chapt35 - Chapter 35: Interference Problem 1 The speed of...

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Chapter 35: Interference Problem 1 The speed of yellow light (from a sodium lamp) in a certain liquid is measured to be 1.92 × 10 8 m/s. What is the index of refraction of this liquid for the light? Answer 1.56 Solution The index of refraction is equal to the speed of light in vacuum divided by the speed of light in the liquid: n = c v l = 3 × 10 8 m s 1.92 × 10 8 m s = 1.56 Problem 3 How much faster, in meters per second, does light travel in sapphire than in diamond? See Table 33-1. Answer 4.55 × 10 7 m/s Solution The difference in the speed of light in sapphire and diamond is v s v d = c n s c n d = 3 × 10 8 m/s ( ) 1 1.77 1 2.42 = 4.55 × 10 7 m/s Problem 4 The wavelength of yellow sodium light in air is 589 nm. (a) What is its frequency? (b) What is its wavelength in glass whose index of refraction is 1.52? (c) From the results of (a) and (b) find its speed in this glass. Answer (a) 5.09 × 10 14 Hz. (b) 388 nm. (c) 1.97 × 10 8 m/s.
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35-2 Chapter 35: Interference Solution (a) Since the speed of light in air is about the same as c, its speed in a vacuum, the frequency is f = c λ = 3 × 10 8 m s 5.89 × 10 7 m = 5.09 × 10 14 Hz (b) The frequency of the light does not change. If v is the speed of the light in the glass then the wavelength in glass is g = v f but n = c v so g = c nf = 3 × 10 8 m s 1.52 5.09 × 10 14 Hz = 388 nm (c) The speed of the light in the glass is v = g f = 388 nm 5.09 × 10 14 Hz = 1.97 × 10 8 m s Problem 9 Suppose that the two waves in Fig. 35-4 have wavelength 500 nm in air. In wavelengths, what is their phase difference after traversing media 1 and 2 if (a) n 1 = 1.50, n 2 = 1.60, and L = 8.50 μ m; (b) n 1 = 1.62, n 2 = 1.72, and L = 8.50 μ m; and (c) n 1 = 1.59, n 2 = 1.79, and L = 3.25 μ m? (d) Suppose that in each of these three situations the waves arrive at a common point after emerging. Rank the situations according to the brightness the wave produces at the common point. Answer (a) 1.70; (b) 1.70; (c) 1.30; (d) all tie. n 1 n 2 L Figure 35-4 Solution (a) The number of wavelengths difference in the two paths is δ = L 2 L 1 = n 2 L a n 1 L a = L a n 2 n 1 ( ) = 8.50 μ m 500 nm 1.6 1.5 ( ) = 1.7
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Chapter 35: Interference 35-3 (b) The difference is δ = L λ a n 2 n 1 ( ) = 8.50 μ m 500 nm 1.72 1.62 ( ) = 1.7 (c) The difference is = L a n 2 n 1 ( ) = 3.25 m 500 nm 1.79 1.59 ( ) = 1.3 (d) The brightness will be the same for all three since the intensity is proportional to the square of the cosine of half the phase difference where the phase difference is Φ = 2 π δ . cos 2 1 2 2 π 1.7 ( ) = cos 2 1 2 2 1.3 ( ) Problem 13 Two waves of light in air, of wavelength 600.0 nm, are initially in phase. They then travel through plastic layers as shown in Fig. 35-37, with L 1 = 400 μ m, L 2 = 3.50 *m, n 1 = 1.40, and n 2 = 1.60. (a) In wavelengths, what is their phase difference after they both have emerged from the layers? (b) If the waves later arrive at some common point, what type
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chapt35 - Chapter 35: Interference Problem 1 The speed of...

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