University:
California State University, Northridge
Course:
PHYS 420 | Modern Optics
Academic year:

2013
Views:

384

Pages:

12
Author:

Royce Delacruz

28-1 Superposition and Interference One important aspects of the wave is superposition/interference of two waves. Same wavelength from same light source Destructive Constructive Constructive Figure 28-1 Constructive and Destructive Interference (a) In phase, (a) 360º () out of phase, (c) 180º (/2) out of phase. The conditions for two wave Interference: 1) Monochromatic light; 2) Coherence: the two wave must have constant phase difference. Incoherence: wave source whose relative phases vary randomly. Such as 2 different light sources. Laser light is very good coherent light source. Largely used for the interferometer. We only limit our discussion on Monochromatic and coherence waves. That is, you can imagine that Monochromatic waves from the same light source. The interference of two waves The conditions for constructive and destructive interferences are: l2  l1  m l2  l1  m  (constructive)  2 (destructive), m  0,  1,  2    Destructive Constructive Destructive Constructive Figure 28-2 Two Radio Antennas Transmitting the Same Signal Example 28-1 Two friend tune their radios to the same frequency and pick up a signal transmitted simultaneously by a pair of antennas. The friend at p0 received a strong signal. The friend at point Q1 received a very weak signal. Find the wavelength of the radio wave, if d=7.5 km, L= 14.0 km, and y=1.88 km. Assuming Q1 is the first point of minimum signal as one moves away from center point P0 in the y direction. Example 28-1 Two May Not Be Better than One Solution: 1) Calculate l1 and l2: d l1  L2  (  y ) 2 2  (14.0km) 2  ( 7.5km  1.88) 2  14.1 km 2 d l2  L2  (  y ) 2 2  (14.0km) 2  ( 7.5km  1.88) 2  15.1 km 2 2) For “first-order” destructive interference: 1 l2  l1   2   2(l2  l1 )  2(15.1km  14.1km)  2.0 km 28-2 Young’s Two-Slit Experience Figure 28-3 Young’s Two-Slit Experiment Figure 28-4 Huygens’s Principle Figure 28-5 Path Difference in the Two-Slit Experiment Fringes at the screen of infinite distance Condition for Bright Fringe (constructive Interference) in the two-slit experiment d sin   m m  0,1,2    (28  1) M=0, central bright fringe. M=1, first bright fringe. Condition for Dark Fringe (destructive Interference) in the two-slit experiment d sin   m   2 M=1, first dark fringe. m  0,1,2    (28  2) Exercise 28-1 Red light ( = 752 nm) passes through a pair of slits with a separation of 6.20x10-5 m. Find the angles corresponding to (a) the first bright fringe and (b) the second dark fringe above the central fringe. Figure 28-6 The Two-Slit Pattern Solution: (a) For the first bright fringe, m=1, d sin   m   6.20 105 (meter )  sin   7.52 107 (meter ),   0.695 (b) For the second dark fringe, m=2, d sin   m    2   / 2  1.5 2 6.20 105 sin   1.5  (7.52 107 ) ,   1.04

Physical Optics: Interference and Diffraction

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