Two-Source Interference of Waves

HomeworkProb. # 1, 2, 4, 5, 9, 10- p. 631-2.

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Interference of WavesReview

What is interference?o Nature of the superposition of waveso Interference is the property of waveso If a phenomenon exhibits interference it must be a

wave.How interference becomes a phenomenon of

interest?o To distinguish between wave and particle behaviouro Light was proven as a wave because of interference

(Thomas Young two-slit interference experiment)

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Two-Source Interference of Waves

Mathematically, it is the superposition of two (or more) waves.

In Physics, some conditions must apply: Interference can be observed only if:

the sources are coherent

The waves have same amplitude (destructive condition)

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Two-Source Interference of Waves Condition for coherence:

same wavelength and frequency In-phase or the phase difference between

the waves (sources) is constant in time.Constant phase difference alters the orientation of the interference pattern.

http://www.youtube.com/watch?v=kO2yFC7_k2s&feature=player_embedded

http://www.youtube.com/watch?v=ovZkFMuxZNc

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Two-Source Interference Examples

Light interference is achieved using the double slit set-up.

Two sound sourcesTwo water waves source

Light double slit experiment

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Two-Point Source Interference Pattern

d2

d1d4

d3

|di+1 – di| = Δd = ?

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Two-Source Interference• Condition for constructive interference:

Δd = nλ n = 0, 1, 2, 3…

------------------------------------------------------------------

• Condition for destructive interference: Δd = (n + ½) λ n= 0, 1, 2, 3…

Resultant amplitude = 0

Path difference Wavelength

Resultant amplitude = 2A (A is the wave amplitude, source)

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Two-Source Interference

• If Δd = mλ where n < m < (n + ½)

0 < resultant amplitude < 2A

(Each of waves has amplitude A)

http://ngsir.netfirms.com/englishhtm/Interference2.htm

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Light interference Two-slit experimentTwo-Slit Interference for Light

Fringe width

Bright fringe

Dark fringe

The slits separation distance and slit width are very small

Central maximum

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Two-Slit Interference – Two relations

What is the condition for constructive/destructive interference at point P?What is the relationship between the location of point P and the two-slit set-up?

Slit width, w, slit separation distance, d, and their relative values to the wavelength are important for interference (w < d = kλ, k < 200)

sin 𝜃 ≈ 𝜃 ≈ tan , for small angles ( in rads)𝜃 𝜃

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Path difference and Fringe Width

• Condition for constructive interference at point P:

• The location sn of the nth bright fringe is given by:

D: distance of screen from the slit set-up λ: wavelength of light

• The fringe width is given by:

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Questions 1. Under which condition a

greater amount of interference is produced:

a) moving the two sources closer together

b) moving the two sources further apart

Explain your answer.

2. When the frequency is increased (sources at fixed distance):

a) less interference is produced

b) more interference is produced

Explain your answer.

The two sources are assumed in phase.(hint: you may draw wavefronts emerging from the sources)

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Answers1. A greater amount of

interference is produced when the two sources are moved further apart, as evidenced by greater number of destructive interference paths. When the amount of interference decreases, the width of constructive interference zone increases.

2. When the frequency is increased, more interference is produced since the wavelengths will decrease, generating more wavefronts between the two sources (this is equivalent to moving the point sources further apart).

http://ngsir.netfirms.com/englishhtm/Interference2.htm

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Problem # 8. p.632Light with λ=644 nm (in air) is incident normally on two narrow parallel slits,

1mm apart. A screen is placed a distance 1.2 m from the slits .a) Determine the distance on the screen between the central maximum and

the fifth bright spot.b) If the experiment were repeated in water (n=1.33), how would the answer

change?

Solutionc) y5 = 5λL/d = 5(644×10-9)(1.2)/10-3 = 3.864×10-3m

or y5 = 3.864 mm

b) n = ca/cw = λaf/λwf = n λw= λair/n

λw= 644x10-9/1.33 = 484×10-9 m

y5 = 5λwL/d = 5(484×10-9)(1.2)/10-3 = 2.905 mm.

The 5th fringe will be closer to the central maximum.

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Phase difference between waves

• If the phase difference φ between the waves stays constant in time, the sources are coherent.

• If φ is present, conditions for constructive/ destructive interference :

Δd = nλ + (φ/2π)λ constructive Δd = (n + ½)λ + (φ/2π)λ destructive

n = 0, 1 ,2, 3, …

Note:

Study of waves interference becomes easier with phasors (out of scope) http://ngsir.netfirms.com/applets/interference/larger/Interference.htm