1. Chapter 37 - Interference and Diffraction
A PowerPoint Presentation by
Paul E. Tippens, Professor of Physics
Southern Polytechnic State University
© 2007

2. Objectives: After completing this module, you should be able to:
Define and apply concepts of constructive interference, destructive interference, diffraction, and resolving power.
Describe Young’s experiment and be able to predict the location of dark and bright fringes formed from the interference of light waves.
Discuss the use of a diffraction grating, derive the grating equation, and apply it to the solution of optical problems.

3. Diffraction of Light
Diffraction is the ability of light waves to bend around obstacles placed in their path.
Ocean
Beach
Fuzzy Shadow
Light rays
Water waves easily bend around obstacles, but light waves also bend, as evidenced by the lack of a sharp shadow on the wall.

4. A new set of waves is observed emerging from the gap to the wall.
Water Waves
A wave generator sends periodic water waves into a barrier with a small gap, as shown below.
A new set of waves is observed emerging from the gap to the wall.

5. Interference of Water Waves
An interference pattern is set up by water waves leaving two slits at the same instant.

6. Young’s Experiment
In Young’s experiment, light from a monochromatic source falls on two slits, setting up an interference pattern analogous to that with water waves.
Light source S1 S2

7. The Superposition Principle
The resultant displacement of two simul-taneous waves (blue and green) is the algebraic sum of the two displacements.
The composite wave is shown in yellow.
Constructive Interference
Destructive Interference
The superposition of two coherent light waves results in light and dark fringes on a screen.

8. Young’s Interference Pattern
Constructive
Bright fringe
Destructive
Dark fringe
Constructive
Bright fringe

9. Conditions for Bright Fringes
Bright fringes occur when the difference in path Dp is an integral multiple of one wave length l. l l l p1 p2 p3 p4
Path difference
Dp = 0, l , 2l, 3l, …
Bright fringes: Dp = nl, n = 0, 1, 2, . . .

10. Conditions for Dark Fringes
Dark fringes occur when the difference in path Dp is an odd multiple of one-half of a wave length l/2. l l p1
n = odd
n = 1,3,5 … p2 p3 p3
Dark fringes:

11. Analytical Methods for Fringesx y
d sin q s1 s2 d q p1 p2
Dp = p1 – p2
Dp = d sin q
Path difference determines light and dark pattern.
Bright fringes: d sin q = nl, n = 0, 1, 2, 3, . . .
Dark fringes: d sin q = nl/2 , n = 1, 3, 5, . . .

12. Analytical Methods (Cont.)x y
d sin q s1 s2 d q p1 p2
From geometry, we recall that:
So that . . .
Bright fringes:
Dark fringes:

13. The third dark fringe occurs when n = 5
Example 1: Two slits are 0.08 mm apart, and the screen is 2 m away. How far is the third dark fringe located from the central maximum if light of wavelength 600 nm is used? x y
d sin q s1 s2 q
n = 1, 3, 5
x = 2 m; d = 0.08 mm
l = 600 nm; y = ?
d sin q = 5(l/2)
The third dark fringe occurs when n = 5
Dark fringes:

14. Example 1 (Cont. ): Two slits are 0
Example 1 (Cont.): Two slits are 0.08 mm apart, and the screen is 2 m away. How far is the third dark fringe located from the central maximum if l = 600 nm?
x = 2 m; d = 0.08 mm
l = 600 nm; y = ? x y
d sin q s1 s2 q
n = 1, 3, 5
y = 3.75 cm

15. The Diffraction Grating
A diffraction grating consists of thousands of parallel slits etched on glass so that brighter and sharper patterns can be observed than with Young’s experiment. Equation is similar.
d sin q q d
d sin q = nl
n = 1, 2, 3, …

16. The Grating Equation 3l The grating equation: 2l l 1st order
d = slit width (spacing)
l = wavelength of light
q = angular deviation
n = order of fringe
2nd order 2l 4l 6l

17. To find slit separation, we take reciprocal of 300 lines/mm:
Example 2: Light (600 nm) strikes a grating ruled with 300 lines/mm. What is the angular deviation of the 2nd order bright fringe?
To find slit separation, we take reciprocal of 300 lines/mm:
300 lines/mm
n = 2
Lines/mm  mm/line

18. Angular deviation of second order fringe is:
Example (Cont.) 2: A grating is ruled with 300 lines/mm. What is the angular deviation of the 2nd order bright fringe?
300 lines/mm
n = 2
l = 600 nm
Angular deviation of second order fringe is:
q2 = 21.10

19. A compact disk acts as a diffraction grating
A compact disk acts as a diffraction grating. The colors and intensity of the reflected light depend on the orientation of the disc relative to the eye.

20. Interference From Single Slit
When monochromatic light strikes a single slit, diffraction from the edges produces an interference pattern as illustrated.
Pattern Exaggerated
Relative intensity
The interference results from the fact that not all paths of light travel the same distance some arrive out of phase.

21. Single Slit Interference Pattern
Each point inside slit acts as a source.
a/2 a 1 2 4 3 5
For rays 1 and 3 and for 2 and 4:
First dark fringe:
For every ray there is another ray that differs by this path and therefore interferes destructively.

22. Single Slit Interference Pattern1 2 4 3 5
First dark fringe:
Other dark fringes occur for integral multiples of this fraction l/a.

23. Example 3: Monochromatic light shines on a single slit of width 0
Example 3: Monochromatic light shines on a single slit of width 0.45 mm. On a screen 1.5 m away, the first dark fringe is displaced 2 mm from the central maximum. What is the wavelength of the light? q
x = 1.5 m y
a = 0.35 mm
l = ?
l = 600 nm

24. Diffraction for a Circular Opening
Circular diffraction D
The diffraction of light passing through a circular opening produces circular interference fringes that often blur images. For optical instruments, the problem increases with larger diameters D.

25. Resolution of Images
Consider light through a pinhole. As two objects get closer the interference fringes overlap, making it difficult to distinguish separate images. d1
Clear image of each object d2
Separate images barely seen

26. Resolution Limit
Images are just resolved when central maximum of one pattern coincides with first dark fringe of the other pattern. d2
Resolution limit
Separate images
Resolution Limit

27. Resolving Power of Instruments
The resolving power of an instrument is a measure of its ability to produce well-defined separate images.
Limiting angle D q
For small angles, sin q  q, and the limiting angle of resolution for a circular opening is:
Limiting angle of resolution:

28. Resolution and Distanceq so p D
Limiting angle qo
Limiting Angle of Resolution:

29. Example 4: The tail lights (l = 632 nm) of an auto are 1
Example 4: The tail lights (l = 632 nm) of an auto are 1.2 m apart and the pupil of the eye is around 2 mm in diameter. How far away can the tail lights be resolved as separate images? q so p
Eye D
Tail lights
p = km

30. Summary Young’s Experiment:
Monochromatic light falls on two slits, producing interference fringes on a screen. x y
d sin q s1 s2 d q p1 p2
Bright fringes:
Dark fringes:

31. Summary (Cont.) The grating equation: d = slit width (spacing)
l = wavelength of light
q = angular deviation
n = order of fringe

32. Interference from a single slit of width a:
Summary (Cont.)
Interference from a single slit of width a:
Pattern Exaggerated
Relative Intensity

33. Summary (cont.) The resolving power of instruments. p D soq so p D
Limiting angle qo
Limiting Angle of Resolution:

34. CONCLUSION: Chapter 37 Interference and Diffraction