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Electromagnetism & LightInterference & Diffraction
Interferenceand Diffraction
33-1 Phase Difference and Coherence
33-2 Interference in Thin Films
33-3 Two-Slit Interference Pattern
33-4 Diffraction Pattern of a Single Slit
33-5 Using Phasors to Add Harmonic Waves
33-6 Fraunhofer and Fresnel Diffraction
33-7 Diffraction and Resolution
33-8 Diffraction Gratings
Interference and diffraction are the important phenomena that distinguishwaves from particles.* Interference is the formation of a lasting intensity pat-tern by two or more waves that superpose in space. Diffraction is the bendingof waves around corners that occurs when a portion of a wavefront is cut offby a barrier or obstacle.
In this chapter, we will see how the pattern of the resulting wave can be cal-culated by treating each point on the original wavefront as a point source,according to Huygens’s principle, and calculating the interference patternresulting from these sources.
33C H A P T E R
Have you ever wondered
if the phenomenon that produces
the bands that you see in
the light reflected off a soap bubble
has any practical applications?
(See Example 33-2.)
?
1141
WHITE LIGHT IS REFLECTED OFF A SOAPBUBBLE. WHEN LIGHT OF ONEWAVELENGTH IS INCIDENT ON A THINSOAP-AND-WATER FILM, LIGHT ISREFLECTED FROM BOTH THE FRONT ANDTHE BACK SURFACES OF THE FILM.IFTHE ORDER OF MAGNITUDE OF THETHICKNESS OF THE FILM IS ONEWAVELENGTH OF THE LIGHT, THE TWOREFLECTED LIGHT WAVES INTERFERE.IFTHE TWO REFLECTED WAVES ARE OUT OF PHASE, THE REFLECTED WAVESINTERFERE DESTRUCTIVELY, SO THE NETRESULT IS THAT NO LIGHT ISREFLECTED. IF WHITE LIGHT, WHICHCONTAINS A CONTINUUM OFWAVELENGTHS, IS INCIDENT ON THETHIN FILM, THEN THE REFLECTED WAVESWILL INTERFERE DESTRUCTIVELYONLY FOR CERTAIN WAVELENGTHS,AND FOR OTHER WAVELENGTHS THEYWILL INTERFERE CONSTRUCTIVELY.THIS PROCESS PRODUCES THECOLORED FRINGES THAT YOU SEE INTHE SOAP BUBBLE. (Aaron Haupt/Photo Researchers.)
180°
* Before you study this chapter, you may wish to review Chapter 15 and Chapter 16, where the general topics of inter-ference and diffraction of waves are first discussed.
*
*
Diffraction - Single Slit
Neil Alberding (SFU Physics) Physics 121: Optics, Electricity & Magnetism Spring 2010 3 / 1
Name ______________________ Date(YY/MM/DD) ______/_________/_______ St.No. __ __ __ __ __-__ __ __ __ Section_________Group #________
UNIT 31: INTERFERENCE AND DIFFRACTION
Interference of two circular waves, snapshots of absolute value of (real,scalar) wave field for different wave lengths and distances of point sources. (Wikimedia Commons)
OBJECTIVES
1. Understand the creation of double-slit interference and single-slit diffraction patterns.
2. Measure slit separation using double-slit interference of He-Ne laser light.
3. Compare single slit diffraction patterns to double-slit patterns.
4. Examine some 2-D diffraction patterns.
5. Construct and calibrate a portable diffraction-grating spectrometer and use it to examine the spectra of several light sources.
© 2008 by S. Johnson. Adapted from PHYS 131 Optics lab #4
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Your opinion is very important to us.
Diffraction - Huygen’s Principle
Huygen’s Principle1 Each point on a wave front is the source of a spherical wavelet
that spreads out at the wave speed.2 At a later time, the shape of the wavefront is the tangent line to all
of the wavelets.
Neil Alberding (SFU Physics) Physics 121: Optics, Electricity & Magnetism Spring 2010 2 / 1
Diffraction - Single Slit
Neil Alberding (SFU Physics) Physics 121: Optics, Electricity & Magnetism Spring 2010 3 / 1
Young’s Double-Slit Experiment
Neil Alberding (SFU Physics) Physics 121: Optics, Electricity & Magnetism Spring 2010 4 / 1
Young’s Double-Slit Experiment
NotesThe slit-width (a) and slit-separation (d) are similar in size to thewavelength of light (�)The wave fronts arrive at the two slits from the same source inabout the same time - they are in phase (�⇥ = 0).Each slit acts like a point-source by Huygen’s principle.
Neil Alberding (SFU Physics) Physics 121: Optics, Electricity & Magnetism Spring 2010 5 / 1
Analyzing Young’s Double-Slit Experiment
Neil Alberding (SFU Physics) Physics 121: Optics, Electricity & Magnetism Spring 2010 6 / 1
Analyzing Young’s Double-Slit Experiment
Constructive interference occurswhen
�r = d sin�m = m⇥,m = 0,1,2,3, . . .
In practice, the angle is small andsin� � �
�m = m⇥d
Neil Alberding (SFU Physics) Physics 121: Optics, Electricity & Magnetism Spring 2010 7 / 1
Analyzing Young’s Double-Slit Experiment
Constructive interference occurswhen
�r = d sin�m = m⇥,m = 0,1,2,3, . . .
In practice, the angle is small andsin� � �
�m = m⇥d
Neil Alberding (SFU Physics) Physics 121: Optics, Electricity & Magnetism Spring 2010 7 / 1
Analyzing Young’s Double-Slit Experiment
Using some simple trigonometry:
ym =m�L
d,m = 0,1,2,3, . . .
Similarly, we can get the dark fringe positions:
y⇥m =✓m +
12
◆ �Ld,m = 0,1,2, . . .
And we can get the fringe spacing
�y = ym+1 � ym =(m + 1)�L
d� m�L
d=�Ld
Neil Alberding (SFU Physics) Physics 121: Optics, Electricity & Magnetism Spring 2010 8 / 1
Analyzing Young’s Double-Slit Experiment
Using some simple trigonometry:
ym =m�L
d,m = 0,1,2,3, . . .
Similarly, we can get the dark fringe positions:
y⇥m =✓m +
12
◆ �Ld,m = 0,1,2, . . .
And we can get the fringe spacing
�y = ym+1 � ym =(m + 1)�L
d� m�L
d=�Ld
Neil Alberding (SFU Physics) Physics 121: Optics, Electricity & Magnetism Spring 2010 8 / 1
Analyzing Young’s Double-Slit Experiment
Using some simple trigonometry:
ym =m�L
d,m = 0,1,2,3, . . .
Similarly, we can get the dark fringe positions:
y⇥m =✓m +
12
◆ �Ld,m = 0,1,2, . . .
And we can get the fringe spacing
�y = ym+1 � ym =(m + 1)�L
d� m�L
d=�Ld
Neil Alberding (SFU Physics) Physics 121: Optics, Electricity & Magnetism Spring 2010 8 / 1
Young’s Double-Slit Fringe Intensity
Idouble = 4I1 cos2✓⇥d�L
y◆
Neil Alberding (SFU Physics) Physics 121: Optics, Electricity & Magnetism Spring 2010 9 / 1
Phase Difference & Path Length
A phase difference due to a path-‐length difference is observed for monochromaMc visible light. Which phase difference requires the least (minimum) path-‐length difference?
(A) 90°
(B) 180°
(C) 270°
(D) The answer depends on the wavelength of the light.
Two-slit Pattern
A two-‐slit interference paUern is formed using monochromaMc laser light that has a wavelength of 450 nm. What happens to the distance between the first maximum and the central maximum as the two slits are moved closer together?
(A) The distance increases.
(B) The distance decreases.
(C) The distance remains the same.
Multiple Slits
If you have more than two slits, the maxima get brighter and beUer separated.Using Phasors to Add Harmonic Waves S E C T I O N 3 3 - 5 | 1155
As the number of sources is increased, the principal maxima become sharper andmore intense, and the intensities of the secondary maxima become negligiblecompared to those of the principal maxima.
*CALCULATING THE SINGLE-SLIT DIFFRACTION PATTERNWe now use the phasor method for the addition of harmonic waves to calculate theintensity pattern shown in Figure 33-11. We assume that the slit of width is di-vided into equal intervals and that there is a point source of waves at the mid-point of each interval (Figure 33-23). If is the distance between two adjacentsources and is the width of the opening, we have Because the screen onwhich we are calculating the intensity is far from the sources, the rays from thesources to a point on the screen are approximately parallel. The path-length dif-ference between any two adjacent sources is and the phase difference isrelated to the path-length difference by
If is the amplitude due to a single source, the amplitude at the central maxi-mum, where and all the waves are in phase, is (Figure 33-24).
We can find the amplitude at some other point at an angle by using the pha-sor method for the addition of harmonic waves. As in the addition of two, three, orfour waves, the intensity is zero at any point where the phasors representing thewaves form a closed polygon. In this case, the polygon has sides (Figure 33-25).At the first minimum, the wave from the first source just below the top of the open-ing and the wave from the source just below the middle of the opening are outof phase. In this case, the waves from the source near the top of the opening differ
180°
N
u
Amax ! NA0u ! 0A0
d !d sinul
2p
dd sinu,P
d ! a>N.ad
Na
I/I0
0
Four sources
Three sources
Two sources
dλ
dλ sin θ
–F I G U R E 3 3 - 2 2 Plot of relative intensityversus for two, three, and four coherentsources that are equally spaced and in phase.
sinu
θ
a
F I G U R E 3 3 - 2 3 Diagram for calculatingthe diffraction pattern far away from a narrowslit. The slit width is assumed to contain alarge number of in-phase, equally spacedpoint sources separated by a distance Therays from the sources to a point far away areapproximately parallel. The path-lengthdifference for the waves from adjacent sourcesis d sinu.
d.
a
Screen
Amax = NA0
A0
Nsources
F I G U R E 3 3 - 2 4 A single slit is represented by sources, each of amplitude At thecentral maximum point, where the waves from the sources add in phase, giving a resultantamplitude Amax ! NA0 .
u ! 0,A0 .N
360°Nδ =
F I G U R E 3 3 - 2 5 Phasor diagram forcalculating the first minimum in the single-slitdiffraction pattern. When the waves fromthe sources completely cancel, the phasors form a closed polygon. The phasedifference between the waves from adjacentsources is then When is verylarge, the waves from the first and last sourcesare approximately in phase.
Nd ! 360°>N.
NN
Diffraction GratingThe Diffraction Grating
If one extends the double slit to largenumber of slits very closely spaced, onegets what is called a diffraction grating.d sin�. Maxima are still at
d sin�m = m⇥,m = 0,1,2,3, . . .
The difference is that the fringes arethinner and brighter.
Neil Alberding (SFU Physics) Physics 121: Optics, Electricity & Magnetism Spring 2010 1 / 15
Diffraction GratingLines of high intensity occur only where the wavefronts from all the slits interfere construcMvely. Therefore the maxima are very intense and very narrow.
The angle from the middle of the graMng to the maxima is given byd sinθm = mλ, m = 0,1,2,3,...
The distance from the central maximum to the next maximum is given byym =L tanθm
The Diffraction Grating
Lines of high intensity occur onlywhere the wavefronts from all the slitsinterfere constructively. Therefore themaxima are very intense and verynarrow.The angle from the middle of thegrating to the maxima is given by
d sin�m = m⇥,m = 0,1,2,3, . . .
The distance from the centralmaximum to the next maximum isgiven by
ym = L tan�m
Neil Alberding (SFU Physics) Physics 121: Optics, Electricity & Magnetism Spring 2010 2 / 15
Diffraction GratingThe Diffraction Grating
The angles to the maxima are notsmall. Therefore, the small angleapproximation cannot be used. Thedistance on the screen to the brightlines is given by
ym = L tansin�1
✓m�d
◆�
The distances to the maxima providea good way of determiningwavelengths of light.Diffraction gratings are essentialcomponents of optical spectrometers.
Neil Alberding (SFU Physics) Physics 121: Optics, Electricity & Magnetism Spring 2010 3 / 15
Reflection Diffraction GratingsReflection Gratings
Many common gratings are actuallyreflection gratings rather thantransmission gratings.A mirror with thousands of narrowparallel grooves makes a gratingwhich reflects light instead oftransmitting it, but the math is thesame.A CD is an excellent example.
Neil Alberding (SFU Physics) Physics 121: Optics, Electricity & Magnetism Spring 2010 6 / 15
Diffraction GratingSingle Slit Diffraction
A wave front passes through a narrowslit (width a). Note that narrow isimportant.Each point on the wave-front emits aspherical waveOne slit becomes the source of manyinterfering wavelets.A single slit creates a diffractionpattern on the screen.
Neil Alberding (SFU Physics) Physics 121: Optics, Electricity & Magnetism Spring 2010 10 / 15
Single SlitSingle Slit Diffraction
It is rather strange to talkabout thousands of slits beforetalking about 1. However,thousands are actually a littleeasier.A single slit diffraction patterninvolves a wide centralmaximum flanked by weakersecondary maxima and darkfringes.It would appear that we haveonly one light source in thiscase, so how do weunderstand the interference?We have to go back toHuygen’s principle.
Neil Alberding (SFU Physics) Physics 121: Optics, Electricity & Magnetism Spring 2010 9 / 15
Single SlitWhy the Wide Central Maximum?
Wavelets from any part of the slithave to travel approximately the samedistance to reach the center of thescreen.A set of in-phase wavelets thereforeproduce constructive interference atthe center of the screen.
Neil Alberding (SFU Physics) Physics 121: Optics, Electricity & Magnetism Spring 2010 11 / 15
Single SlitWhy the Dark Bands?
Consider the path-lengths well awayfrom the centre axisFor any wavelet it is possible to find apartner which is a/2 away.If the path difference betweenpartners happens to be �/2 then thispair will create total destructiveintereference. A dark band will becreated.For any given wavelength there willbe an angle for which this condition istrue! There will always be darkbands, as long as a is greater than �and the slit is narrow.
Neil Alberding (SFU Physics) Physics 121: Optics, Electricity & Magnetism Spring 2010 12 / 15
Single SlitThe Mathematics of the Dark Bands
The path difference between 1 and 2 is
�r12 =a2
sin�1 =⇥2
What about the other angles for destructiveinterference? The general formula becomes
a sin�p = p⇥,p = 1,2,3, . . .
The small angle approximation means wecan write
�p = p⇥a,p = 1,2,3, . . .
But if a is small then �p is large and thesmall angle approximation is not valid.
Neil Alberding (SFU Physics) Physics 121: Optics, Electricity & Magnetism Spring 2010 13 / 15
Single SlitThe Width of the Bands
It can be useful to express the fringeposition in distance rather than angle.The position on the screen is given byyp = L tan�p. This leads to
yp =p⇥L
a,p = 1,2,3, . . .
The width of the central maximum is giveby twice the distance to the first darkfringe
w =2⇥L
aIt is important to note that: 1) the widthgrows if the screen is farther away 2) Athinner slit makes a wider centralmaximum.
Neil Alberding (SFU Physics) Physics 121: Optics, Electricity & Magnetism Spring 2010 14 / 15
Circular ApertureCircular Aperture Diffraction
�1 =1.22⇥
DAnd the width of the central maximum is
w = 2y1 = 2L tan�1 �2.44⇥L
D
Neil Alberding (SFU Physics) Physics 121: Optics, Electricity & Magnetism Spring 2010 15 / 15
Wave vs. Ray Models of LightThe Wave and Ray Models of Light
The factor that determines how mucha wave spreads out is �/aWith water or sound we seediffraction in our everyday livesbecause the wavelength is roughlythe same as the macroscopicopenings and structures we seearound us.We will only notice the spreading oflight with apertures of roughly thesame scale as the wavelength of light.
Neil Alberding (SFU Physics) Physics 121: Optics, Electricity & Magnetism Spring 2010 16 / 15
Waves or Rays?The Wave and Ray Models of Light
Sometimes we treat light like a stream of particles, sometimes like awave and sometimes like a ray. Does light travel in a straight line ornot? The answer depends on the circumstances.
Choosing a Model of LightWhen light passes through openings < 1mm in size, diffractioneffects are usually important. Treat light as a wave.When light passes through openings > 1mm in size, treat it as aray.
Neil Alberding (SFU Physics) Physics 121: Optics, Electricity & Magnetism Spring 2010 17 / 15
Diffraction Grating
When a diffracMon graMng is illuminated by white light, the first-‐order maximum of green light
(A) is closer to the central maximum than the first-‐order maximum of red light,
(B) is closer to the central maximum than the first-‐order maximum of blue light,
(C) overlaps the second-‐order maximum of red light,
(D) overlaps the second-‐order maximum of blue light.
The Diffraction Grating
The angles to the maxima are notsmall. Therefore, the small angleapproximation cannot be used. Thedistance on the screen to the brightlines is given by
ym = L tansin�1
✓m�d
◆�
The distances to the maxima providea good way of determiningwavelengths of light.Diffraction gratings are essentialcomponents of optical spectrometers.
Neil Alberding (SFU Physics) Physics 121: Optics, Electricity & Magnetism Spring 2010 3 / 15
Green is between red and blue
(A) Green is closer to the central maximum than the first-‐order maximum of red light,
Single Slit Diffraction
A single-‐slit diffracMon paUern is formed using monochromaMc laser light that has a wavelength of 450 nm. What happens to the distance between the first maximum and the central maximum as the slit is made narrower?
(A) The distance increases.
(B) The distance decreases.
(C) The distance remains the same.
Class 2 Laser Safety
< 1 mW
Do not stare into the beam (0.25 s max)
Do not point at anybody.
Keep the laser on the table. Beam horizontal.
Do not point or reflect beam across the room
OpMcal experiments should be carried out on an opMcal table with all laser beams travelling in the horizontal plane only, and all beams should be stopped at the edges of the table. Users should never put their eyes at the level of the horizontal plane where the beams are in case of reflected beams that leave the table.
Watches and other jewelry that might enter the opMcal plane should not be allowed in the laboratory. All non-‐opMcal objects that are close to the opMcal plane should have a maUe finish in order to prevent specular reflecMons.
Adequate eye protecMon should always be required for everyone in the room if there is a significant risk for eye injury.
High-‐intensity beams that can cause fire or skin damage (mainly from class 4 and ultraviolet lasers) and that are not frequently modified should be guided through opaque tubes.
Alignment of beams and opMcal components should be performed at a reduced beam power whenever possible.
Bonus Exam Question
A resistor, an inductor, and a capacitor walk into a bar.
The inductor says to the capacitor, "I'm just as reac>ve as you are!"
The resistor says, "That's nothing! I can shiB BOTH your phases by 45 degrees with one ohm >ed behind my back!!”
Just then a big bad baJery walks in. "Mr. Capacitor and Mr. Resistor. Why don't you just shake hands and I'll see how you discharge.”
Mr, Capacitor and Mr. Resistor agree to the test. Big bad baJery renders his verdict. "Five seconds. Sounds kinda puny to me. But now you know.."
The barkeeper, Mr. Generator, chimes in. "Well, I guess I'll have to excite you both and see if you're all telling the truth."
What frequency is Mr. Generator?