7: Interference(Chapter 35)

Phys130, A01Dr. Robert MacDonald

Huygens’ Principle• Each point on a wave front

serves as a source of new spherical wavelets.

• After a time t, the new position of the wave front is the surface tangent to these wavelets.

We can use this model to study how a wave:

• bends around edges

• interferes with itself or other waves

• reflects and refracts (!).

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Fig. 35.2: Huygens’ Principle fora plane wave in vaccuum.

Huygens & Refraction

Fig. 35-3

λ2 = v2 T

λ1 = v1 Td

http://www.walter-fendt.de/ph14e/huygenspr.htm

Diffraction & InterferenceWaves bend (“diffract”) at at barrier edges.

Small slits act like sources of circular waves!

A barrier with 2 slits will produce two sets of circular waves.

• Made by the same incoming wave. ! coherent.

! Interference! (Just like two speakers, two ninja sound-guns, etc.)

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Young’s Double SlitExperiment (1801)

Fig. 35-8Fig. 35-9

Young’s Double Slit ExperimentIn 1800 Thomas Young passed light through a pair of closely-spaced slits.

He saw exactly the sort of interference pattern on the screen that you’d expect for two overlapping waves. (So light is a wave! Huzzah!)

We’ll use this experiment as an example to study some of the properties of two-source interference.

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Recall: Interference & PhaseHow two waves interfere at some point depends on their relative phase.

Remember that phase is kx – ωt + ϕ0.

So the relative phase of the two waves at some point depends on:

• how each wave was created (their ϕ0 values).

• how far each wave traveled (path difference).

• what happened to each wave along the way (reflections etc).

For Young’s experiment, only the path length differs.

Path Length DifferenceThe interference pattern is caused by the difference in path length from the two slits to each point on the screen. Each point P can be identified by the angle θ in the diagram.

In practice, the slit-screen distance R is far larger than the slit separation d, so the rays from the two slits are pretty much parallel.

The path length difference is given by: ∆L = d sinθ.

Then the phase difference will be! ϕ = (d sinθ)•(2π)/λ.

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Actual geometry

Approximate geometry

Fig. 35-10

Phase Diff and InterferenceTwo waves will interfere constructively when their phase difference is ! ϕ = 2πm (m = 0, ±1, ±2, etc.).

Two waves will interfere destructively when their phase difference is! ϕ = 2π(m + 1⁄2) (m = 0, ±1, ±2, etc.).

If their phase difference doesn’t look like either of these, then the interference is neither constructive nor destructive; it’s something in between.

(These statements are always true, for any interference.)

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Young: Fringe Locations

The light bands (“bright fringes”) occur at angles where the phase difference is 2πm, which gives:! d sinθm = mλ (m = 0, ±1, ±2, ...)

The dark bands (“dark fringes”) occur at angles where the phase difference is 2π(m + 1⁄2), which gives:! d sinθm = (m + 1⁄2)λ (m = 0, ±1, ±2, ...)

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Constructiveinterference

Destructiveinterference

Remember: for two-slit inteference, the phase difference is (d sinθ)•(2π)/λ.

Light from a helium arc lamp (λ = 502 nm) is passed through a double slit, and the resulting diffraction pattern is viewed on a screen 1.20 m away. The centre of the 20th fringe is found to be 10.6 mm from the central bright fringe.

What is the separation of the two slits?

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Example: double slit Double-Slit and White LightThe width of the diffraction pattern depends on the wavelength of light.

If you use a single colour (e.g. green, wavelength 550 nm) you get a nice neat pattern like the one on the left (which is often very small).

If you use white light, though, you get a pattern like the one on the right.

• Does the order of colours make sense?

• Why is the central fringe white?

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Two source intensity

I0 = intensity of each source (e.g. light at each slit).Intensity I at the screen is given by:

where ϕ is the phase difference we found earlier:! ϕ = (d sinθ)•(2π)/λ

Two identical radio antennas, 10.0 m apart, are transmitting at a frequency of f = 60.0 MHz.

The signal intensity from each antenna at the distance we’re at is I0 = 0.020 W/m2.

What is the intensity in the direction θ=4.0º (0.0700 rad)?

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Example: Antenna Array

Phase Flips!When the wave meets the end of the string and reflects, it will sometimes be flipped over after the reflection, depending on whether the end was loose or fixed.

It turns out this also happens if you attach the string to another string, which is lighter or heavier than the first.

• If the second rope is lighter, you don’t get an inversion (not flipped), just like a loose end.

• If the second rope is heavier, the reflection is inverted (flipped), just like a fixed end.

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(Remember that waves travel faster on light ropes than on heavier ropes.)

Flipping the wave is the same as shifting by a half cycle, or changing the phase by π.

Light does the same sort of thing.

• Reflecting off a lower index n: No flip.

• Reflecting off a higher index n: Flip!! Add π to the phase difference for each flip.

Fig. 35.16

Interference in Thin FilmsWhen we see a soap bubble or oil slick, we see light reflected from the top (or closer) surface, but also light reflected from the bottom surface.

Light both reflects and refracts at the top surface — it splits into two coherent waves.

• One reflects directly back to the eye.

• One refracts into the film, reflects, then comes back out to reach the eye.

For a thin film, both rays come from basically the same spot.

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“Giant Soap Bubble” by flickr user sunnyUKhttp://flickr.com/photos/sunnyuk/2794253120

Recall: Interference & Phase

For the two rays we get from light on a thin film:

• Ray 2 (into the film) travels farther.

• Ray 2 travels through a different medium (different λ!)

• Both rays are reflected, so each one might be flipped over (adding π to the relative phase).

Remember that phase is kx – ωt + ϕ0.So the relative phase of the two waves at some point depends on:• how each wave was created (their ϕ0 values).• how far each wave traveled (path difference).• what happened to each wave along the way.

Let’s assume (for simplicity) the light is striking the film at normal incidence (θ = 0).

The phase difference between r1 and r2 includes:

• Any “phase flips” from the reflections.

• Phase difference from the travel ray 2 spends in the film.

If the film’s thickness is L, then for θ = 0 the path difference is 2L (straight in and out).

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Fig. 35.15

We’ll draw the diagrams with huge angles like this just so we can see what’s going on.

So the phase difference you get from this extra travel is! 2π(2L)/λ,where λ is the wavelength in the medium:! λ = nλ0 (λ0 = wavelength in vaccum).

Add π for each phase flip.

• Depends on what n1, n2, and n3 are!

e.g. For a thin film surrounded by air, you have n1 = n3 = 1, and n2 > 1.

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Fig. 35.17

Reflecting off a higher n:

Flip! (Add π.)

Reflecting off a lower n:No flip!

Total phase difference between r1 and r2:! 2π(2L)/λ + π

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Two glass plates are separated at one end by a piece of paper of thickness h, and are in contact at the other end. With light of wavelength λ0 = 500nm, the interference fringes are found to be 1.25 mm apart.

How thick is the paper?

Many lenses (glasses, cameras, etc.) have anti-reflective coatings. These are just thin films designed to cancel out the reflected light using destructive interference.

• Only works completely for a single wavelength!

• Usually designed to cancel green (mid-spectrum).

Then red and blue light does reflect, at least a little bit. So these lenses appear to have a purple-ish coating.

You can also use a coating to make an extra-reflective surface, using constructive interference in a film.

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Anti-reflective coatings

The Michelson InterferometerFig. 35.21

LIGO

L1

L2

Laser InterferometerGravitational-wave Observatoryhttp://www.ligo.caltech.edu

Shown: LIGO Hanford Observatory, Hanford, Washington.