1. Diffraction

2. b S S’ A B According to geometrical optics region AB of Screen SS’ to be illuminated and remaining portion will be dark

4. If observation are made carefully and width of the slit is not very large with respect to wavelength then the light intensity in the region AB is not uniform and there is also some intensity inside the geometrical shadow. This spreading out of a wave when it passes through a narrow opening is known as diffraction pattern.

7. Bending of light

10. Diffraction is the apparent ``bending'' of light waves around obstacles in its path This bending is due to Huygen's principle, which states that all points along a wave front act as if they were point sources. Thus, when a wave comes against a barrier with a small opening, all but one of the effective point sources are blocked, and the light coming through the opening behaves as a single point source, so that the light emerges in all directions, instead of just passing straight through the slit.

12. Definition: The phenomenon of bending of light round the corner of an obstacle and their spreading into the region of geometrical shadow is called diffraction and the distribution of light intensity resulting in dark and bright fringes is called diffraction pattern.

13. The diffraction phenomenon usually devided into two categories: 1)Fresnel diffraction 2)Fraunhafer diffraction 1) The source of light and the screen are at finite distance from the diffracting aperture. S S’ Point source

14. 2) The source and the screen are at infinite distance from the aparture Point source ff f

15. FRAUNHOFER DIFFRACTION PATTERN AT ASINGLE SLIT Incident plane wave Long narrow slit Lens Screen Diffraction pattern f

16. A single slit placed between a distant light source and a screen produces a diffraction pattern –It will have a broad, intense central band –The central band will be flanked by a series of narrower, less intense secondary bands Called secondary maxima –The central band will also be flanked by a series of dark bands Called minima

18. Δ A A1A1 A2A2 A1’A1’ b B B1B1 B2B2  P f 

19. s L1 L2 A B A1 A1 ’ P C O b   B1

20. Slit consists of large number of equally spaced points sources and that each point on the alit is a source of Huygens Secondary wavelets which interfere with the other wavelets. Let the point sources be at A, A 1,A 2,A 3,A 4 …. Distance between two consecutive points be  Number of points be n b = (n-1)  Interference at P on the screen which is at focal plane of L 2. P receives the parallel rays making an angle  with normal to the slit If n goes to infinity then  goes to zero such that n  tends to b

21. So there is a path difference between the rays coming from A and A 1. But for an incident plane wave the points A,A1,A2,A3,A4…. are in phase So the additional path traversed by the disturbance emanating From point A and A 1 will be A 1 A’ 1 AA’ 1 is the foot of the perpendicular drawn on A 1 B 1. If diffracted rays makes an angle  with the normal to the slit Then, A 1 A’ 1 =  sin  The corresponding phase diference will be  = (2  / )  sin 

22. If the field at P due to the disturbance emanating from the Point A is acos(  t) then field due to the disturbance Emanatinfg from A 1 will be acos(  t-  ) Now the difference between the phases of the disturbance reaching from the points A 1 and A 2 will also be  So the resultant field at P will be

23. Resultant of n simple harmonic waves of equal amplitude And periods and phases increasing in arithmetic progressions φ 2 φ 3 φ (n-1) φ O A B C D E P  a R

24. Resolving the amplitude parallel and perpendicular to OA Multiply equa(1) with 2sinφ/2

27. Squaring and adding equation (3) and (4) Similarly

28. Dividing (4) by (3) we have If we have large number of vibration as

29. Then resultant vibration will be

30. s L1 L2 A B A1 A1 ’ P C O b   B1

31. If we simplify it (Ghatak,superposition of waves)

32. Where amplitude R of the resultant wave will be

33. According to the theory of composition of n simple harmonic Motion of equal amplitude and common phase difference between successive vibrations, the resltant amplitude at P is

35. For large value of n sin  /n =  /n Resultant intensity at P which is proportional to the square Of the resultant amplitude R is given by,

36. ………………….(1) Here I 0 represents intensity at  = 0. Equation (1) gives the variation of intensity with 

38. Position of maxima and minima PRINCIPLE MAXIMA The resultant amplitude is given by

39. R will be maximum if the negative terms vanish. This is possible when So maximum value of resultant amplitude is R and is A So I max = I 0 = A 2

40. MINIMA From equation (1) again, intensity is minimum when sin  =0 and   0 ………..(2)

41. So the condition for minima is Where n=1,2,3…. Gives the condition of 1 st, 2 nd, 3 rd, minima. Here n  0, because n = 0 for  = 0 corresponds principle maxima. so first minima will occur at 2 nd minima at

42. Secondary maxima Apply the method of finding maxima and minima

43. In order to determine the position of secondary maxima Differentiate the above equation with respect to 

44. So either or

45. Bur here sin  =0 or  = p  gives the position of primary minima Except when  = 0. Thus the position of secondary maxima is given by the equation  = tan  (secondary maxima) Here for  = 0 corresponds to the central maximum. The other roots can be found by determining the points of intersections of the curves y =  and y = tan  by Graphical method.

46.  y =  y = tan 

47. The first value of  =0 gives the principle maxima. The remaining values of  are near  = 3  /2, 5  /2, 7  /2,…… The more exact values are = 1.430 , 2.46 , 3.47 , 4.471  So the direction of secondary maxima are approx

49. INTENSITY CALCULATIONS for principle maxima(central maxima)  = 0 I max = I 0 = A 2

50. first secondary maxima (I 1 is 4.5% of I 0 )

51. 2 ND Secondary maxima ( I 2 is 1.61% of I 0 )

52. Central maxima Secondary maxima 0  22 33 -- -2  -3  3  /2

53.  22 -- -2  33 -3 