Interference, Diffraction & Polarization

1. Interference, Diffraction & Polarization

2. light as waves • so far, light has been treated as if it travels in straight lines • ray diagrams • refraction, reflection • To describe many optical phenomena, we have to treat light as waves. • Just like waves in water, or sound waves, light waves can interact and form interference patterns. Remember c = f  interference, diffraction & polarization

3. interference constructive interference destructive interference at any point in time one can construct the total amplitude by adding the individual components interference, diffraction & polarization

4.   +   = destructive interference waves ½ out of phase demo: interference Interference III + = constructive interference waves in phase interference, diffraction & polarization

5. r1=r2 r1 r2 positive constructive interference negative constructive interference destructive interference if r2-r1 = n then constructive interference occurs if r2-r1 = (n+½) then destructive interference occurs Interference in spherical waves maximum of wave minimum of wave interference, diffraction & polarization

6. interference, diffraction & polarization

7. light as waves it works the same for light waves, sound waves, and *small* water waves interference, diffraction & polarization

8. double slit experiment • the light from the two sources is incoherent (fixed phase with respect to each other • in this case, there is • no phase shift between • the two sources • the two sources of light must have identical wave lengths interference, diffraction & polarization

9. Young’s interference experiment there is a path difference: depending on its size the waves coming from S1 or S2 are in or out of phase interference, diffraction & polarization

10. Young’s interference experiment If the difference in distance between the screen and each of the two slits is such that the waves are in phase, constructive interference occurs: bright spot difference in distance must be a integer multiple of the wavelength: d sin = m, m=0,1,2,3… m = 0: zeroth order, m=1: first order, etc. if the difference in distance is off by half a wavelength (or one and a half etc.), destructive interference occurs (d sin = [m+1/2], m=0,1,2,3…) path difference demo interference, diffraction & polarization

11. distance between bright spots tan=y/L L if  is small, then sin     tan  so: d sin = m, m=0,1,2,3… converts to dy/L = m difference between maximum m and maximum m+1: ym+1-ym= (m+1)L/d-mL/d= L/d ym=mL/d demo interference, diffraction & polarization

12. distance is equal so 1/2 difference: destructive int. question • two light sources are put at a distance d from a screen. Each source produces light of the same wavelength, but the sources are out of phase by half a wavelength. On the screen exactly midway between the two sources … will occur • a) constructive interference • b) destructive interference +1/2 interference, diffraction & polarization

13. question • two narrow slits are illuminated by a laser with a wavelength of 600 nm. the distance between the two slits is 1 cm. a) At what angle from the beam axis does the 3rd order maximum occur? b) If a screen is put 5 meter away from the slits, what is the distance between the 0th order and 3rd order maximum? • use d sin = m with m=3 • =sin-1(m/d)=sin-1(3x600x10-9/0.01)=0.01030 • b) Ym = mL/d • m=0: y0 =0 • m=3: y3 = 3x600x10-9x5/0.01 = 9x10-4 m = 0.9 mm interference, diffraction & polarization

14. other ways of causing interference • remember equivalent to: n1>n2 n1<n2 1 2 2 1 interference, diffraction & polarization

15. phase changes at boundaries If a light ray travels from medium 1 to medium 2 with n1<n2, the phase of the light ray will change by 1/2. This will not happen if n1>n2. n1>n2 1 2 1 2 n1<n2 1/2 phase change no phase change In a medium with index of refraction n, the wavelength changes (relative to vacuum) to /n interference, diffraction & polarization

16. thin film interference n=1 The two reflected rays can interfere. To analyze this system, 4 steps are needed: n=1.5 n=1 • Is there phase inversion at the top surface? • Is there phase inversion at the bottom surface • What are the conditions for constructive/destructive interference? • what should the thickness d be for 3) to happen? interference, diffraction & polarization

17. thin film analysis • top surface? • bottom surface? • conditions? • d? 1 2 n=1 n=1.5 • top surface: n1<n2 so phase inversion 1/2 • bottom surface: n1>n2 so no phase inversion • conditions: • constructive: ray 1 and 2 must be in phase • destructive: ray 1 and 2 must be out of phase by 1/2 • if phase inversion would not take place at any of the surfaces: • constructive: • 2d=m (difference in path length=integer number of wavelengths) • due to phase inversion at top surface: 2d=(m+1/2) • since the ray travels through film: 2d=(m+1/2)film =(m+1/2)/nfilm • destructive: 2d=mfilm =m/nfilm n=1 interference, diffraction & polarization

18. Note The interference is different for light of different wavelengths interference, diffraction & polarization

19. n1<n2 in both cases question • Phase inversion will occur at • top surface • bottom surface • top and bottom surface • neither surface na=1 nb=1.5 nc=2 • constructive interference will occur if: • 2d=(m+1/2)/nb • 2d=m/nb • 2d=(m+1/2)/nc • 2d=m/nc note: if destructive 2d=(m+1/2)/nb this is used e.g. on sunglasses to reduce reflections interference, diffraction & polarization

20. another case 1 2 The air gap in between the plates has varying thickness. Ray 1 is not inverted (n1>n2) Ray 2 is inverted (n1<n2) where the two glasses touch: no path length difference: dark fringe. if 2t=(m+1/2) constructive interference if 2t=m destructive interference. interference, diffraction & polarization

21. question Given h=1x10-5 m 30 bright fringes are seen, with a dark fringe at the left and the right. What is the wavelength of the light? 2t=m destructive interference. m goes from 0 (left) to 30 (right). =2t/m=2h/m=2x1x10-5/30=6.67x10-7 m=667 nm interference, diffraction & polarization

22. newton’s rings demo spacing not equal interference, diffraction & polarization

23. quiz (extra credit) • Two beams of coherent light travel different paths arriving • at point P. If constructive interference occurs at point P, • the two beams must: • travel paths that differ by a whole number of • wavelengths • b)travel paths that differ by an odd number of half • wavelengths interference, diffraction & polarization

24. question • why is it not possible to produce an interference pattern • in a double-slit experiment if the separation of the slits • is less than the wavelength of the light used? • the very narrow slits required would generate different • wavelength, thereby washing out the interference pattern • the two slits would not emit coherent light • the fringes would be too close together • in no direction could a path difference as large as one wavelength be obtained interference, diffraction & polarization

25. diffraction In Young’s experiment, two slits were used to produce an interference pattern. However, interference effects can already occur with a single slit. This is due to diffraction: the capability of light to be “deflected” by edges/small openings. In fact, every point in the slit opening acts as the source of a new wave front interference, diffraction & polarization

26. interference, diffraction & polarization

27. interference pattern from a single slit pick two points, 1 and 2, one in the top top half of the slit, one in the bottom half of the slit. Light from these two points interferes destructively if: x=(a/2)sin=/2 so sin=/a we could also have divided up the slit into 4 pieces: x=(a/4)sin=/2 so sin=2/a 6 pieces: x=(a/6)sin=/2 so sin=3/a Minima occur if sin = m/a m=1,2,3… In between the minima, are maxima: sin = (m+1/2)/a m=1,2,3… AND sin=0 or =0 interference, diffraction & polarization

28. slit width a a if >a sin=/a > 1 Not possible, so no patterns if <<a sin=m/a is very small diffraction hardly seen <a : interference pattern is seen interference, diffraction & polarization

29. the diffraction pattern The intensity is not uniform: I=I0sin2()/2 =a(sin)/  a a a a a a interference, diffraction & polarization

30. question light with a wavelength of 500 nm is used to illuminate a slit of 5m. At which angle is the 5th minimum in the diffraction pattern seen? sin = m/a  = sin-1(5x500x10-9/(5x10-6))=300 interference, diffraction & polarization

31. diffraction from a single hair instead of an slit, we can also use an inverse image, for example a hair! demo interference, diffraction & polarization

32. double slit interference revisited The total response from a double slit system is a combination of two single-source slits, combined with a diffraction pattern from each of the slit due to diffraction minima asin=m, m=1,2,3… maxima asin=(m+1/2), m=1,2,3… and =0 a: width of individual slit due to 2-slit interference maxima dsin=m, m=0,1,2,3… minima dsin=(m+1/2), m=0,1,2,3… d: distance between two slits interference, diffraction & polarization

33. double-slit experiment a d if >d, each slit acts as a single source of light and we get a more or less prefect double-slit interference spectrum if <d the interference spectrum is folded with the diffraction pattern. interference, diffraction & polarization

34. 7th 7th question A person has a double slit plate. He measures the distance between the two slits to be d=1 mm. Next he wants to determine the width of each slit by investigating the interference pattern. He finds that the 7th order interference maximum lines up with the first diffraction minimum and thus vanishes. What is the width of the slits? 7th order interference maximum: dsin=7 so sin=7/d 1st diffraction minimum: asin=1 so sin=/a sin must be equal for both, so /a=7/d and a=d/7=1/7 mm interference, diffraction & polarization

35. diffraction grating consider a grating with many slits, each separated by a distance d. Assume that for each slit >d. We saw that for 2 slits maxima appear if: d sin = m, m=0,1,2,3… This condition is not changed for in the case of n slits. d Diffraction gratings can be made by scratching lines on glass and are often used to analyze light instead of giving d, one usually gives the number of slits per unit distance: e.g. 300 lines/mm d=1/(300 lines/mm)=0.0033 mm interference, diffraction & polarization

36. separating colors d sin = m, m=0,1,2,3… for maxima (same as for double slit) so  = sin-1(m/d) depends on , the wavelength. cd’s can act as a diffraction grating (DVD’s work even better because their tracks are more closely spaced.) interference, diffraction & polarization

37. question • If the interference conditions are the same when using a double slit or a diffraction grating with thousands of slits, what is the advantage of using the grating to analyze light? • a) the more slits, the larger the separation between maxima. • b) the more slits, the narrower each of the bright spots and thus easier to see • c) the more slits, the more light reaches each maximum and the maxima are brighter • d) there is no advantage interference, diffraction & polarization

38. question An diffraction grating has 5000 lines per cm. The angle between the central maximum and the fourth order maximum is 47.20. What is the wavelength of the light? d sin = m, m = 0,1,2,3… d = 1/5000 = 2x10-4 cm = 2x10-6 m m = 4, sin(47.2)=0.734 so  = d sin/m = 2x10-6x0.734/4 = 3.67x10-7 m = 367 nm interference, diffraction & polarization

39. polarization • We saw that light is really an electromagnetic wave with electric and magnetic field vectors oscillating perpendicular to each other. In general, light is unpolarized, which means that the E-field vector (and thus the B-field vector as long as it is perpendicular to the E-field) could point in any direction E-vectors could point anywhere: unpolarized propagation into screen interference, diffraction & polarization

40. polarized light • light can be linearly polarized, which means that the E-field only oscillates in one direction (and the B-field perpendicular to that) • The intensity of light is proportional to the square of amplitude of the E-field. I~Emax2 interference, diffraction & polarization

41. How to polarize? • absorption • reflection • scattering interference, diffraction & polarization

42. polarization by absorption • certain material (such as polaroid used for sunglasses) only transmit light along a certain ‘transmission’ axis. • because only a fraction of the light is transmitted after passing through a polarizer the intensity is reduced. • If unpolarized light passes through a polarizer, the intensity is reduced by a factor of 2 interference, diffraction & polarization

43. polarizers and intensity polarization axis direction of E-vector For unpolarized light, on average, the E-field has an angle of 450 with the polarizer. I=I0cos2=I0cos2(45)=I0/2  If E-field is parallel to polarization axis, all light passes If E-field makes an angle  pol. axis only the component parallel to the pol. axis passes: E0cos So I=I0cos2 interference, diffraction & polarization

44. question • unpolarized light with intensity I0 passes through a linear polarizer. It then passes through a second polarizer (the second polarizer is usually called the analyzer) whose transmission axis makes and angle of 300 with the transmission axis of the first polarized. What is the intensity of the light after the second polarizer, in terms of the intensity of the initial light? After passing through the first polarizer, I1=I0/2. After passing through the second polarizer, I2=I1cos230=0.75I1=0.375I0 interference, diffraction & polarization

45. polarization by reflection n1 • If unpolarized light is reflected, than the reflected light is partially polarized. • if the angle between the reflected ray and the refracted ray is exactly 900 the reflected light is completely polarized • the above condition is met if for the angle of incidence the equation tan=n2/n1 • the angle =tan-1(n2/n1) is called the Brewster angle • the polarization of the reflected light is (mostly) parallel to the surface of reflection n2 interference, diffraction & polarization

46. horizontal vertical direction of polarization of reflected light question • Because of reflection from sunlight of the glass window, the curtain behind the glass is hard to see. If I would wear polaroid sunglasses that allow … polarized light through, I would be able to see the curtain much better. • a) horizontally • b) vertically interference, diffraction & polarization

47. sunglasses wearing sunglasses will help reducing glare (reflection) from flat surfaces (highway/water) without with sunglasses interference, diffraction & polarization

48. polarization by scattering • certain molecules tend to polarize light when struck by it since the electrons in the molecules act as little antennas that can only oscillate in a certain direction interference, diffraction & polarization