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Chapter 35: Interference 35-55 THINK The index of refraction of oil is greater than that of the air, but smaller than that of the water. EXPRESS Let the indices of refraction of the air, oil, and water be n1, n2, and n3, respectively. Since n1 n2 and n2 n3 , there is a phase change of rad from both surfaces. Since the second wave travels an additional distance of 2L, the phase difference is 2 (2 L) 2 where 2 / n2 is the wavelength in the oil. The condition for constructive interference is 2 (2 L) 2m , 2 or 2 L m , m 0, 1, 2,... n2 ANALYZE (a) For m 1, 2,..., maximum reflection occurs for wavelengths 2n2 L 2 1.20 460 nm 1104 nm , 552 nm, 368nm... m m We note that only the 552 nm wavelength falls within the visible light range. (b) Maximum transmission into the water occurs for wavelengths for which reflection is a minimum. The condition for such destructive interference is given by 1 2L m 2 n2 4n2 L 2m 1 which yields = 2208 nm, 736 nm, 442 nm … for the different values of m. We note that only the 442 nm wavelength (blue) is in the visible range, though we might expect some red contribution since the 736 nm is very close to the visible range. LEARN A light ray reflected by a material changes phase by rad (or 180°) if the refractive index of the material is greater than that of the medium in which the light is traveling. Otherwise, there is no phase change. Note that refraction at an interface does not cause a phase shift. 35-81 THINK There is a small difference between the wavelength in air and the wavelength in vacuum. EXPRESS Let 1 be the phase difference of the waves in the two arms when the tube has air in it, and let 2 be the phase difference when the tube is evacuated. If is the wavelength in vacuum, then the wavelength in air is /n, where n is the index of refraction of air. This means 4 n 1 L 2n 2 1 2 2 L L M N b g O P Q where L is the length of the tube. The factor 2 arises because the light traverses the tube twice, once on the way to a mirror and once after reflection from the mirror. Each shift by one fringe corresponds to a change in phase of 2 rad, so if the interference pattern shifts by N fringes as the tube is evacuated, then b g 4 n 1 L 2 N . ANALYZE Solving for n, we obtain c c h h 60 500 109 m N n 1 1 100030 . . 2L 2 5.0 102 m LEARN The interferometer provides an accurate way to measure the refractive index of the air (and other gases as well).