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Today Ch.36 (Diffraction) Next week, Dec.6, Review This week off. hours: Th: 2:00-3:15pm; F: 1:00-3:00pm Next week off. hours: Tu:2-3:15pm,W:1-3pm,Th:1-3pm.

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Presentation on theme: "Today Ch.36 (Diffraction) Next week, Dec.6, Review This week off. hours: Th: 2:00-3:15pm; F: 1:00-3:00pm Next week off. hours: Tu:2-3:15pm,W:1-3pm,Th:1-3pm."— Presentation transcript:

1 Today Ch.36 (Diffraction) Next week, Dec.6, Review This week off. hours: Th: 2:00-3:15pm; F: 1:00-3:00pm Next week off. hours: Tu:2-3:15pm,W:1-3pm,Th:1-3pm Webct homework is due by Dec.12. Check your Midterm Exams Grades on elearning! Final Exam: (Ch.21-25, 27-29, 32,33,35,36) Secs.511-515: December 9, Friday: 12:30-2:30 pm Secs.521-525, 528: December 12, Monday, 8-10 am

2 Lecture 23 (Ch. 35) Interference 1. Interference and superposition principle 2. Condition of constructive and distractive interference: phase difference and path difference 4.Thomas Youngs double-slit experiment 5.Interference in thin films 6.Applications

3 Interference and superposition principle Let us consider a superposition of two electric fields oscillating in the same direction, with the same frequency and given initial phase shift. The same result can be easily obtained from the phasers diagram.

4 Total intensity does depend on (in general I t I 1 +I 2 ! ). This phenomenon is called interference. Monochromatic waves of the same frequency with fixed relative phase are called coherent waves. Coherence time:, coherence length If phase is random interference is absent: Phase difference:

5 General case Prove it is using

6 Path difference S1S1 S2S2 r1r1 r2r2 Taking into account that In particular case when

7 Thomas Youngs double slits experiment, 1800 Thomas Young (1773 – 1829)

8 Max and min positions (bright and dark stripes)

9 Example. A very thin sheet of plastic (n=1.6) covers one slit of a double-slit apparatus illuminated by 640nm light. The center point of the screen, instead of being a maximum, is dark. What is the minimum thickness of the plastic sheet? d

10 Example. In a two-slits interference experiment, the slits are 0.2mm apart, and a screen is at a distance of 1m. The third bright fringe (not counting the central bright fringe) is found to be displaced 9.49mm from the central fringe. Find the wavelength of the light used. NB:R>>d, R>>y 3

11 Intensity distribution 4

12 Max and min positions (bright and dark stripes)

13 Example A radio station operating at a frequency of 1500kHz has two identical vertical dipole antennas spaced 400m apart, oscillating in phase. At distances much greater then 400m, in what directions is the intensity greatest in the resulting radiation pattern? If intensity produced by each antenna 400km away along y axis is 2mW/m 2 what is the total intensity produced by two antennas at this point? y

14 Phase change in the reflected wave

15 In the case of a normal coincidence: Interference in the thin films n { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/4/1428986/slides/slide_15.jpg", "name": "In the case of a normal coincidence: Interference in the thin films n

16 Example The walls of a soap bubble have about the same refractive index as a plain water, n=1.33. In the point where the wall is 120nm thick what colors of incoming white light are the most strongly reflected? (orange) t

17 Example Light with wavelength 648nm in air is incident perpendicularly from air on a film 8.75 μm thick and with refractive index 1.35. Part of the light is reflected from the first surface of the film and is reflected back at the second surface. The second surface of the film is again in contact with air. 1. Find the total phase difference in terms of at the exit of the film between the rays reflected from the first and second surfaces. 2. Will you see constructive or distructive interference in the reflected light? Explain. 3. Suppose the second surface of the film is in a contact with a plastic material with refractive index 1.85. How does it change the answers on the previous questions? Explain. 2. Constractive since the total phase shift is multiple to 2 3. Reflection from each surface gives the phase shift. Total phase shift now consists of an even number of. Interference will be destractive.

18 NB: if light is produced by a thermal source the typical coherence length is of an order of 1μm. Hence for films much thicker then 1μm an interference is impossible.

19 Interference at the thin wedge of air NB: Interference between the light reflected from the upper and lower surfaces of the glass plate is neglected due to the larger thickness of the plate. Example A monochromatic light with wavelength in the air 500nm is at normal incidence on the top glass plate (see the figure). What is the spacing of interference fringes? NB: the fringe at the line of contact is dark, because of phase shift produced by reflection from the second plate.

20 Newtons rings NB: Interference between the light reflected from the upper and lower surfaces of the lens is neglected due to the larger thickness of the lens. NB: the fringe at the line of contact is dark, because of phase shift produced by reflection from the glass plane. R R r The positions of max or min: rmrm

21 Example: Newtons rings can be seen when a plano-convex lens is placed on a flat glass surface (see the figure). The radius of curvature of the convex surface is 95.2cm. The lens is illuminated R R r rmrm from above with red light, λ=580nm. 1) Is the spot in the center bright or dark? 2) Formulate the condition for the constructive interference. 3) Find the diameter of the first bright ring. 4) Would the first ring be closer or further from the center for the blue light compare to the red light? 1. The spot at the center is dark, because of phase shift produced by reflection from the glass plane. 2. 3.t 0 = λ 0 /4 4. Closer, since λ for blue light is shorter and hence r is smaller.

22 Nonreflective and reflective coatings Nonreflecting film corresponds to destructive interference in the light reflected from the upper and lower surfaces of the film. Reflecting film corresponds to the constructive interference in reflection. NB: Minimum in reflection corresponds to maximum in light transmitted through the film into the glass. Maximum in reflection corresponds to minimum in light transmitted through the film into the glass. Such quarter wavelength films are used to cover lenses in cameras and solar cells to increase an amount of light through lens and solar cells. They are used also to make planes invisible for radars (which typically use 2cm wavelength). NB: If n film >n glass (or other substrate) then quarter wavelength film is the reflective film (because the phase shift at the second interface becomes zero).

23 Example. A plastic film with index of refraction 1.85 is put on the surface of a car window to increase the reflectivity and thus to keep the interior of the car cooler. The window glass has index of refraction 1.52. (a) What minimum thickeness is required if light with wavelength 550 nm in air reflected from two sides of the film is to interfere constructively? (b) It is found to be difficult to manufacture and instol coatings as thin as calculated in part (a). What is the next greatest thickness for which there will be also constructive interference? We need constructive interference in reflection:.

24 The Michelson interferometer Compensator plate D is made of the same glass as C and has the same thickness. It induces the same phase shift in ray 1 as those enquired by ray 2 in C.

25 1887 V=3x10 4 m/s


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