NURJULIANA JUHARI School of Microelectronic Engineering

1. CHAPTER 4 WAVE OPTICS Prepared by • NURJULIANA JUHARI • School of Microelectronic Engineering

2. OBJECTIVES • Scope of studies includes knowledge, application and comprehension about: • Transverse wave, sine wave and phase angles • Superposition of waves • Interference • Electromagnetic wave • Polarization

3. Transverse Wave • All light are classified as transverse waves. • Transverse wave are those in which each small part of the wave vibrate along a line perpendicular to the direction of propagation and all parts are vibrating in the same plane. • General appearance wave in SHM as shown in Fig 11H.

5. The distance between two similar points of any two consecutive wave forms is called wavelength, λ. • One wavelength is equal to the distance between two wave crests and teo wave troughs. • Amplitude is letter a. • The frequency is the number of waves passing by or arriving at any given point of second.

6. The speed of waves, ν is given by the wave equation • Where f is frequency and λ is wavelength • The length of wave times the number of waves per second equals the distance the travel waves will travel in 1s.

7. SINE WAVES • Motion of along the wave have displacement, y given by the sine or cosine of some uniformity function. • The displacement y of any point on the wave is then given by

8. To make the wave move to the right with a velocity, v then introduce time t as follows; • Any particle of wave such as P in the Fig 11I will carry out SHM and will occupy successive positions P,P’,P”,P”’,etc, as the wave moves. • Time for complete vibration of any point is same as another.

9. Period T and its reciprocal the frequency v are given by the wave equation (11q) • If we put several variables in Eq (11s), we obtain useful equation for wave motion in general:

11. PHASE ANGLES • In wave motion instantaneous displacement and direction of propagation are described by specifying the position of the graph point on the circle of refernce (Fig 11J) • Angle θ measured counterclockwise from the +x axis, called phase angle. • The position of the mass point P is given by the projection of the graph point p1 on the y axis.

12. From the right triangle PpC on the diagram • Graph point moving constant speed, v the angular speed  ω is constant and can write for any angle θ, • Substitution in Eq (11v) gives

13. At time t=0 the graph is at +P0 and the mass point is at P0 • Later time t when the mass point is at P, the graph point is at p and modify Eq(11w) by adding angle α as follows: • The angle α is constant and is called initial phase angle. • As the point p is moves, the angle wt increases at a uniform rate and always measured from the starting angle α. Angle in radian.

15. Example • A given point is vibrating with SHM with a period of 5s and an amplitude of 3cm. If the initial phase angle is π/3 rad (600). Find • The initial displacement • The displacement after 12s • Make a graph.

16. ANS a) b) After 12 s, sub in Eq 11x

17. c) The total phase angle of 4π + π/3 is 924, and measured from +x axis places the graph point 240 below the –x axis on the circle of reference. The angle gives

19. Another expression the equation for simple harmonic waves is in terms of angular frequency ω=2πv and the propagation number k=2π/λ. Eq 11u becomes

20. The addition of constant quantity is parentheses is of little physical significance, since such a constant can be eliminated by suitable adjusting the zero of the time scale. • The equation when written • Will describe the wave Fig 11I • If t=T/4 and T/2, respectively, instead of t=0

21. THE SUPERPOSITION OF WAVES Addition of SHM Along The Same Line • Superposition- the resultant displacement of any point merely the sum of the displacement due to each wave separately or resultant displacement y is merely the sum of y1 and y2. • Considering first effect of superimposing two sine waves of same freq. • The problem finding resultant motion when particles execute two SHM in same time.

22. Displacement due to the two waves are taken along the same line, y direction. • According to Eq 11x, separate displacemet is write as follows: • ω is the same for both wave, assume them have same frequency. Hence

23. Expression for the sine of difference two angles is used, this can be written • Since a’s and α’s are constant, we justified in setting

24. Provided that constant values of A and θ can be found which satisfy these equation. Squaring and adding Eq 12c, we have • Dividing the lower equation (12c) by upper one we obtain

25. Eq 12d and 12e show that value of A and θ exist which satisfy Eq (12c), rewrite Eq 12b,substituting the right hand members of Eq 12c. This gives • Which has the form of the sine of difference of two angles and can be expressed as

26. This equation is same as either of original equation for separate SHM but contains new amplitude A and new phase constant θ. • The amplitude and phase constant of resultant motion can calculate by Eqs 12d and 12e respectively. • The resultant amplitude A depends, according to Eq 12d, upon amplitudes a1 and a2 of the component motion and upon their difference phase δ=α1- α2,

27. When bring together beam of light, as done in Michelson interferometer, the intensity of light at any point will proportional to the square of the resultant amplitude. By Eq 12d, in case a1=a2 • Phase difference δ=0,2π, 4π,…, this gives 4a2, or 4 times intensity of either beam. • If δ= π,3π,5π,…the intensity is zero. • These modification of intensity is obtained by combining waves are referred to interference effect.

28. Exercise I • A wave is specified by y= 6 sin 2π(4t-5x+2/3). Calculate • Amplitude • Wavelength • Frequency • initial phase angle • Displacement at time, t=0 and x=0

29. Additional of Sample Harmonic At Right Angles • Consider the effect when two sine waves of the same frequency but having displacement in two perpendicular directions act simultaneously at a point. • Direction as y and z may express two component motion as

30. These are to be added according to the superposition principle, to find the path of resultant motion. • One does eliminating t from 2 equation • Multiplying Eq (12r) by sin α2 and Eq (12s) by sin α1 and subtracting the first equation from the second gives

31. Similarly, multiplying Eq (12r) by cos α2 and Eq (12s) by cos α1 and subtracting the second from the list, we obtain

32. Eliminate t from Eq (12t) and (12u) by squaring and adding these equation. Then gives as the equation for the resultant path

34. Fig 12K the heavy curves are graphs of this equation for various values of the phase difference δ=α1- α2. • For special cases where thy degenerate into straight lines, these curve are all ellipses. • The principle axes of ellipse are in general inclined to the y and z axes but coincide with them when δ= π/2,3π/2,5π/2,… as can readily be seen from Eq (12v). In this case

35. Equation of an ellipse with semiaxes a1, and a2 coinciding with the y and z axes, respectively. When δ= 0, 2π,4π,.. We have • Representing a straight line passing through the origin, with slope a1/a2. If δ= π,3π,5π • a straight line with the same slope, but opposite sign

36. Exercise 2 Two waves are travelling together along the same line are given by Y1=5 sin (ωt+π/2) and Y2=7 sin (ωt+π/3). Calculate: • Resultant amplitude • Initial phase angle of the resultant • resultant equation of the motion

37. INTERFERNCE OF TWO BEAMS OF LIGHT • Modification of intensity obtained by the superposition of two or more beams of light is called interference. • If the resultant intensity is zero or in general less than we expect from the separate intensities, we have destructive interference, while if it greater we have constructive interference.

39. Huygens Principle • When waves pass through aperture, they always spread to some extent into the region which is not directly exposed to the oncoming waves. This phenomena called diffraction. • Huygens proposed the rule that each point on a wave front may be regarded as a new source of waves.

40. Fig 13A, let a set of plane waves approach the barrier AB from the left, and let the barrier contain an opening S of width somewhat smaller than the wavelength. • At all points except S the waves will either reflected or absorbed, but S free to produce disturbance behind the screen. • From experimentally, wave spread out from S in form of semicircles.

41. If slit is made is very narrow, broadening of this patch is observed, its breadth increasing as the slit is narrowed further. • This evidence that light not always travel in straight line. • Screen CE replace with photographic plate, Fig 13B is obtained. • Light is most intense in the forward direction, but intensity is decreased slowly as angle increased

42. If the slit is small compared with the wavelength light, the intensity does not come to zero even when the angle observation becomes 90°. • This brief will be sufficient for understanding the interference phenomena

43. YOUNG EXPERIMENT • Fig 13C, sunlight allowed to pass through a pinhole S at considerable distance away, through S1 and S2 • Two sets of spherical waves emerging from holes, form symmetrical pattern of varying intensity on the screen AC

44. To get good result perhaps replace pinholes by narrow slit • Circular lines represents crests of waves with same phase or with phases differing by a multiple of 2π. • A close examination of the light on the screen will reveal evenly spaced light dark bands or fringes, see Fig 13D.

47. The straight vertical filament S acts as the source and the first slit. • Double slit for observer made from small photographic plates about 1 to 2 in.square. • The lamp is now viewed by the holding the double slit D close to the eye E and looking at the lamp filament. • If slits are close, widely fringes appear whereas slits farther apart , give narrow fringes.

48. A piece of red glass F placed adjacent to and above another of green glass in front of the lamp will show red waves produce wider fringes than green, due to greater wavelength. • Accurate result: (i) use monochomatic light (ii) use sodium arc (iii) use mercury arc plus a filter (iv) suitable filter consists combination of didymium glass, to absorb yellow lines and a light yellow glas to absorb the blue and violet lines.

49. Interference Fringes From A Double Source

50. Two waves arrive at P having traversed different distances S2P and S1P • Hence superimposed with a phase different given by • Assume that S1 and S2 in the same phase, because these slit were taken to be equidistant from the source slit S. • Amplitude is same S1 and S2 are of equal width