2. Chapter 8 Wave Optics (2) (May 13, 2005)

3. A brief summary to the last lecture 1. Young’s double-slit experiments 20~100cm1~5m What are the coherent conditions of lights in Young’s double slit experiment? link1link1

4. (1) Monochromatic ( 单色的 ) light; (2) lights from the two small openings are in phase and from the same source. (3) The openings and the distance between the two openings are not too large (<1mm) compared with the wavelength of the incident light.

5. Constructive interference d P O S0S0  L  S1S1 S2S2 B x

6. Destructive interference The spacing of two bright or dark fringes:

7. 2. Lloyd’s mirror L What does Lloyd’s mirror interference explain?

8. 3. light path length (also called Optical length, or optical path) optical length = refractive index times the geometrical distance (nL)  is the optical length difference.

9. 4. light interference for thin film e i i2i2 O A B C Light path –length difference is

10. For normal incidence, the incident angle is zero, so we have For the destructive interference, we have m = 1, 2, 3,…

11. 8.1.5 Equal thickness interference, Generally speaking, the abrupt phase change occurs at one of the surface of the wedge. So it is easy to get the difference of light path length. m = 0, 1, 2, … Glass plate e Zero-order dark fringe Incident ray Interfering rays The condition for destructive interference is simpler Air wedge

12. Distance between two dark fringes:

13. Example 2: two microscope slides each 7.5cm long are in contact along one pair of edges while the other edges are held apart by a piece of paper 0.012mm thick. Calculate the spacing of interference fringes under illumination by light of 632nm wavelength at near normal incidence. Solution: let the air thickness e corresponding the m th -order dark fringe and e 1 to the (m+1) th - order dark fringe. As the refractive index of air is 1, we can write out: 2e = m,2e 1 = (m+1)

14. e B It is easy to find the spacing of two neighbor fringes by deleting m from above equations. We have From similar triangles, we know 0.012mm Zero-order dark fringe e1e1 x 7.5cm A C

15. Newton’s rings If the convex surface of a lens with large radius is placed in contact with a plane glass plate, a thin film of air is formed between the two surfaces. The thickness of this film is very small at the point of contact, gradually increasing as one proceeds outward. The loci (locus, 轨迹 ) of points of equal thickness are circles concentric with point of contact. e R r R-e O C

16. At such a case, the difference of optical-length is Where e is the thickness of air film, /2 is from the half-wavelength lost for the two rays considered. e R r R-e O C

17. The condition for bright fringes is The condition for dark fringes is

18. On the other hand, we could also calculate the radii of bright and dark rings. e R r R-e O C As R>>e, e 2 can be dropped. So we have:

19. The radius for kth bright ring is The radius for kth dark ring is

20. 8.2 Diffraction ( 衍射 ) of light There are two kinds of diffractions; (1) Light emitted from a small source is diffracted by a barrier and the diffraction pattern shows on a screen which is not far from the barrier. The diffraction is called Fresnel diffraction. (2) The parallel rays pass through a barrier and the diffraction pattern shows on a screen which is at a great distance. The diffraction is called Fraunhofer diffraction. We discuss the second one only.

21. 8.2.1 Fraunhofer Single slit diffraction( 单缝衍射 ) θ a/ 2 a/2 sin θ P f2f2 Destructive interference: Back to slice 22

22. Consider two narrow strips ( 带 ), one just below the top edge of the slip and one just below its center. The difference in length to point P is (a/2) sinθ, where a is the slit width. Suppose this path difference happens to be equal to λ/2:

23. Then lights from these two strips arrive at point P with half-cycle (π) phase difference, and cancellation occurs. Similarly, light from two strips just below these two will also arrive a half-cycle out of phase; and in fact, lights from every strip in the top half cancels out those from the corresponding strip in the bottom half, resulting in complete cancellation. or

24. It is easy to find that when the optical length difference (a/2)sinθ is equal to the integer multiple of λ/2, the dark fringe will occur. Therefore, the dark fringe positions are determined by the equation. m = 1, 2, 3, … For example, if the slit width is equal to ten wavelengths, dark fringes occur at

25. Midway between the dark fringes are bright fringes. We also note that sinθ = 0 is a bright band, since then light from the entire slit arrives at P in phase. Thus the central bright fringe is twice as wide as the others, (see figure). Finally we have:

26. m = 1, 2, 3, …, dark fringes m = 1, 2, 3, …, bright fringes y m is the distance from the m th -dark fringe to the center bright fringe.

27. 8.2.2 Diffraction by a circular hole 11 D The condition for the first-order dark fringe is obtained as:

28. According to Huygen’s principle we must add up the radiation or wavelets originating from all points inside the open circle in order to get the resultant amplitude. The diffraction pattern consists of a number of concentric fringes with a maximum intensity in the center. This phenomenon is quite important. The resolution ( 分辨率 ) of many instrument like telescope and microscope depends on this.

29. 8.2.3 The diffraction grating ( 衍射光栅 ) Suppose that, instead of a single slit, or two slits side by side in Yong’s experiments, we have a very large number of parallel slits, with the same width and spaced at regular intervals. Such an arrangement, known as diffraction grating, was first constructed by Fraunhofer. d = a + b d is called grating spacing ( 光 栅间距 ) or grating constant. ymym d d d a b L 

30. Let’s assume that the slits are so narrow that the diffracted beam from each spreads out over a sufficiently wide angle for it to interfere with all other diffracted beams. Consider first the light proceeding from elements of infinitesimal width at the upper edge of each opening and traveling in a direction making an angle  with that of the incident beam. A lens at the right of the grating forms in its focal plane a diffraction pattern similar to that which would appear on a screen. First-order maximum when AB=λ, second order maximum when AB=2λ. a b d θ1θ1 λ A B θ2θ2 2λ A B

31. It is easy to find that the diffraction grating equation (bright fringes) is d sinθ = ±mλ (m = 0, 1, 2, 3, …) The results are involved in interferences and diffractions. It is known that for the interference of the double-slit, the fringes are equally bright. But when we consider the diffraction by a slit, the final pattern actually observed is a combination of both effects.

32. The interference pattern locates the position of each bright fringe, and the diffraction pattern from one slit modifies the intensity of each bright fringe. Diffraction modifies the interference pattern of a grating in just the same way to the double-slit. The interference pattern determines the position of each bright fringe, and the diffraction pattern determines its intensity. On the other hand, a grating produces much sharper bright fringes than a double slit.

33. If a sharp bright fringe of interference occurs at the position where the first dark fringe of diffraction happens to be, the sharp bright fringe will be missing. This phenomenon is called the order-missing phenomenon of a grating.

34. 8.2.4 the resolving power of optical instruments Consider a situation in which two distant point sources of light, such as two stars, illuminate a small hole. Each beam of light produces its own diffraction pattern on the screen.. When the two stars have a relatively large angular separation and the hole is reasonably wide, the two diffraction patterns are will separated. We can easily identify two patterns as being distinct.

35. However, if the angular separation of the two stars is sufficiently small, their diffraction pattern overlap, and identifying the two patterns as being distinct becomes a matter of judgment. Resolving power is the angular separation of the two objects. A condition for identifying diffraction patterns as being distinct was given by the Rayleigh criterion ( 准则 ). Rayleigh criterion is that the center of one diffraction pattern must come no closer than the first dark fringe of the other pattern

36. Therefore ， Rayleigh’s formula for the resolving power of a circular `aperture of diameter D is ： We know that the aperture often is an aperture of a lens in the optical instrument. For example, in a microscope the aperture is the objective ( 物镜 ). So D is the diameter of the objective. When it is bigger, the theta will be small and we have higher resolving power! It is the same reason for having a big D of a telescope.

37. Example 1: A coherent beam of light from a hydrogen discharge tube falls normally on a diffraction grating of 8000 lines per centimeter. Calculate the angular deviation of each line in the first-order spectrum. Do any lines of the second- order spectrum overlap the first-order spectrum? ( For the hydrogen discharge tube, λ red = 656.3 nm; λ blue = 486.1 nm, λ violet 1 = 434.0 nm, λ violet 2 = 410.1 nm.) SOLUTION: Since the grating has 8000 lines per centimeter, the grating spacing is given by:

38. We can calculate the deviation of each component in turns by using the diffraction grating equation. According to the grating equation for bright fringes which is

39. It is necessary to calculate the angles of the first order fringes and the second order fringes for the given lights and to compare them to determine whether they have overlap. The general idea is to compare the biggest angle θ 1max of the first-order fringes and the smallest angle θ 2min of second-order fringes. If θ 1max > θ 2min,the overlap happens. Otherwise, no overlap occurs. grating screen Zeroth-order Violet 2 Violet 1 Blue Red 1 th order Violet 2 Violet 1 31.7° 40.1° (1). First-order Violet 2

40. Violet 1 Blue Red Second-order for violet 2 which should have the smallest diffracting angle in the second-order diffractions. So Violet 2

41. It is obvious to see that the biggest angle for the first-order is less than the smallest angle in the second-order. Therefore there is no overlap happening in such a case. From the example we know that diffraction gratings can be used to distinguish very approximate wavelengths, such as violet 1 and violet 2. The two wavelengths appear as the same color by human eye, but they can be distinguished after single-slit diffraction.

42. 8.3 light polarization ( 偏振 ) Interference and diffraction happens to all kinds of waves, longitudinal and transverse waves. However, a phenomenon, called polarization, occurs on with transverse waves.

43. Idea of polarization: recall mechanical waves on a string. Displacements of points on the string are perpendicular to the length of the string, and to the direction of propagation of the wave, so the wave is transverse. In the three dimensional space, if the string is along x- axis, the string can be moved along y or z direction, so they are in xy or xz plane. It is also possible that the displacements are the superposition of the above two vibrations.

44. The wave having only y-direction displacements is said to be linearly polarized in the y-direction and same for z-direction. If the string moves like a rotating helix, the wave can be said to be circularly or elliptically polarized based on whether they have the same amplitudes in different directions. Filters: filter is an object which can permit only waves with a certain polarization direction to pass. Mechanical filters: vertical or horizontal slits

45. Light is actually electromagnetic wave which is transverse wave. It has the vector of electric field strength E and vector of magnetic induction B. (E can cause exposing and physiological functions, so we consider E only). Natural lights are non polarized and their light vectors point to all directions, have the same magnitudes and are in the plane perpendicular to the direction of light propagation.

47. linearly polarized light (also called polarized light ) plane of vibration for polarized light plane of polarization (contain the line of propagation and perpendicular to the plane of vibration. partial polarized light circularly polarized lights elliptically polarized light

49. Maluss’ Law

50. Brewster’s Law

52. And the optical activity of matter explained on page 250-251, experiments will use such a theory.