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PHY 1371Dr. Jie Zou1 Chapter 37 Interference of Light Waves.

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Presentation on theme: "PHY 1371Dr. Jie Zou1 Chapter 37 Interference of Light Waves."— Presentation transcript:

1 PHY 1371Dr. Jie Zou1 Chapter 37 Interference of Light Waves

2 PHY 1371Dr. Jie Zou2 Outline Conditions for interference Young’s double-slit experiment Set-up and observation Qualitative and quantitative explanation Intensity distribution of the double-slit interference pattern The derivation

3 PHY 1371Dr. Jie Zou3 Interference (revisited) Interference: The combination of separate waves in the same region of space to produce a resultant wave is called interference. Light waves also interfere with each other, when the electromagnetic fields that constitute the individual waves combine.

4 PHY 1371Dr. Jie Zou4 Conditions for interference of light waves For sustained interference in light waves to be observed, the following conditions must be met: The sources must be coherent-that is, they must maintain a constant phase with respect to each other. The sources should be monochromatic-that is, of a single wavelength. A common method for producing two coherent light sources is to use one monochromatic source to illuminate a barrier containing two small openings, e.g. two slits.

5 PHY 1371Dr. Jie Zou5 Young’s double-slit experiment: Set-up and observation Set-up: Slits S 1 and S 2 serve as a pair of coherent light sources. Observation: The light from S 1 and S 2 produces on a viewing screen a visible pattern of bright and dark parallel bands called fringes.

6 PHY 1371Dr. Jie Zou6 Young’s double-slit experiment: Explanation The light waves from the two slits overlap as they spread out, filling what we expect to be shadowed regions with light and producing interference fringes. Diffraction: This divergence of light from its initial line of travel when encountering a barrier or opening is called diffraction. When the light from S 1 and S 2 both arrive at a point on the screen such that constructive interference occurs at that location, a bright fringe appears. When the light from the two slits combines destructively at any location on the screen, a dark fringe results.

7 PHY 1371Dr. Jie Zou7 (a) Two waves start in phase and strike the screen at the central point P. Equal path length and in phase at P. Constructive interference and thus bright fringe at P. (b) Two waves start in phase. Path difference is one wavelength to reach point Q. In phase at Q. Constructive interference and a second bright fringe. (c) Two waves start in phase. Path difference is half a wavelength at R. Out of phase at R. Destructive interference and a dark fringe.

8 PHY 1371Dr. Jie Zou8 Young’s double-slit experiment: Quantitative explanation Path difference  = r 2 –r 1 = d sin  (assuming that L >> d) The value of  determines whether the two waves are in phase at point P. Condition for bright fringes or constructive interference:  = d sin  bright = m, (m = 0,  1,  2,…(order number)) Condition for dark fringes or destructive interference:  = d sin  dark = (m+1/2), (m = 0,  1,  2,…) Positions of the bright and dark fringes on the screen: y = L tan   L sin  (small angle approximation) Positions of the bright fringes: y bright = ( L/d) m, (m = 0,  1,  2,…) Positions of the dark fringes: y dark =( L/d) (m+1/2), (m = 0,  1,  2,…)

9 PHY 1371Dr. Jie Zou9 Example 37.1 Measuring the wavelength of a light source A viewing screen is separated from a double-slit source by 1.2 m. The distance between the two slits is 0.030 mm. The second-order bright fringe (m=2) is 4.5 cm from the center line. (A) Determine the wavelength of the light. (B) Calculate the distance between adjacent fringes.

10 PHY 1371Dr. Jie Zou10 Intensity distribution of the double-slit interference pattern The total magnitude of the electric field at point P is the superposition of the two waves: E 1 = E 0 sin  t and E 2 = E 0 sin(  t+  ). Superposition: E P =E 1 +E 2 =E 0 [sin  t+sin(  t+  )] =2E 0 cos(  /2)sin(  t+  /2). Intensity: I  E P 2 = 4E 0 2 cos 2 (  /2)sin 2 (  t+  /2). Time-averaged light intensity: I =I max cos 2 (  /2), where  = (2  / )  =(2  / )dsin   (2  / )dy/L. So, I  I max cos 2 [(  d/ L)y].

11 PHY 1371Dr. Jie Zou11 Plot of I versus dsin  (L >> d and  is small) The interference pattern consists of equally spaced fringes of equal intensity (when L >>d and  is small.

12 PHY 1371Dr. Jie Zou12 Homework Ch. 37, P. 1197, Problems: #2, 13, 14, 15, 18.

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