1. Chapter 34 The Wave Nature of Light; Interference
Chapter 34 Opener. The beautiful colors from the surface of this soap bubble can be nicely explained by the wave theory of light. A soap bubble is a very thin spherical film filled with air. Light reflected from the outer and inner surfaces of this thin film of soapy water interferes constructively to produce the bright colors. Which colors we see at any point depends on the thickness of the soapy water film at that point and also on the viewing angle. Near the top of the bubble, we see a small black area surrounded by a silver or white area. The bubble’s thickness is smallest at that black spot, perhaps only about 30 nm thick, and is fully transparent (we see the black background). We cover fundamental aspects of the wave nature of light, including two-slit interference and interference in thin films.

2. Units of Chapter 34
Waves versus Particles; Huygens’ Principle and Diffraction
Huygens’ Principle and the Law of Refraction
Interference – Young’s Double-Slit Experiment
Intensity: Double-Slit Interference Pattern
Interference in Thin Films
Michelson Interferometer
Luminous Intensity

3. 34-1 Waves versus Particles; Huygens’ Principle and Diffraction
Huygens’ principle: every point on a wave front acts as a point source; the wave front as it develops is tangent to all the wavelets.
Figure Huygens’ principle, used to determine wave front CD when wave front AB is given.

4. 34-1 Waves versus Particles; Huygens’ Principle and Diffraction
Huygens’ principle is consistent with diffraction:
Figure Huygens’ principle is consistent with diffraction (a) around the edge of an obstacle, (b) through a large hole, (c) through a small hole whose size is on the order of the wavelength of the wave.

5. 34-2 Huygens’ Principle and the Law of Refraction
Huygens’ principle can also explain the law of refraction.
As the wavelets propagate from each point, they propagate more slowly in the medium of higher index of refraction.
This leads to a bend in the wave front and therefore in the ray.

6. 34-2 Huygens’ Principle and the Law of Refraction
Figure Refraction explained, using Huygens’ principle. Wave fronts are perpendicular to the rays.

7. 34-2 Huygens’ Principle and the Law of Refraction
The frequency of the light does not change, but the wavelength does as it travels into a new medium:

8. 34-2 Huygens’ Principle and the Law of Refraction
Highway mirages are due to a gradually changing index of refraction in heated air.
Figure (a) A highway mirage. (b) Drawing (greatly exaggerated) showing wave fronts and rays to explain highway mirages. Note how sections of the wave fronts near the ground move faster and so are farther apart.
Hotter air = less dense, smaller “n”, like diverging lens!

9. Young’s Double-Slit Experiment
If light is a wave, interference effects will be seen, where one part of a wave front can interact with another part.
One way to study this is to do a double-slit experiment:
Figure (a) Young’s double-slit experiment. (b) If light consists of particles, we would expect to see two bright lines on the screen behind the slits. (c) In fact, many lines are observed. The slits and their separation need to be very thin.

10. 34-3 Interference – Young’s Double-Slit Experiment
If light is a wave, there should be an interference pattern.
Figure If light is a wave, light passing through one of two slits should interfere with light passing through the other slit.

11. Young’s Double-Slit Experiment
The interference occurs because each point on the screen is not the same distance from both slits. Depending on the path length difference, the wave can interfere constructively (bright spot) or destructively (dark spot).
Figure How the wave theory explains the pattern of lines seen in the double-slit experiment. (a) At the center of the screen the waves from each slit travel the same distance and are in phase. (b) At this angle θ, the lower wave travels an extra distance of one whole wavelength, and the waves are in phase; note from the shaded triangle that the path difference equals d sin θ. (c) For this angle θ, the lower wave travels an extra distance equal to one-half wavelength, so the two waves arrive at the screen fully out of phase. (d) A more detailed diagram showing the geometry for parts (b) and (c).

12. Young’s Double-Slit Experiment
We can use geometry to find the conditions for constructive and destructive interference:
and

13. Young’s Double-Slit Experiment
Between the maxima and the minima, the interference varies smoothly.
Figure (a) Interference fringes produced by a double-slit experiment and detected by photographic film placed on the viewing screen. The arrow marks the central fringe. (b) Graph of the intensity of light in the interference pattern. Also shown are values of m for Eq. 34–2a (constructive interference) and Eq. 34–2b (destructive interference).

14. Young’s Double-Slit Experiment
Conceptual Example 34-1: Interference pattern lines.
(a) Will there be an infinite number of points on the viewing screen where constructive and destructive interference occur, or only a finite number of points? (b) Are neighboring points of constructive interference uniformly spaced, or is the spacing between neighboring points of constructive interference not uniform?
Solution. a. Due to the fact that sin θ cannot be greater than 1, the maximum value of m is equal to the (truncated) value of d/λ.
b. The spacing increases with θ, although for small θ it is close to being uniform.

15. Example 34-2: Line spacing for double-slit interference.
A screen containing two slits mm apart is 1.20 m from the viewing screen. Light of wavelength λ = 500 nm falls on the slits from a distant source. Approximately how far apart will adjacent bright interference fringes be on the screen?
Figure Examples 34–2 and 34–3. For small angles θ (give θ in radians), the interference fringes occur at distance x = θl above the center fringe (m = 0); θ1 and x1 are for the first-order fringe (m = 1), θ2 and x2 are for m = 2.
Solution: Using the geometry in the figure, x ≈ lθ for small θ, so the spacing is 6.0 mm.

16. Conceptual Example 34-3: Changing the wavelength.
(a) What happens to the interference pattern in the previous example if the incident light (500 nm) is replaced by light of wavelength 700 nm?
(b) What happens instead if the wavelength stays at 500 nm but the slits are moved farther apart?
Solution: a. As the wavelength increases, the fringes move farther apart.
b. Increasing the slit spacing causes the fringes to move closer together.

17. Young’s Double-Slit Experiment
Since the position of the maxima (except the central one) depends on wavelength, the first- and higher-order fringes contain a spectrum of colors.
Figure First-order fringes are a full spectrum, like a rainbow.

18. Example 34-4: Wavelengths from double-slit interference.
White light passes through two slits 0.50 mm apart, and an interference pattern is observed on a screen 2.5 m away. The first-order fringe resembles a rainbow with violet and red light at opposite ends. The violet light is about 2.0 mm and the red 3.5 mm from the center of the central white fringe. Estimate the wavelengths for the violet and red light.
Solution: Using the geometry of the previous examples and the small-angle approximation, the wavelength for the first-order fringe is given by dx/l = 400 nm for the violet light and 700 nm for the red light.

19. 34-4 Intensity in the Double-Slit Interference Pattern
The electric fields at the point P from the two slits are given by .
Figure Determining the intensity in a double-slit interference pattern. Not to scale: in fact l >> d and the two rays become essentially parallel.
where

20. 34-4 Intensity in the Double-Slit Interference Pattern
The two waves can be added using phasors, to take the phase difference into account:
Figure Phasor diagram for double-slit interference pattern.

21. 34-4 Intensity in the Double-Slit Interference Pattern
The time-averaged intensity is proportional to the square of the field:

22. 34-4 Intensity in the Double-Slit Interference Pattern
This plot shows the intensity as a function of angle.
Figure Intensity I as a function of phase difference δ and position on screen y (assuming y << l).

23. Example 34-5: Antenna intensity.
Two radio antennas are located close to each other, separated by a distance d. The antennas radiate in phase with each other, emitting waves of intensity I0 at wavelength λ. (a) Calculate the net intensity as a function of θ for points very far from the antennas.
Figure 34-15: Example 34–5.The two dots represent the antennas.
Solution: a. The points of constructive and destructive interference are the same as those for the double slit; eq or 34-7 can be used to find the intensity as a function of angle.
b. When d = λ, I = I0 cos2(π sin θ). I is a maximum when sin θ = +1, 0, or -1, and a minimum when sin θ = +1/2 or -1/2.
c. When d = λ/2, I is a maximum when θ = 0 and 180°, and a minimum when θ = 90° and 270°.

24. Example 34-5: Antenna intensity.
Two radio antennas are located close to each other, separated by a distance d. The antennas radiate in phase with each other, emitting waves of intensity I0 at wavelength λ. (b) For d = λ, determine I and find in which directions I is a maximum and a minimum. (c) Repeat part (b) when d = λ/2.
Figure 34-15: Example 34–5.The two dots represent the antennas.
Solution: a. The points of constructive and destructive interference are the same as those for the double slit; eq or 34-7 can be used to find the intensity as a function of angle.
b. When d = λ, I = I0 cos2(π sin θ). I is a maximum when sin θ = +1, 0, or -1, and a minimum when sin θ = +1/2 or -1/2.
c. When d = λ/2, I is a maximum when θ = 0 and 180°, and a minimum when θ = 90° and 270°.

25. 34-5 Interference in Thin Films
Path lengths can differ, & waves interfere, if waves travel through different media.
For very thin film – a few wavelengths thick – light reflects from both bottom & top of layer, causing interference.
Seen in soap bubbles and oil slicks.
Figure Thin film interference patterns seen in (a) a soap bubble, (b) a thin film of soapy water, and (c) a thin layer of oil on wet pavement.

26. 34-5 Interference in Thin Films
The wavelength of the light will be different in the oil and the air, and the reflections at points A and B may or may not involve phase changes.
Figure Light reflected from the upper and lower surfaces of a thin film of oil lying on water. This analysis assumes the light strikes the surface nearly perpendicularly, but is shown here at an angle so we can display each ray.

27. 34-5 Interference in Thin Films
A similar effect takes place when a shallowly curved piece of glass is placed on a flat one. When viewed from above, concentric circles appear that are called Newton’s rings.
Figure Newton’s rings. (a) Light rays reflected from upper and lower surfaces of the thin air gap can interfere. (b) Photograph of interference patterns using white light.

28. 34-5 Interference in Thin Films
A beam of light reflected by a material with index of refraction greater than that of the material in which it is traveling, changes phase by 180° or ½ cycle.
Figure (a) Reflected ray changes phase by 180° or ½ cycle if n2 > n1, but (b) does not if n2 < n1.

29. 34-5 Interference in Thin Films
Example 34-6: Thin film of air, wedge-shaped.
A very fine wire 7.35 x 10-3 mm in diameter is placed between two flat glass plates. Light whose wavelength in air is 600 nm falls (and is viewed) perpendicular to the plates and a series of bright and dark bands is seen. How many light and dark bands will there be in this case? Will the area next to the wire be bright or dark?
Figure (a) Light rays reflected from the upper and lower surfaces of a thin wedge of air interfere to produce bright and dark bands. (b) Pattern observed when glass plates are optically flat; (c) pattern when plates are not so flat. See Example 34–6.
Solution: The path lengths are different for the rays reflected from the upper and lower surfaces; in addition, the ray reflected from the lower surface undergoes a 180° phase change. Dark bands will occur when 2t = (m + ½)λ. At the position of the wire, t is 24.5 wavelengths. This is a half-integer; the area next to the wire will be bright, and there will be 25 dark bands between it and the other edge. Including the band next to the wire, there will also be 25 light bands.

30. 34-5 Interference in Thin Films
Example 34-7: Thickness of soap bubble skin.
A soap bubble appears green (λ = 540 nm) at the point on its front surface nearest the viewer. What is the smallest thickness the soap bubble film could have? Assume n = 1.35.
Figure 34-21: Example 34–7.The incident and reflected rays are assumed to be perpendicular to the bubble’s surface. They are shown at a slight angle so we can distinguish them.
Solution: The path length difference is twice the thickness of the film; the ray reflecting from the first surface undergoes a 180° phase change, while the other ray does not. The smallest thickness where the green light will be bright is when 2t = λ/2n, or t = 100 nm. Constructive interference then occurs for every additional 200 nm in thickness.

31. 34-5 Interference in Thin Films
Problem Solving: Interference
Interference occurs when two or more waves arrive simultaneously at the same point in space.
Constructive interference occurs when the waves are in phase.
Destructive interference occurs when the waves are out of phase.
An extra half-wavelength shift occurs when light reflects from a medium with higher refractive index.

32. 34-5 Interference in Thin Films
Example 34-8: Nonreflective coating.
What is the thickness of an optical coating of MgF2 whose index of refraction is n = 1.38 and which is designed to eliminate reflected light at wavelengths (in air) around 550 nm when incident normally on glass for which n = 1.50?
Figure Example 34–8. Incident ray of light is partially reflected at the front surface of a lens coating (ray 1) and again partially reflected at the rear surface of the coating (ray 2), with most of the energy passing as the transmitted ray into the glass.
Solution: To eliminate reflection, the two rays should interfere destructively. Both rays change phase on reflection, so the net result is no phase difference due to reflection. The condition for destructive interference is then 2t = λ/2n, or t = 99.6 nm.

33. 34-6 Michelson Interferometer
The Michelson interferometer is centered around a beam splitter, which transmits about half the light hitting it and reflects the rest. It can be a very sensitive measure of length.
Figure Michelson interferometer.

34. 34-7 Luminous Intensity
The intensity of light as perceived depends not only on the actual intensity but also on the sensitivity of the eye at different wavelengths.
Luminous flux: 1 lumen = 1/683 W of 555-nm light
Luminous intensity: 1 candela = 1 lumen/steradian
Illuminance: luminous flux per unit area

35. 34-7 Luminous Intensity Example 34-9: Lightbulb illuminance.
The brightness of a particular type of 100-W lightbulb is rated at 1700 lm. Determine (a) the luminous intensity and (b) the illuminance at a distance of 2.0 m.
Solution: a. Assume the light output is uniform in all directions. Then the luminous intensity is 1700 lm/4π sr = 135 cd.
b. At d = 2.0 m, the illuminance is 1700 lm/(4πd2) = 34 lm/m2.

36. Summary of Chapter 34
The wave theory of light is strengthened by the interference and diffraction of light.
Huygens’ principle: every point on a wave front is a source of spherical wavelets.
Wavelength of light in a medium with index of refraction n:
Young’s double-slit experiment demonstrated interference.

37. Summary of Chapter 34
In the double-slit experiment, constructive interference occurs when
and destructive interference when
Two sources of light are coherent if they have the same frequency and maintain the same phase relationship.

38. Summary of Chapter 34
Interference can occur between reflections from the front and back surfaces of a thin film.
Light undergoes a 180° phase change if it reflects from a medium of higher index of refraction.