1. Chapter 35
Interference

2. Learning Goals for Chapter 35
Looking forward at …
what happens when two waves combine, or interfere, in space.
how to understand the interference pattern formed by the interference of two coherent light waves.
how to calculate the intensity at various points in an interference pattern.
how interference occurs when light reflects from the two surfaces of a thin film.
how interference makes it possible to measure extremely small distances.

3. Principle of superposition
The term interference refers to any situation in which two or more waves overlap in space.
When this occurs, the total wave at any point at any instant of time is governed by the principle of superposition:
When two or more waves overlap, the resultant displacement at any point and at any instant is found by adding the instantaneous displacements that would be produced at the point by the individual waves if each were present alone.

4. Wave fronts from a disturbance
Interference effects are most easily seen when we combine sinusoidal waves with a single frequency and wavelength.
Shown is a “snapshot” of a single source S1 of sinusoidal waves and some of the wave fronts produced by this source.

5. Constructive and destructive interference
Shown are two identical sources of monochromatic waves, S1 and S2.
The two sources are permanently in phase; they vibrate in unison.
Constructive interference occurs at point a (equidistant from the two sources).

6. Conditions for constructive interference
The distance from S2 to point b is exactly two wavelengths greater than the distance from S1 to b.
The two waves arrive in phase, and they reinforce each other.
This is called constructive interference.

7. Can light waves interfere like sound waves?
Superposition principle: When two or more waves overlap, the resultant displacement at any instant is the sum of the displacements of each of the individual waves. ?

8. Coherence
In order to see interference of light the light must be coherent
Two or more light beams of the same frequency are coherent if they are monochromatic (single l) and have a fixed phase relationship.
Light bulbs and the sun produce incoherent light—individual pulses of light do not have a fixed phase relationship with each other.
Incoherent light: Pulses are randomly emitted on average about every 10-8 s. When they arrive at a screen, the pulse’s interference pattern changes too rapidly to see. We would just see a bright spot.
So light can interfere, but you need coherent, monochromatic light.
Using one source of light (with two or more slits) produces coherence; lasers also produce coherent light.
Coherent light from a laser. The pulses have a fixed phase relationship. So when they hit the screen, we would always see the same interference pattern.

9. Conditions for destructive interference
The distance from S1 to point c is a half-integral number of wavelengths greater than the distance from S2 to c.
The two waves cancel or partly cancel each other.
This is called destructive interference.

10. Constructive and destructive interference
Shown are two identical sources of monochromatic waves, S1 and S2, which are in phase.
The red curves show all positions where constructive interference occurs; these curves are called antinodal curves.
Not shown are the nodal curves, which are the curves that show where destructive interference occurs.

11. Constructive and destructive interference
Two sources are in phase
Two sources are in phase
Figure 35.2 at the right shows two coherent wave sources. Two sources are in phase.
Constructive interference occurs when the path difference is an integral number of wavelengths.
Destructive interference occurs when the path difference is a half-integral number of wavelengths.

12. Constructive and destructive interference
The concepts of constructive interference and destructive interference apply to these water waves as well as to light waves and sound waves.

13. Two-source interference of light
Shown below is one of the earliest quantitative experiments to reveal the interference of light from two sources, first performed by Thomas Young.
The interference of waves from slits S1 and S2 produces a pattern on the screen.

14. Young’s Double Slit Experiment
As Huygen’s principle suggests, each slit S1 and S2 acts as a separate source of coherent light (like the loudspeakers for sound waves).

15. Two-source interference of light
(b) shows the actual geometry of Young’s experiment.
If the distance R to the screen is much greater than the distance d between the slits, we can use the approximate geometry shown in (c).

16. Analyzing Two-Slit Interference
Far-field (Fraunhofer) assumption
Just like with the loudspeakers, the path difference is the critical thing: path difference = Dr
Path difference =
m = 0, 1, 2, 3, …
For constructive interference (bright spots),
tan(theta) = y/L but for small angles tan(theta) = theta = y/L
(For small angles)
Bright fringes

17. For destructive interference (dark spots),
Path difference D r = d sin q = (m + ½) l m = 0, 1, 2, 3,…
Dark fringes
Young’s experiment seemed to prove light was a wave. Newton’s idea of light as a particle was dead (until 1905, when Einstein brought it back).

18. Two sources S1 and S2 oscillating in phase emit sinusoidal waves.
Q35.1
Two sources S1 and S2 oscillating in phase emit sinusoidal waves.
Point P is 7.3 wavelengths from source S1 and 4.3 wavelengths from source S2. As a result, at point P there is
A. constructive interference.
B. destructive interference.
C. neither constructive nor destructive interference.
D. not enough information given to decide
Answer: A

19. Two sources S1 and S2 oscillating in phase emit sinusoidal waves.
Q35.1
Two sources S1 and S2 oscillating in phase emit sinusoidal waves.
Point P is 7.3 wavelengths from source S1 and 4.3 wavelengths from source S2. As a result, at point P there is
A. constructive interference.
B. destructive interference.
C. neither constructive nor destructive interference.
D. not enough information given to decide
Answer: A

20. Example: Double Slit Interference
Light from a helium-neon laser (l = 633 nm) illuminates two slits placed 0.40 mm apart. A viewing screen is placed 2.0 m behind the slits.
What are the distances Dy2 between the two m = 2 bright fringes and Dy2’ between the two m = 2 dark fringes?

21. Example: Measuring the Wavelength of Light
A screen is placed 1.0 m behind a pair of slits, which are spaced 0.30 mm apart. When this system is illuminated by a certain frequency of monochromatic light, ten bright fringes are found to span a distance of 1.65 cm on the screen (center to center).
What is the wavelength of the light?

22. Electric field in interference patterns
To find the intensity at any point in a two-source interference pattern, we have to combine the two sinusoidally varying fields (from the two sources) at a point P.
If the two sources are in phase, then the waves that arrive at P differ in phase by an amount ϕ that is proportional to the difference in their path lengths,
If we further assume the amplitudes of the two waves are both approximately equal to E at point P, the combined amplitude is:

23. Young’s Double Slit Intensity Distribution
Choose the phase of the 1st slit at the screen at t = 0 to be 0. Then:
Total Electric field at point P:
Trig. Identity:
With  = (t + ), = t, get:

24. Double Slit Intensity Distribution: “Phasor” Treatment
Recall harmonic oscillation with amplitude A and angular frequency  can be represented as projection on x or y axis of a rotating vector (phasor) of magnitude (length) A rotating about origin.
Light
We can use this concept to add oscillations with the same frequency, but different phase constant  by “freezing” this rotation in time and treating the 2 oscillations as fixed vectors, the so-called “phasor method”
Phasor is not a physical quantity, it is a geometric entity that helps us to describe and analyze physical quantities that varies sinusoidally with time

25. Phasor diagram for superposition
To add the two sinusoidal functions with a phase difference, we can use the same phasor representation that we used for simple harmonic motion (Chapter 14) and for voltages and currents in ac circuits (Chapter 31).
Each sinusoidal function is represented by a rotating vector (phasor) whose projection on the horizontal axis at any instant represents the instantaneous value of the sinusoidal function.

26. Double Slit Intensity Distribution: “Phasor” Treatment
Use phasor diagram to do the addition E1 + E2

27. So, ETOT has an “oscillating” amplitude:
From chapter 32 : The average energy transfer over one cycle is ½ the maximum when E and B peak.
where
If sources are in phase, then
Hw 5
Therefore

28. Interference in thin films

29. Phase shifts during reflection
Reflection Phase Change: When light traveling in a transparent medium (or vacuum) enters another transparent material with a larger index of refraction, the reflected wave has a “phase change” of p radians. If the new index of refraction is smaller, there is no phase change.

30. Thin film interference
But wave 2 has also traveled about 2t farther than wave 1. So the conditions for interference are:
constructive: 2t = (m +1/2) ln
destructive: 2t = m ln
where m = 0, 1, 2, 3, 4, …
No phase change
Note: ln = l/n
The conditions are reversed for waves 3 and 4:
constructive: 2t = m ln
destructive: 2t = (m +1/2) ln
where m = 0, 1, 2, 3, 4, …

31. Thin-film Interference Examples
Iridescence in seaweeds
Iridescence in a Soap Bubble
Iridaea, an iridescent seaweed found in tidal pools along the central and northern coast of California. Alternating surface layers are about 200 nm thick. If the light emerging from the various reflections is in phase, iridescence is seen.
Iridescence from a shell
Iridescence from Peacock Feathers

32. Nonreflective coatings

33. Example: Designing an Antireflection Coating
Magnesium fluoride (MgF2) is often used as an antireflection coating on lenses. The index of refraction of MgF2 is n = 1.39.
What is the thinnest film thickness of MgF2 that works as an antireflection coating for light with l = 510 nm (near the center of the visible spectrum)?
Destructive interference: 2t = (m +1/2) ln
where m = 0, 1, 2, 3, 4, … ln = l/n
HW 10

34. Light Beam Splitting: The Half-Silvered Mirror
One way of making a mirror is by placing a flat glass plate in a vacuum chamber and evaporating a reflective metal (e.g., silver or chromium) on its surface. If this process is halted before a solid reflective coating is achieved, a “partially-silvered” mirror is created, which reflects and transmits fractions of the incident light. This is useful because the reflected and transmitted beams are coherent, (i.e., have a definite common phase) even when the incident light has a time-varying random phase.
f = 00 *
900
Half-silvered
Mirror.
f = 900
You will recall that we previously stated that reflection from a surface produces a “phase-change” of 1800 in the reflected light. This is true for a 1800 reflection. However, when light is reflected at an angle other than 1800, the phase change in the reflected light is the same as the reflection angle. Thus, light reflected at 900 has a 900 change in phase (a quarter wavelength).

35. Michelson interferometer

36. Example: Measuring the Wavelength of Light
An experimenter uses a Michaelson interferometer to measure one of the wavelengths of light emitted by electrically excited neon atoms. She slowly moves the mirror M2 and uses a photo-detector and a computer to determine that 10,000 new bright central spots have appeared. She then determines with a micrometer that the mirror has moved mm.
What is the wavelength of the light?