CHAPTER 26 INTERFERENCE AND DIFFRACTION

Transcription

1 CHAPTER 26 INTERFERENCE AND DIFFRACTION INTERFERENCE CONSTRUCTIVE DESTRUCTIVE YOUNG S EXPERIMENT THIN FILMS NEWTON S RINGS DIFFRACTION SINGLE SLIT MULTIPLE SLITS RESOLVING POWER 1

2 IN PHASE OUT OF PHASE 2

3 OR THE BOOKS DISCRIPTION CONSTRUCTIVE INTERFERENCE 3

4 DESTRUCTIVE INTERFERENCE 4

5 When two waves arrive at a point where path differences from the sources differ by one wavelength, two wavelengths, three wavelengths, etc. there will be constructive interference. Or mathematically: = h =0,±1,±2,±3,. 5

6 When two waves arrive at a point where path differences from the sources differ by h, h, h,. (Or half integer number of wavelengths) there will be destructive interference. Or mathematically: = + Where h =0,±1,±2,±3,. 6

7 Many ways to produce two rays of coherent light. 7

8 Consider path differences h = 8

9 Constructive Interference = h =0,±1,±2,±3,. Destructive Interference = + h =0,±1,±2,±3,. 9

10 We can obtain the distance a bright line is below (or above) the center of the screen. Use 10

11 If tan Therefore = tan And for the bright line tan For small angles tan sin Therefore sin 11

12 Use this with = Or = To get h =0,±1,±2,±3, This is the equation that Young used to measure the wave length of light. 12

13 As was said many ways of producing two rays of coherent light. We have been discussing 13

14 Can also get two coherent rays from thin films. 14

15 15

16 We will discuss the interference between rays coming from the two surfaces on opposite sides of a space filled with liquid (thin film) or air. First: we need to know that when light reflects from a surface sometimes it undergoes a phase shift of For example if a ray is in air =1.00 and reflects at the surface with water =1.33 back into the air it does not undergo a phase shift. But if a ray is in glass =1.33 and reflects at the surface with air =1.00 back into the glass it undergoes a phase shift. 16

17 Rule If ray goes from low n material and reflects at surface of high n material no phase shift. If ray goes from high n material and reflects at surface with low n material 180 degree phase shift. > No phase shift. But > (half wavelength) phase shift 17

18 Now consider we have a film as shown Assume perpendicular. Film thickness where shown. 18

19 No phase shift at either surface or phase shift at both surfaces Constructive interference if 2 = =0,1,2,3, But if one surface has phase shift and other does not Destructive interference. Or No phase shift at either surface or phase shift at both surfaces Destructive interference if 2 = + =0,1,2,3, 19

20 Example 26.3 Glass Air Find separation of interference fringes. One reflection (top) no phase shift other (bottom) phase shift. 20

21 Equation for destructive interference 2 = =0,1,2,3, Where glass touches destructive interference. From figure So = h = h Therefore 2 h = = = 2h 2h 21

22 = =

23 EXAMPLES OF INTERFERENCE PHENOMONO Newton s Rings 23

24 24

25 Nonreflective Coatings 25

26 DIFFRACTION Light goes in straight line. Therefore would expect sharp shadow of object on a screen. But 26

27 Consider light going through single slit. This shows a ray from the top of the slit and one from half way down the slit (middle). Or looking only at the slit 27

28 The difference in path distance to the screen is h.= 2 sin If the path difference is one half wave length destructive interference. Therefore the requirement for a dark fringe is 2 sin = ± 2 sin = ± 28

29 If you divide the slit into 4 regions Becomes a/4 Becomes / sinθ If the path difference is one half wave length destructive interference. Therefore the requirement for a dark fringe is 4 sin = ± 2 sin = ±2 29

30 Divide the slit into more sections and can show The requirements for dark fringes are sin = = ±1,±2,±3, 30

31 If the wavelength of light is much smaller than the slit width (and normally it is) then Therefore And 1 1 sin Therefore for dark fringes = ±1,±2,±3, 31

32 Also for Use for distance to screen = Then tan = And for the h dark band tan = 32

33 If the screen is a distance from the slit such that And it normally is Then 1 And tan Therefore 33

34 Then using = ±1,±2,±3, = ±1,±2,±3, Solve for = ±1, ±3, ±5, This is equation in the book. 34

35 Multiple Slits Path difference will be sin 35

36 Constructive interference occurs for sin = = 0,±1,±2,±3,. More complicated than two slit diffraction 36

37 Eight Slit 37

38 More 16 slits 38

39 DIFFRACTION GRATINGS Prisms refract light different amounts depending on the wavelength of the light. Note short wavelengths (blue) are refracted more than he long wavelengths (red). 39

40 Diffraction Gratings will do much the same. 40

41 The equation for the maxima is the same as for multiple slits. sin = = 0,±1,±2,±3,. Note sin = Thus long wavelengths diffracted more than short wavelengths. (Opposite to prisim.) 41

42 X-RAY DIFFRACTION First experiments in

43 Crystal is uniform arrangement of atoms forming planes for diffraction. 43

44 Gives pattern 44

45 Consider the atoms as all lined up on the sites and represented by the dots shown. 45

46 Look at top row for example Path lengths are equal when = 46

47 Therefore when the two angles are equal get constructive interference from diffraction from rows. Then considering two rows one beneath the other. 47

48 Additional constructive interference when path lengths are equal to integer times wavelength of x-rays. 2 = =1,2,3,.. When both conditions are satisfied 1. = And 2. 2 = When these conditions are not satisfied radiation interferes destructively. Therefore obtain diffraction pattern characteristic of crystal. 48

49 Scattering and thus diffraction can occur from various planes within the crystal. 49

50 CIRCULAR APERTURES Light travelling through a small circular aperture is diffracted also. 50

51 The angular locations of the dark rings are given by: sin =1.22 sin =2.23 sin =3.24 OBSERVING OBJECTS THROUGH VARIOUS SIZE APERTURES. 51

52 52

53 If the aperture in (a) were smaller 3 and 4 would appear as one object. The larger the aperture the better the resolving power. Lord Rayleigh proposed a criterion for telescopes (and other optical instruments) to give a measure for their capability. Two objects can only be resolved if the center of the second just falls at a distance of the center of the first dark ring of the first. The limit of resolution, for a telescope (or other optical instrument): =