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Waves (in general) sine waves are nice other types of waves (such as square waves, sawtooth waves, etc.) can be formed by a superposition of sine waves.

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Presentation on theme: "Waves (in general) sine waves are nice other types of waves (such as square waves, sawtooth waves, etc.) can be formed by a superposition of sine waves."— Presentation transcript:

1 Waves (in general) sine waves are nice other types of waves (such as square waves, sawtooth waves, etc.) can be formed by a superposition of sine waves - this is called Fourier Series. This means that sine waves can be considered as fundamental.

2 Waves (in general) E = E o sin (  ) where  is a phase angle which describes the location along the wave   = 90 degrees is the crest    = 270 degrees is the trough

3 Waves (in general) E = E o sin (  ) where  is a phase angle in a moving wave,  changes with both – time ( goes 2  radians in time T) and –distance (goes 2  radians in distance ) so  = (2  / )*x +/- (2  /T)*t – where 2  /T =  and –where 2  / = k and so phase speed: v = distance/time = /T = f =  /k

4 Waves (in general) For nice sine waves: E = E o sin(kx +/-  t) For waves in general, can break into component sine waves; this is called spectral analysis

5 Light and Shadows Consider what we would expect from particle theory: sharp shadows light dark

6 Light and Shadows Consider what we would expect from wave theory: shadows NOT sharp light dark dim crest

7 Light and Shadows What DOES happen? look at a very bright laser beam going through a vertical slit. (A laser has one frequency unlike white light.)

8 Double Slit Experiment We will consider this situation but only after we consider another: the DOUBLE SLIT experiment:

9 Double Slit Experiment Note that along the green lines are places where crests meets crests and troughs meet troughs. crest on crest followed by trough on trough

10 Double Slit Experiment Note that along the dotted lines are places where crests meets troughs and troughs meet crests. crest on crest followed by trough on trough crest on trough followed by trough on crest

11 Double Slit Experiment Further explanations are in the Introduction to the Computer Homework Assignment on Young’s Double Slit, Vol 5, #3. crest on crest followed by trough on trough crest on trough followed by trough on crest

12 Double Slit Experiment Our question now is: How is the pattern of bright and dark areas related to the parameters of the situation:, d, x and L? d SCREEN L x bright

13 Young’s Double Slit Formula λ/d = sin(  ) ≈ tan(  ) = x/L The two (black) lines from the two slits to the first bright spot are almost parallel, so the two angles are almost 90 degrees, so the two  ’s are almost equal. d λ L x bright

14 Double slit: an example n = d sin(  ) = d x / L d = 0.15 mm = 1.5 x m x = ??? measured in class L = ??? measured in class n = 1 (if x measured between adjacent bright spots) = d x / L = (you do the calculation)

15 Interference: Diffraction Grating The same Young’s formula works for multiple slits as it did for 2 slits. d lensbright s1 s2 s3 s4 s5 s2 = s1 + s3 = s2 + = s1 + 2 s4 = s3 + = s1 + 3 s5 = s4 + = s1 + 4 λ

16 Interference: Diffraction Grating With multiple slits, get MORE LIGHT and get sharper bright spots. d lensbright s1 s2 s3 s4 s5 s2 = s1 + s3 = s2 + = s1 + 2 s4 = s3 + = s1 + 3 s5 = s4 + = s1 + 4

17 Interference: Diffraction Grating With 5 slits, get cancellation when  s = 0.8 ; with two slits, only get complete cancellation when  s = 0.5. d lensbright s1 s2 s3 s4 s5 s2 = s1 +.8 s3 = s2 +.8 = s s4 = s3 +.8 = s s5 = s4 +.8 = s dark

18 Diffraction Grating: demonstrations look at the white light source (incandescent light due to hot filament) look at each of the gas excited sources (one is Helium, one is Mercury)

19 Interference: Thin Films Before, we had several different parts of a wide beam interfering with one another. Can we find other ways of having parts of a beam interfere with other parts?

20 Interference: Thin Films We can also use reflection and refraction to get different parts of a beam to interfere with one another by using a thin film. air film water reflected red interferes with refracted/reflected/refracted blue.

21 Interference: Thin Films Blue travels an extra distance of 2t in the film. air film water reflected red interferes with refracted/reflected/refracted blue. t

22 Interference: Thin Films Also, blue undergoes two refractions and reflects off of a different surface. air film water reflected red interferes with refracted/reflected/refracted blue. t

23 Interference: Thin Films When a wave encounters a new medium: –the phase of the refracted wave is NOT affected. –the phase of the reflected wave MAY BE affected.

24 Interference: Thin Films When a wave on a string encounters a fixed end, the reflected wave must interfere with the incoming wave so as to produce cancellation. This means the reflected wave is 180 degrees (or /2) out of phase with the incoming wave.

25 Interference: Thin Films When a wave on a string encounters a fixed end, the reflected wave must interfere with the incoming wave so as to produce cancellation. This means the reflected wave is 180 degrees (or /2) out of phase with the incoming wave.

26 Interference: Thin Films When a wave on a string encounters a free end, the reflected wave does NOT have to destructively interfere with the incoming wave. There is NO phase shift on this reflection.

27 Interference: Thin Films When a wave on a string encounters a free end, the reflected wave does NOT have to destructively interfere with the incoming wave. There is NO phase shift on this reflection.

28 Interference: Thin Films When light is incident on a SLOWER medium (one of index of refraction higher than the one it is in), the reflected wave is 180 degrees out of phase with the incident wave. When light is incident on a FASTER medium, the reflected wave does NOT undergo a 180 degree phase shift.

29 Interference: Thin Films If n a < n f < n w, BOTH red and blue reflected rays will be going from fast to slow, and no difference in phase will be due to reflection. air film water reflected red interferes with refracted/reflected/refracted blue. t

30 Interference: Thin Films If n a n w, there WILL be a 180 degree phase difference ( /2)due to reflection. air film water reflected red interferes with refracted/reflected/refracted blue. t

31 Interference: Thin Films There will ALWAYS be a phase difference due to the extra distance of 2t/. air film water reflected red interferes with refracted/reflected/refracted blue. t

32 Interference: Thin Films When t= /2 the phase difference due to path is 360 degrees (equivalent to no difference) air film water reflected red interferes with refracted/reflected/refracted blue. t

33 Interference: Thin Films When t= /4 the phase difference due to path is 180 degrees. air film water reflected red interferes with refracted/reflected/refracted blue. t

34 Interference: Thin Films Recall that the light is in the FILM, so the wavelength is not that in AIR: f = a /n f. air film water reflected red interferes with refracted/reflected/refracted blue. t

35 Interference: Thin Films reflection: no difference if n f < n w ; 180 degree difference if n f > n w. distance: no difference if t = a /2n f 180 degree difference if t = a /4n f Total phase difference is sum of the above two effects.

36 Interference: Thin Films Total phase difference is sum of the two effects of distance and reflection For minimum reflection, need total to be 180 degrees. –anti-reflective coating on lens For maximum reflection, need total to be 0 degrees. –colors on oil slick

37 Thin Films: an example An oil slick preferentially reflects green light. The index of refraction of the oil is 1.65, that of water is 1.33, and or course that of air is What is the thickness of the oil slick?

38 Thin Films: an example Since we have preferentially reflected green light, the TOTAL phase difference must be 0 degrees (or 360 degrees).

39 Thin Films: an example Since we have preferentially reflected green light, the TOTAL phase difference must be 0 degrees (or 360 degrees). Since we have n f > n w, we have 180 degrees due to reflection.

40 Thin Films: an example Since we have preferentially reflected green light, the TOTAL phase difference must be 0 degrees (or 360 degrees). Since we have n f > n w, we have 180 degrees due to reflection. Therefore, we need 180 degrees due to extra distance, so need t = a /4n f where a = 500 nm, n f = 1.65, and so t = 500 nm / 4(1.65) = 76 nm.

41 Michelson Interferometer Split a beam with a Half Mirror, the use mirrors to recombine the two beams. Mirror Half Mirror Screen Light sourc e

42 Michelson Interferometer If the red beam goes the same length as the blue beam, then the two beams will constructively interfere and a bright spot will appear on screen. Mirror Half Mirror Screen Light sourc e

43 Michelson Interferometer If the blue beam goes a little extra distance,  s, the the screen will show a different interference pattern. Mirror Half Mirror Screen Light sourc e ss

44 Michelson Interferometer If  s = /4, then the interference pattern changes from bright to dark. Mirror Half Mirror Screen Light sourc e ss

45 Michelson Interferometer If  s = /2, then the interference pattern changes from bright to dark back to bright (a fringe shift). Mirror Half Mirror Screen Light sourc e ss

46 Michelson Interferometer By counting the number of fringe shifts, we can determine how far  s is! Mirror Half Mirror Screen Light sourc e ss

47 Michelson Interferometer If we use the red laser ( =632 nm), then each fringe shift corresponds to a distance the mirror moves of 316 nm (about 1/3 of a micron)! Mirror Half Mirror Screen Light sourc e ss

48 Michelson Interferometer We can also use the Michelson interferometer to determine the index of refraction of a gas (such as air). Put a cylinder with transparent ends into one of the beams.

49 Michelson Interferometer Evacuate the cylinder with a vacuum pump Slowly allow the gas to seep back into the cylinder and count the fringes. Mirror Half Mirror Screen Light sourc e cylinder

50 Michelson Interferometer In vacuum, # v v = 2L. In the air, # a a = 2L. Since v a # v. Knowing v and L, can calculate # v.

51 Michelson Interferometer Knowing v and L, can calculate # v. By counting the number of fringe shifts, we can determine  #. Since  # = # a - # v, we can calculate # a. Now knowing L and # a, we can calculate a.

52 Michelson Interferometer We now know v and a, so: with v f = c and a f = v a, we can use n a = c/v a = v f / a f = v / a.

53 Michelson Interferometer an example If L = 6 cm, and if v = 632 nm, and if 50 fringes are counted when air is let back into the cylinder, then: # v = 2L/ v = 2 *.06 m / 632 x m = 189,873 # a = # v +  # = 189, = 189,923 a = 2L/# a = 2 *.06 m / 189,923 = nm n a = v / a = nm / nm =


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