1. Two-Source Interference of Waves
Homework
Prob. # 1, 2, 4, 5, 9, 10- p

2. Interference of Waves Review What is interference?
Nature of the superposition of waves
Interference is the property of waves
If a phenomenon exhibits interference it must be a wave.
How interference becomes a phenomenon of interest?
To distinguish between wave and particle behaviour
Light was proven as a wave because of interference (Thomas Young two-slit interference experiment)

3. 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)

4. 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.

5. Two-Source Interference Examples
Two water waves source
Two sound sources
Light double slit experiment
Light interference is achieved using the double slit set-up.

6. Two-Point Source Interference Patternd4 d3 d2 d1
|di+1 – di| = Δd = ?

7. 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)

8. Two-Source Interference
If Δd = mλ where n < m < (n + ½)
 0 < resultant amplitude < 2A
(Each of waves has amplitude A)

9. Two-Slit Interference for Light
Light interference  Two-slit experiment
Fringe width
Central maximum
Bright fringe
Dark fringe
The slits separation distance and slit width are very small

10. Two-Slit Interference – Two relations
Slit width, w, slit separation distance, d, and their relative values to the wavelength are important for interference (w < d = kλ, k < 200)
sin𝜃 ≈ 𝜃 ≈ tan𝜃, for small angles (𝜃 in rads)
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?

11. 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:

12. Questions
Under which condition a greater amount of interference is produced:
moving the two sources closer together
moving the two sources further apart
Explain your answer.
When the frequency is increased (sources at fixed distance):
less interference is produced
more interference is produced
Explain your answer.
The two sources are assumed in phase.
(hint: you may draw wavefronts emerging from the sources)

13. Answers
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.
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).

14. Problem # 8. p.632
Light 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 .
Determine the distance on the screen between the central maximum and the fifth bright spot.
If the experiment were repeated in water (n=1.33), how would the answer change?
Solution
y5 = 5λL/d = 5(644×10-9)(1.2)/10-3 = 3.864×10-3m
or y5 = mm
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 = mm.
The 5th fringe will be closer to the central maximum.

15. 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)