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. θsindL =Δ

integer 0,1,2...m :veConstructi =,=Δ λmL

integer 0,1,2...m )21

( :eDestructiv =,+=Δ λmL

λθ md =sin :veConstructi

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

⎟⎠⎞

⎜⎝⎛=dmλ

θ arcsin

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

1 2 0 02 cos 2 cosE E E E Eφβ ⎛ ⎞= + = = ⎜ ⎟2⎝ ⎠

r r

⎟⎠

⎞⎜⎝

⎛2

==φ2

20

2

0

cos4E

E

I

I

θλπ

λπφ sindL ⎟

⎠⎞

⎜⎝⎛2

=Δ⎟⎠⎞

⎜⎝⎛2

=

The ReturnOf the Phasor!

Phase Difference2π=360°π=180°

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

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

ExampleExampleIn a double slit experiment, we can measure the wavelength of the light if we know the distances between the slits and the angular separation of the fringes. If the separation between the slits is 0.5mm and the first order maximum of the interference pattern is at an angle of 0.059o from the center of the pattern, what is the wavelength and color of the light 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 coherent 600nm light. At what distance above the central maximum is the average intensity on the screen 75% of the maximum?

I/I0=4cos22 ; I/Imax=cos22 =0.75 => =2cos–1 (0.75)1/2=60o=π/3 rad=(2πd/λ)sinθ => θ= sin-1(λ/2πd)22rad (small!)y=Lθ48m

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. If the car is positioned at the second maximum, what is the wavelength of 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=0.625m travels through glass (n=1.46) a distance of 1mm. A second beam, parallel to the first one and originally in phase

with it, travels the same distance through sapphire (n=1.77).

•How many wavelengths are there of each beam inside the material? In glass, λg=0.625m/1.46= 0.428 m and Ng=D/ λg=2336.45In sapphire, λs=0.625m/1.77= 0.353 m (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 Δ=ΔNx360o=0.41x360o=148o

•How thick should the glass be so that the beams are exactly out of phase at the exit (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 m would make the waves OUT of phase.For example, 1.008 mm makes them in phase, and 1.010 mm makes them OUT of phase.

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 the speed of propagation in the rope of the right is faster than on the left. (Low impedance.)

The reflected pulse is DOWN if the speed of propagation in the right is slower than on the 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 180 degrees OUT of phase, and UP means 360° or IN PHASE.

Thin FilmsFirst reflected lightray comes from firstinterface, second from second. These rays interfere with each 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.

22number odd

2 :phaseIn n

Lλ×=

2integer2 :phase-Anti

nL

λ×=

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 (due to 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

Air

Soap: n>1

180°

0°

Reflective CoatingsReflective CoatingsTo make mirrors that reflects light of only a given wavelength, a coating of a specific thickness 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) and CeO2 (n=2.35), and are called “dielectric films”. What thickness is necessary for reflecting IR light with λ=1064nm?

n=2.35n=1.38

First ray: Δ=180deg=πSecond ray: Δ=2L(2π/(λn)=π=> L= λCeo2λ/n)/nmThird ray? If wafer has the same thickness (and is of the same material), Δ=4L(2π/(λn)=2π: destructive!

Choose MgFe2 wafer so that Δ2n1L1+2n2L2) (2π/λ)= π2n2L2 (2π/λ)=3πL2= λ2n2386 nm

We can add more layers to keep reflecting the light, until no light is transmitted: all the light is either absorbed or reflected.

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

λ= ⇒ =

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

λ= = ⇒

×

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

λ= = ⇒

×

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 right and half goes upwards. The two halves are bounced back towards the half plated mirror, interfere, and the interference can be seen by the observer at the bottom. The observer will see light if the two distances travelled d1 and d2 are equal, and will see darkness if they differ by half a wavelength.

Michelson-Morley ExperimentMichelson won the Nobel prize in 1907, "for his optical precision instruments and the spectroscopic and metrological investigations carried out with their aid"

"The interpretation of these results is that there is no displacement of the interference bands. ... The result of the hypothesis of a stationary ether is thus shown 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 in the project).

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

Mirrors are suspended with wires and will movedetecting ripples inthe gravitational field due to 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