SM_chapter37

SM_chapter37 - 37 Interference of Light Waves CHAPTER...

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37 Interference of Light Waves CHAPTER OUTLINE 37.1 Conditions for Interference 37.2 Young’s Double-Slit Experiment 37.3 Light Waves in Interference 37.4 Intensity Distribution of the Double-Slit Interference Pattern 37.5 Change of Phase Due to ReF ection 37.6 Interference in Thin ±ilms 37.7 The Michelson Interferometer ANSWERS TO QUESTIONS Q37.1 (a) Two waves interfere constructively if their path difference is zero, or an integral multiple of the wavelength, according to δ λ = m , with m = 0123 , , , ,. .. . (b) Two waves interfere destructively if their path difference is a half wavelength, or an odd multiple of 2 , described by δλ =+ m 1 2 , with m = Q37.2 The light from the fl ashlights consists of many different wavelengths (that’s why it’s white) with random time differences between the light waves. There is no coherence between the two sources. The light from the two fl ashlights does not maintain a constant phase relationship over time. These three equivalent statements mean no possibility of an interference pattern. *Q37.3 (i) The angles in the interference pattern are controlled by / d , which we estimate in each case: (a) 0.45 µ m/400 m 1.1 × 10 3 (b) 0.7 m/400 m 1.6 × 10 3 (c) and (d) 0.7 m/800 μ m 0.9 × 10 −3 . The ranking is b > a > c = d. (ii) Now we consider L / d : (a) 4 m (0.45 m/400 m) 4.4 mm (b) 4 m (0.7 m/400 m) 7 mm (c) 4 m(0.7 m/800 m) 3 mm (d) 8 m(0.7 m/800 m) 7 mm. The ranking is b = d > a > c. *Q37.4 Yes. A single beam of laser light going into the slits divides up into several fuzzy-edged beams diverging from the point halfway between the slits. *Q37.5 Answer (c). Underwater, the wavelength of the light decreases according to water air water = n . Since the angles between positions of light and dark bands are proportional to , the underwater fringe separations decrease. Q37.6 Every color produces its own pattern, with a spacing between the maxima that is characteristic of the wavelength. With white light, the central maximum is white. The F rst side maximum is a full spectrum with violet on the inner edge and red on the outer edge on each side. Each side maximum farther out is in principle a full spectrum, but they overlap one another and are hard to distinguish. Using monochromatic light can eliminate this problem. *Q37.7 With two F ne slits separated by a distance d slightly less than , the equation d sin θ = 0 has the usual solution = 0. But d sin = 1 has no solution. There is no F rst side maximum. d sin = (1/2) has a solution. Minima fl ank the central maximum on each side. Answer (b). 353
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354 Chapter 37 Q37.8 As water evaporates from the “soap” bubble, the thickness of the bubble wall approaches zero. Since light refl ecting from the front of the water surface is phase-shifted 180 ° and light refl ecting from the back of the soap F lm is phase-shifted 0 ° , the refl ected light meets the conditions for a minimum. Thus the soap F lm appears black, as in the textbook illustration accompanying this question.
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This note was uploaded on 01/19/2012 for the course PHY 232 taught by Professor Williams,frank during the Spring '11 term at Ohio State.

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SM_chapter37 - 37 Interference of Light Waves CHAPTER...

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