since the screen is so faraway, the rays striking thescreen are approximatelyparallel making an angle θwith the horizontal axis:

∆L = d sin θ = m λ

Bright fringes:

m = 0, ± 1, ± 2, ...

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Diffraction Grating

The condition for bright fringes from a diffraction grating isidentical to the condition for constructive inteference from adouble slit:

• Overall intensity patterndepends on the numberof slits.

• The larger the numberof slits, the narrowerthe peaks.

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Diffraction Grating

A diffraction grating will produce an intensity pattern on thescreen for each color:

• The different colors will have different angles anddifferent places on the screen.

• Diffraction gratings are widely used to analyze the colorsin a beam of light.

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Diffraction and CDs

Light reflected from the arcs in a CD actsas sources of Huygens waves:

• The reflected waves exhibitconstructive interference atcertain angles.

• Light reflected from a CD hasthe colors “separated”.

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