1. Difference of Optical Path Length Interference Two waves One wave Many waves Diffraction

2. Intensity in single-slit diffraction Maximum at center Minimum Maximum

3. Central maximum  Diameter d light  Width of the bright fringe at the center:

4. Two objects are just resolved when the maximum of one is at the minimum of the other. not resolved resolvedjust resolved

5. Double-slit Interference and Diffraction Combined Each slit Combined

6. R Double-slit Interference and Diffraction Combined

7. The width of the central fringe is dependent on λ/a. missing fringes if m=d/a the mth maximum the first minimum

8. A slit of width a = 0.5 mm is illuminated by a monochromic light. Behind the slit there placed a lens (f = 100 cm), and the first maximum fringe is observed at a distance of 1.5 mm from the central bright fringe on the focal plane (screen). Find the wavelength of the light, and the width of the central bright fringe. Example

9. A typical astronomical telescope has a lens with a dimension of 10 m. Find the minimum resolving angle. Considering we use this telescope to observe the moon surface. What is the separation of two objects can just be resolved? (take wavelength of 540 nm) Example

10. Two slits with separation of d = 5.5 mm is illuminated by a monochromic light. There are 21 fringes in the central maximum. What is the width of each slit? The missing fringe is 11 th. a=d/11=0.5mm

11. Chapter 43 Gratings and Spectra

12. 1-slit 2-slit Multiple slits? Gratings Diffraction Diffraction Interference N is increasing, the maximum become sharper

13. The location of principal maxima is determined by: Principal maxima d is slip separation, also is called grating constant (grating spacing). N is increasing, the  is decreasing, the maxima become sharper

14. The location of principal maxima is determined by: Principal maxima d is slip separation, also is called grating constant (grating spacing). N is increasing, the δθ is decreasing, the maxima become sharper

15. Secondary maxima There are some minima between two principal maxima. Superposition of some phases Secondary maxima

16. 3 slips There are 2 minima between two principal maxima. 4 slips Δφ=0, 2π, …. Δφ=π/2 Δφ=3π/2 Δφ=πΔφ=π Δφ=0, 2  Δφ=4π/3 1 23 Δφ=2π/3 1 23 There are 3 minima between two principal maxima. If the grating has N slips, there are N-1 minima between two principle maxima. If the grating has N slips, there are N-2 secondary maxima between two principle maxima. N is increasing, the δθ is decreasing, the maxima become sharper N=5

17. Homework: P977-978 18, 24, 27,29 P979 7

18. Multiple slits Gratings N is increasing, the maximum become sharper two slits

19. 3 slips There are 2 minima between two principal maxima. 4 slips There are 3 minima between two principal maxima. If the grating has N slips, there are N-1 minima between two principle maxima. If the grating has N slips, there are N-2 secondary maxima between two principle maxima. N is increasing, the maxima become sharper

20. Intensity distribution ΔLΔL N is increasing, the maxima become sharper

21. ΔLΔL m th fringe is missing m 1 th fringe is missing

22. ΔLΔL If the grating has N slips, there are N-1 minima between two principle maximum, and there are N-2 secondary maximum Width of the maximum

23. Diffraction gratings Transmission grating Reflection grating Kinds of gratings

24. Example d  Wave length  When    wave  is m th max What is  for (m+1) th max

25. Example: A grating (N=5000) is illuminated by two monochromic lights with wave lengths of 600 and 400 nm respectively. The mth principal maximum of the former light is meet the (m+1)th principal maximum of the later at 3 cm from the central fringe on the screen. The focus length of the lens is 50 cm. Find the grating constant d, and the typical width of the principle fringes.

26.  Diffraction Grating

27.  Spectra of light

28. Dispersion and resolving power Dispersion To display the light with different wavelength The ability of a grating is determined by two intrinsic properties of the grating: Resolving power (1)  (2)the width of the line central 1 st (m=1)2 nd (m=2) 3 rd (m=3)

29. A grating N=9600, with width W=3.00cm, line spectrum of mercury vapor ( =546 nm). Example What is dispersion D  in 3 nd order (m=3) What is resolving power R  in 5 th order (m=5)

30. Dispersion and resolving power Example Dispersion

31. A grating with width W=4.15 cm. To resolve  =415.496 nm and  =415.487 nm in the second order, what is N? At what angle is 2 nd maximum? Whis the width of 2 nd maximum? Example

32. Gratings Spectrum Structure of crystal

33. Constructive interference: in NaCl d 1 st maximum will be at 9 ° Crystal solid =0.17nm X-ray Diffraction As a gratings crystal 2 nd maximum will be at 20 ° 3 rd maximum will be at 31 ° 4 th maximum 5 th maximum =500nm impossible

34. Bragg’s law Wavelength : Laue experiment Laue spots Determination of crystalline structures

35. A X-ray with λ=0.10 nm to 0.14nm, A crystal with a 0 =0.263nm, Example If a diffraction beam is to exist, when  =45 0 a0a0  What is m,  Laue experiment

36. A X-ray with λ=0.110 nm, A crystal with a 0 =0.563nm, Example If a diffraction beam is to exist, 2  =? a0a0

37. a0a0 d A X-ray with λ=0.110 nm, A crystal with a 0 =0.563nm, Example If a diffraction beam is to exist, 2  =? 22

38. X-ray diffraction X-ray sample   Debye method 22 I 2d·cos  =m

39. A X-ray with λ=0.110 nm, the incidence angle is 60 0, assuming the scattering is from the dashed planes, find the unit cell size a 0 Example a0a0 d   

40. Exercises P996-997 7, 11, 17, 19, 27, 33