1. Interference Principle of Superposition- Constructive Interference
If the crest of one wave meets the crest of another wave then the resulting amplitude is the sum of the amplitude of these two waves.
Example of “Constructive Interference”
Wave 1
Wave 2
Resultant
The amplitude of the resultant wave is the sum of the amplitudes of waves one and two above.

2. Interference - 2 Principle of Superposition
If the crest of one wave meets the trough of another wave then the resulting amplitude is the sum of the amplitude of these two waves.
Example of “Destructive Interference”
Wave 1
Wave 2
Resultant
The amplitude of the resultant wave is the sum of the amplitudes of waves one and two above. In this case they cancel each other.

3. Interference of Sound Waves
Compression Waves in the air emitted by a speaker.
High pressure - crest
Low pressure (rarefaction) - trough 

4. Interference of Sound Waves
Sound is emitted coherently from the two speakers. Where is there constructive and destructive interference?
Constructive
Destructive 

5. Loud spots are areas of constructive interference.
Interference of Sound
Q: What did you hear?
A: Loud and quiet spots.
Loud spots are areas of constructive interference.
Quiet spots are areas of destructive interference

6. Light also shows interference!
Thomas Young (1801)
What do you think you would see if you were to arrange for this experiment to be done with light?
Screen
Bright
Dark
Small Slits
Laser Light

7. Interference of Light
Light areas = Constructive Intereference
Dark areas = Destructive Intereference
But why are there lots of bright and dark fringes?

8. Interference of light Crest + Crest
Count the crests to this point from each slit!
The wave from the bottom slit travels exactly one wavelength further than the waves from the top slit, so a crest and crest still meet! –Hence Constructive Interference occurs
Crest + Crest
Crest + Crest

9. Interference of light – Conditions for maxima
Assume P is the first Maximum (bright fringe)
The path difference (Δ) from S1 to P and S2 to P is d2-d1.
But since we know that P is a maximum we know the path difference must be one wavelength
i.e Δ = d2-d1= λ
If P was the second fringe then Δ would be 2 λ and so on.
Thus for a maximum
Δ=n λ
where n=0,1,2,3 etc
n.b. n is often referred to as the ‘order’ of the maxima

10. Interference of light – Conditions for minima
The case for the areas of destructive interference (or minima) is similar to that for maxima except the minima are half way between the maxima.
This happens when the waves from the bottom slit travels ½ λ more than the wave from the top slit and so the waves meet completely out of phase producing destructive interference. (a crest meets a trough)
Thus for minima: Δ=(n+ ½)λ where n=0,1,2,3 etc
The first order minimum occurs between the central maxima and 1st order maxima i.e. when n=0
Minima

11. Double Slit Interference
In the microwave experiment shown below, C is the zero order maximum and D is the first order maximum. AD = 52 cm and BD = 55 cm.
(a) What is the path difference at point D? A B C D
Screen
(b) What is the wavelength of the microwaves?
(c) What is the path difference to the second order maximum?
(d) What is the path difference to the minimum next to C?
(e) What is the path difference to the next minimum?
(f) What is the path difference at point C?
Click the mouse to continue

12. Double Slit Interference
2. In a microwave interference experiment, H is the second minimum, that is
there is one other minimum between H and G. Measurement of distances EH and FH gives: EH = 42.1 cm and FH = 46.6 cm. Calculate the wavelength and frequency of the microwaves used. E F G H
Screen

13. Path Difference and Geometry
Can we relate the fringe pattern to the geometry of the set up? θ S1 S2 d D Δ S
Note because the screen is relatively very far away the rays can be considered to be parallel
1) sinθ=Δ/d
Δ=dsinθ or
n λ=dsinθ
By measuring θ and d you can calculate the wavelength of light.
2) tanθ = S/D
But for such small angles tanθ=sinθ
So sinθ= nλ/d =S/D
Thus if n=1 then
S=λD/d

14. Double Slit Interference
Light of wavelength 4.2x10-7m falls on a double slit and creates a interference pattern on a screen 3m away. The fringe separation is 4.1mm. Calculate the slit separation?
S = λD/d
d = λD/S
d = (4.2x10-7 x 3)/4.1x10-3
d = m

15. Interference of Sound Waves
Conditions for Constructive and destructive interference
Constructive: Path lengths from each speaker differ by an integral number of wavelengths - where the blue circles intersect or the black dotted circles intersect.
Destructive: Path lengths from each speaker differ by /2, 3 /2, 5 /2 etc. - where the blue and black circles intersect 