Lecture 39: FRI 24 APR - Louisiana State University

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Lecture 39: FRI 24 APR Lecture 39: FRI 24 APR Physics 2102 Jonathan Dowling Ch. 35: Interference Ch. 35: Interference Christian Huygens 1629-1695

Transcript of Lecture 39: FRI 24 APR - Louisiana State University

Lecture 39: FRI 24 APRLecture 39: FRI 24 APR

Physics 2102Jonathan Dowling

Ch. 35: InterferenceCh. 35: Interference

Christian Huygens 1629-1695

The Lunatic Fringe:

The waves arriving at the screen fromthe two slits will interfere constructivelyor destructively depending on the differentlength they have travelled. !sindL ="

integer 0,1,2...m :veConstructi =,=! "mL

integer 0,1,2...m )2

1( :eDestructiv =,+=! "mL

!" md =sin :veConstructi

Maximum fringe at θ=0, (central maxima)other maxima at

!"#

$%&

=d

m'( arcsin

Similarly for dark fringes: d sin θ= (m+1/2) λ

1 2 0 02 cos 2 cosE E E E E

!" # $

= + = = % &2' (

! !

!"#

$%&2

=='2

2

0

2

0

cos4E

E

I

I

!"#

"#$ sindL %

&'

()* 2

=+%&'

()* 2

=

The ReturnOf the Phasor!

Phase Difference2π=360°π=180°

ExampleExampleRed laser light (λ=633nm) goes through two slits 1cm apart, andproduces a diffraction pattern on a screen 55cm away. How farapart are the fringes near the center?

For the spacing to be 1mm, we need d~ Lλ/1mm=0.35mm

If the fringes are near the center, we can usesin θ ~ θ, and then mλ=dsinθ~dθ => θ=mλ/d is the angle for eachmaximum (in radians)Δq= l/d =is the “angular separation”.The distance between the fringes is thenΔx=LΔθ=Lλ/d=55cmx633nm/1cm=35 µm

ExampleExampleIn a double slit experiment, we can measure the wavelength of thelight if we know the distances between the slits and the angularseparation of the fringes.If the separation between the slits is 0.5mm and the first ordermaximum of the interference pattern is at an angle of 0.059o fromthe center of the pattern, what is the wavelength and color of thelight used?

d sinθ=mλ => λ=0.5mm sin(0.059o)= 5.15 10-7m=515nm ~ green

θ

ExampleExampleA double slit experiment has a screen 120cm away from the slits,which are 0.25cm apart. The slits are illuminated with coherent600nm light. At what distance above the central maximum is theaverage intensity on the screen 75% of the maximum?

I/I0=4cos2φ/2 ; I/Imax=cos2φ/2 =0.75 => φ=2cos–1 (0.75)1/2=60o=π/3 radφ=(2πd/λ)sinθ => θ= sin-1(λ/2πd)φ=0.0022ο=40 µrad (small!)y=Lθ=48µm

Radio WavesRadio WavesTwo radio towers, separated by 300 m as shown the figure,broadcast identical signals at the same wavelength.A radio in the car, which traveling due north, receives the signals. Ifthe car is positioned at the second maximum, what is the wavelengthof the signals? Where will the driver find a minimum in reception?

dsinθ=mλ => λ= dsinθ/2sin θ= 400/(4002+10002)1/2=0.37λ=55.7m

“Dark” fringes: dsinθ=(m+1/2)λ sinθ=(m+1/2)λ/d=2.5x55.7m/300m =0.464 =>θ= 28o

tanθ=y/1000m => y=1000m x tan(28o)=525m

Interference: ExampleInterference: ExampleA red light beam with wavelength λ=0.625µm travels through glass (n=1.46) adistance of 1mm. A second beam, parallel to the first one and originally in phasewith it, travels the same distance through sapphire (n=1.77).

•How many wavelengths are there of eachbeam inside the material? In glass, λg=0.625µm/1.46= 0.428 µm and Ng=D/ λg=2336.45In sapphire, λs=0.625µm/1.77= 0.353 µm (UV!) and Ns=D/ λs=2832.86

•What is the phase difference in the beams when they come out?The difference in wavelengths is Ns-Ng=496.41.Each wavelength is 360o, so ΔN=496.41 means Δφ=ΔNx360o=0.41x360o=148o

•How thick should the glass be so that the beams are exactly out of phase at theexit (destructive interference!)ΔN=D/ λs− D/ λg= (D/ λ)(n2-n1)=0.31 (D/ λ)=m+1/2A thickness D=(m+0.5) 2.02 µm would make the waves OUT of phase.For example, 1.008 mm makes them in phase, and 1.010 mm makes them OUT ofphase.

Thin Film Interference:

The patterns of colors that one seesin oil slicks on water or in bubblesis produced by interference of thelight bouncing off the two sides ofthe film.

To understand this we need to discuss the phase changes that occur when a wave moves from one mediumto the another where thespeed is different. Thiscan be understood witha mechanical analogy.

Reflection, Refraction and Changes of Phase:

Consider an UP pulse moving in a rope, that reaches a juncturewith another rope of different density. A reflected pulse is generated.

The reflected pulse is also UP if thespeed of propagation in the rope of theright is faster than on the left. (Lowimpedance.)

The reflected pulse is DOWN if the speedof propagation in the right is slower than onthe left. (High impedance.)

The extreme case of ZERO speed on the right corresponds to a ropeanchored to a wall. (Highest impedance.)

If we have a wave instead of a pulse “DOWN” means 180degrees OUT of phase, and UP means 360° or IN PHASE.

Thin FilmsFirst reflected lightray comes from firstinterface, secondfrom second. Theserays interfere witheach other.

How they interfere will depend on the relative indices of refraction.In the example above the first ray suffers a 180 degree phase change(1/2 a wavelength) upon reflection. The second ray does not change phase in reflection, but has to travel a longer distance to come back up. The distance is twice the thickness of the layer of oil. For constructive interference the distance 2L must therefore be ahalf-integer multiple of the wavelength, i.e. 0.5 λ, 1.5 λ,…,(0.5+2n)λ.

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number odd2 :phaseIn

nL

!"=

2

integer2 :phase-Antin

L!

"=

n1

n2

n3

If the film is very thin, then the interference is totally dominated by the180° phase shift in the reflection.

At the top the film is thinnest (dueto gravity it lumps at the bottom),so one sees thefilm dark at the top.

This film is illuminated with white light, therefore we see fringes ofdifferent colors corresponding to the various constructive interferencesof the individual components of the white light, which change as wego down. The thickness increases steeply as we go down, which makesthe width of the fringes become narrower and narrower.

Thin Films: Soap Bubbles

Air: n=1

AirSoap: n>1

180°

Reflective CoatingsReflective CoatingsTo make mirrors that reflects light of only a given wavelength, a coating of a specificthickness is used so that there is constructive interference of the given wavelength.Materials of different index of refraction are used, most commonly MgFe2 (n=1.38) andCeO2 (n=2.35), and are called “dielectric films”. What thickness is necessary forreflecting IR light with λ=1064nm?

n=2.35n=1.38

First ray: Δφ=180deg=πSecond ray: Δφ=2L(2π/(λ/n))=π=> L= λCeo2/4=(λ/n)/4=113nmThird ray? If wafer has the samethickness (and is of the same material), Δφ=4L(2π/(λ/n)=2π: destructive!

Choose MgFe2 wafer so that Δφ=(2n1L1+2n2L2) (2π/λ)= π+ 2n2L2 (2π/λ)=3π => L2= λ/2n2 = 386 nmWe can add more layers to keep reflecting the light, until nolight is transmitted: all the light is either absorbed orreflected.

Semiconductors such a silicon are usedto build solar cells. They are coatedwith a transparent thin film, whose index of refraction is 1.45, in order tominimize reflected light. If the index ofrefraction of silicon is 3.5, what is the minimum width of the coatingthat will produce the least reflection at a wavelength of 552nm?

Both rays undergo 180 phase changes atreflection, therefore for destructive interference (no reflection), the distancetravelled (twice the thickness) should be equal to half a wavelength in the coating

2 95.1L L nmn

!= " =2

n=1.45

Anti-Reflective CoatingsAnti-Reflective Coatings

Radar waves have a wavelength of 3cm.Suppose the plane is made of metal(speed of propagation=0, n is infinite andreflection on the polymer-metal surfacetherefore has a 180 degree phase change).The polymer has n=1.5. Same calculation as in previous example gives,

30.5

4 4 1.5

cmL cm

n

!= = "

#

On the other hand, if one coated a plane with the same polymer(for instance to prevent rust) and for safety reasons wanted to maximizeradar visibility (reflective coating!), one would have

31

2 2 1.5

cmL cm

n

!= = "

#

Anti-Reflective CoatingsAnti-Reflective Coatings

Stealth Fighter

Michelson Interferometers:

As we saw in the previous example, interference is a spectacular wayof measuring small distances (like the thickness of a soap bubble), sincewe are able to resolve distances of the order of the wavelength of thelight (for instance, for yellow light, we are talking about 0.5 of a millionth of a meter, 500nm). This has therefore technological applications.

In the Michelson interferometer, light from asource (at the left, in the picture) hits a semi-plated mirror. Half of it goes through to the rightand half goes upwards. The two halves arebounced back towards the half plated mirror,interfere, and the interference can be seen by theobserver at the bottom. The observer will seelight if the two distances travelled d1 and d2 areequal, and will see darkness if they differ by halfa wavelength.

Michelson-Morley ExperimentMichelson won the Nobel prize in 1907, "for hisoptical precision instruments and the spectroscopicand metrological investigations carried out withtheir aid"

"The interpretation of these results is that there is no displacement of theinterference bands. ... The result of the hypothesis of a stationary ether is thusshown to be incorrect." (A. A. Michelson, Am. J. Sci, 122, 120 (1881))

The largest Michelson interferometer in the world is in Livingston, LA,in LSU owned land (it is operated by a project funded by the NationalScience Foundation run by Caltech and MIT, and LSU collaborates inthe project).

http://www.ligo-la.caltech.edu

Mirrors are suspendedwith wires and will movedetecting ripples inthe gravitational field dueto astronomical events.

Gravitational Waves Interferometry:an International Dream

GEO600 (British-German)Hannover, Germany

LIGO (USA)Hanford, WA and Livingston, LA

TAMA (Japan)Mitaka

VIRGO (French-Italian)Cascina, Italy

AIGO (Australia), Wallingup Plain, 85km north of Perth