1. Chapter 35-Diffraction
Chapter 35 opener. Parallel coherent light from a laser, which acts as nearly a point source, illuminates these shears. Instead of a clean shadow, there is a dramatic diffraction pattern, which is a strong confirmation of the wave theory of light. Diffraction patterns are washed out when typical extended sources of light are used, and hence are not seen, although a careful examination of shadows will reveal fuzziness. We will examine diffraction by a single slit, and how it affects the double-slit pattern. We also discuss diffraction gratings and diffraction of X-rays by crystals. We will see how diffraction affects the resolution of optical instruments, and that the ultimate resolution can never be greater than the wavelength of the radiation used. Finally we study the polarization of light.

2. 34-3 Interference – 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.

3. 34-5 Interference in Thin Films
Another way path lengths can differ, and waves interfere, is if they travel through different media. If there is a very thin film of material – a few wavelengths thick – light will reflect from both the bottom and the top of the layer, causing interference. This can be 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.

4. 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.
If the Path difference 2t=ABC= mλn the two waves reach the eye in phase and (m+1/2) λn if they are out of phase
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.

5. 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.

6. 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.
Phase change
Φ=2mπ for bright fringes
Φ=(2m+1)π for dark fringes

7. 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.

8. 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.

9. 35-1 Diffraction by a Single Slit or Disk
If light is a wave, it will diffract around a single slit or obstacle.
Figure If light is a wave, a bright spot will appear at the center of the shadow of a solid disk illuminated by a point source of monochromatic light.

10. 35-1 Diffraction by a Single Slit or Disk
The resulting pattern of light and dark stripes is called a diffraction pattern.
Figure Diffraction pattern of (a) a circular disk (a coin), (b) razor, (c) a single slit, each illuminated by a coherent point source of monochromatic light, such as a laser.

11. 35-1 Diffraction by a Single Slit or Disk
This pattern arises because different points along a slit create wavelets that interfere with each other just as a double slit would.
Figure Analysis of diffraction pattern formed by light passing through a narrow slit of width D.

12. 35-1 Diffraction by a Single Slit or Disk
The minima of the single-slit diffraction pattern occur when
Figure Intensity in the diffraction pattern of a single slit as a function of sin θ. Note that the central maximum is not only much higher than the maxima to each side, but it is also twice as wide (2λ/D wide) as any of the others (only λ/D wide each).

13. 35-1 Diffraction by a Single Slit or Disk
Example 35-1: Single-slit diffraction maximum.
Light of wavelength 750 nm passes through a slit 1.0 x 10-3 mm wide. How wide is the central maximum (a) in degrees, and (b) in centimeters, on a screen 20 cm away?
Solution: a. The first minimum occurs at sin θ = λ/D = 0.75, or θ = 49°. The full width is twice this, or 98°.
b. The width is 46 cm.

14. 35-1 Diffraction by a Single Slit or Disk
Conceptual Example 35-2: Diffraction spreads.
Light shines through a rectangular hole that is narrower in the vertical direction than the horizontal. (a) Would you expect the diffraction pattern to be more spread out in the vertical direction or in the horizontal direction? (b) Should a rectangular loudspeaker horn at a stadium be high and narrow, or wide and flat?
Solution: a. The pattern will be more spread out in the vertical direction, as the slit is narrower there.
b. A pattern that is wider than it is high is desired, so the speaker should be high and narrow.

15. Problem 8
8. (II) (a) For a given wavelength λ, what is the minimum slit width for which there will be no diffraction minima? (b) What is the minimum slit width so that no visible light exhibits a diffraction minimum?

16. Problem 11

17. 35-2 Intensity in Single-Slit Diffraction Pattern
Light passing through a single slit can be divided into a series of narrower strips; each contributes the same amplitude to the total intensity on the screen, but the phases differ due to the differing path lengths:
Figure Slit of width D divided into N strips of width Δy. .
Phase difference of light from adjacent strips

18. 35-2 Intensity in Single-Slit Diffraction Pattern
Finally, we have the phase difference and the intensity as a function of angle:
and .

19. 35-3 Diffraction in the Double-Slit Experiment
The double-slit experiment also exhibits diffraction effects, as the slits have a finite width. This means the amplitude at an angle θ will be modified by the same factor as in the single-slit experiment:
The intensity is, as usual, proportional to the square of the field.

20. 35-3 Diffraction in the Double-Slit Experiment
The diffraction factor (depends on β) appears as an “envelope” modifying the more rapidly varying interference factor (depends on δ).
Figure (a) Diffraction factor, (b) interference factor, and (c) the resultant intensity plotted as a function of θ for d = 6D = 60λ.

21. 35-3 Diffraction in the Double-Slit Experiment
The diffraction factor (depends on β) appears as an “envelope” modifying the more rapidly varying interference factor (depends on δ).
Figure (a) Diffraction factor, (b) interference factor, and (c) the resultant intensity plotted as a function of θ for d = 6D = 60λ.

22. 35-3 Diffraction in the Double-Slit Experiment
Example 35-4: Diffraction plus interference.
Show why the central diffraction peak shown, plotted for the case where d = 6D = 6λ, contains 11 interference fringes.
Solution: The first minimum in the diffraction pattern occurs where sin θ = λ/D. Since d = 6D, d sin θ = 6λ. This means that the m=6 interference fringes will not appear; the central diffraction peak contains the fringes for m = 0 (one) and m = 1 through 5 (two each for a total of 10), making 11 fringes.

23. 35-4 Limits of Resolution; Circular Apertures
Resolution is the distance at which a lens can barely distinguish two separate objects.
Resolution is limited by aberrations and by diffraction. Aberrations can be minimized, but diffraction is unavoidable; it is due to the size of the lens compared to the wavelength of the light.

24. 35-4 Limits of Resolution; Circular Apertures
For a lens or any circular aperture of diameter D, the central maximum has an angular width:
When a length is free of aberrations
Figure Intensity of light across the diffraction pattern of a circular hole.

25. 35-4 Limits of Resolution; Circular Apertures
The Rayleigh criterion states that two images are just resolvable when the center of one peak is over the first minimum in the diffraction pattern of the other.
Figure The Rayleigh criterion. Two images are just resolvable when the center of the diffraction peak of one is directly over the first minimum in the diffraction pattern of the other. The two point objects O and O’ subtend an angle θ at the lens; only one ray (it passes through the center of the lens) is drawn for each object, to indicate the center of the diffraction pattern of its image.
D is the diameter of the lens or the mirror